H c verma part 1 solution

1.1
SOLUTIONS TO CONCEPTS
CHAPTER – 1
1. a) Linear momentum : mv = [MLT
–1
]
b) Frequency :
T
1
= [M
0
L
0
T
–1
]
c) Pressure :
]L[
]MLT[
Area
Force
2
2
 = [ML
–1
T
–2
]
2. a) Angular speed  = /t = [M
0
L
0
T
–1
]
b) Angular acceleration  = 
 
T
TLM
t
200
[M
0
L
0
T
–2
]
c) Torque  = F r = [MLT
–2
] [L] = [ML
2
T
–2
]
d) Moment of inertia = Mr
2
= [M] [L
2
] = [ML
2
T
0
]
3. a) Electric field E = F/q = ]IMLT[
]IT[
MLT 13
2



b) Magnetic field B = ]IMT[
]LT][IT[
MLT
qv
F 12
1
2




c) Magnetic permeability 0 = ]IMLT[
]I[
]L[]IMT
I
a2B 22
12






4. a) Electric dipole moment P = qI = [IT] × [L] = [LTI]
b) Magnetic dipole moment M = IA = [I] [L
2
] [L
2
I]
5. E = h where E = energy and  = frequency.
h = ]TML[
]T[
]TML[E 12
1
22





6. a) Specific heat capacity = C = ]KTL[
]K][M[
]TML[
Tm
Q 122
22




b) Coefficient of linear expansion =  = ]K[
]R][L[
]L[
TL
LL 1
0
21 



c) Gas constant = R = ])mol(KTML[
]K)][mol[(
]L][TML[
nT
PV 1122
321


 
7. Taking force, length and time as fundamental quantity
a) Density = ]TFL[
TL
F
]L[
]LT/F[
Volume
leration)force/acce(
V
m 24
242
2




b) Pressure = F/A = F/L
2
= [FL
–2
]
c) Momentum = mv (Force / acceleration) × Velocity = [F / LT
–2
] × [LT
–1
] = [FT]
d) Energy = 22
)velocity(
onaccelerati
Force
mv
2
1

= ]FL[]TL[
]LT
F
]LT[
LT
F 22
2
21
2














 



8. g = 2
sec
metre
10 = 36  10
5
cm/min
2
9. The average speed of a snail is 0.02 mile/hr
Converting to S.I. units,
0.02 1.6 1000
3600
 
m/sec [1 mile = 1.6 km = 1600 m] = 0.0089 ms
–1
The average speed of leopard = 70 miles/hr
In SI units = 70 miles/hour =
70 1.6 1000
3600
 
= 31 m/s
click
here

Chapter-I
1.2
10. Height h = 75 cm, Density of mercury = 13600 kg/m
3
, g = 9.8 ms
–2
then
Pressure = hfg = 10  10
4
N/m
2
(approximately)
In C.G.S. Units, P = 10 × 10
5
dyne/cm
2
11. In S.I. unit 100 watt = 100 Joule/sec
In C.G.S. Unit = 10
9
erg/sec
12. 1 micro century = 104
× 100 years = 10
–4
 365  24  60 min
So, 100 min = 10
5
/ 52560 = 1.9 microcentury
13. Surface tension of water = 72 dyne/cm
In S.I. Unit, 72 dyne/cm = 0.072 N/m
14. K = kIa

b
where k = Kinetic energy of rotating body and k = dimensionless constant
Dimensions of left side are,
K = [ML
2
T
–2
]
Dimensions of right side are,
Ia
= [ML
2
]
a
, 
b
= [T
–1
]
b
According to principle of homogeneity of dimension,
[ML
2
T
–2
] = [ML
2
T
–2
] [T
–1
]
b
Equating the dimension of both sides,
2 = 2a and –2 = –b  a = 1 and b = 2
15. Let energy E  M
a
C
b
where M = Mass, C = speed of light
 E = KM
a
C
b
(K = proportionality constant)
Dimension of left side
E = [ML
2
T
–2
]
Dimension of right side
M
a
= [M]
a
, [C]
b
= [LT
–1
]
b
[ML
2
T
–2
] = [M]
a
[LT
–1
]
b
 a = 1; b = 2
So, the relation is E = KMC
2
16. Dimensional formulae of R = [ML
2
T
–3
I
–2
]
Dimensional formulae of V = [ML
2
T
3
I
–1
]
Dimensional formulae of I = [I]
[ML
2
T
3
I
–1
] = [ML
2
T
–3
I
–2
] [I]
 V = IR
17. Frequency f = KL
a
F
b
M
c
M = Mass/unit length, L = length, F = tension (force)
Dimension of f = [T
–1
]
Dimension of right side,
L
a
= [L
a
], F
b
= [MLT
–2
]
b
, M
c
= [ML
–1
]
c
[T
–1
] = K[L]
a
[MLT
–2
]
b
[ML
–1
]
c
M
0
L
0
T
–1
= KM
b+c
L
a+b–c
T
–2b
Equating the dimensions of both sides,
 b + c = 0 …(1)
–c + a + b = 0 …(2)
–2b = –1 …(3)
Solving the equations we get,
a = –1, b = 1/2 and c = –1/2
 So, frequency f = KL
–1
F
1/2
M
–1/2
=
M
F
L
K
MF
L
K 2/12/1


Chapter-I
1.3
18. a) h =
rg
SCos2


LHS = [L]
Surface tension = S = F/I =
2
2MLT
[MT ]
L



Density =  = M/V = [ML
–3
T
0
]
Radius = r = [L], g = [LT
–2
]
RHS =
2
0 1 0
3 0 2
2Scos [MT ]
[M L T ] [L]
rg [ML T ][L][LT ]

 

  


LHS = RHS
So, the relation is correct
b) v =

p
where v = velocity
LHS = Dimension of v = [LT
–1
]
Dimension of p = F/A = [ML
–1
T
–2
]
Dimension of  = m/V = [ML
–3
]
RHS =
1 2
2 2 1/ 2
3
p [ML T ]
[L T ]
[ML ]
 


 

= 1
[LT ]
So, the relation is correct.
c) V = (pr
4
t) / (8l)
LHS = Dimension of V = [L
3
]
Dimension of p = [ML
–1
T
–2
], r
4
= [L
4
], t = [T]
Coefficient of viscosity = [ML
–1
T
–1
]
RHS =
4 1 2 4
1 1
pr t [ML T ][L ][T]
8 l [ML T ][L]
 
 



d) v = )I/mgl(
2
1

LHS = dimension of v = [T
–1
]
RHS = )I/mgl( =
2
2
[M][LT ][L]
[ML ]

= [T
–1
]
LHS = RHS
19. Dimension of the left side =
2 2 2 2
dx L
(a x ) (L L )

 
  = [L
0
]
Dimension of the right side = 





x
a
sin
a
1 1
= [L
–1
]
So, the dimension of
  )xa(
dx
22
≠ 





x
a
sin
a
1 1
So, the equation is dimensionally incorrect.

Chapter-I
1.4
20. Important Dimensions and Units :
Physical quantity Dimension SI unit
Force (F)
]TLM[ 211 
newton
Work (W)
]TLM[ 221 
joule
Power (P)
]TLM[ 321 
watt
Gravitational constant (G)
]TLM[ 231 
N-m
2
/kg
2
Angular velocity ()
]T[ 1
radian/s
Angular momentum (L)
]TLM[ 121 
kg-m
2
/s
Moment of inertia (I)
]LM[ 21
kg-m
2
Torque ()
]TLM[ 221 
N-m
Young’s modulus (Y)
]TLM[ 211 
N/m
2
Surface Tension (S)
]TM[ 21 
N/m
Coefficient of viscosity ()
]TLM[ 111 
N-s/m
2
Pressure (p)
]TLM[ 211 
N/m
2
(Pascal)
Intensity of wave (I)
]TM[ 31 
watt/m
2
Specific heat capacity (c)
]KTL[ 122 
J/kg-K
Stefan’s constant ()
]KTM[ 431 
watt/m
2
-k
4
Thermal conductivity (k)
]KTLM[ 1311 
watt/m-K
Current density (j)
]LI[ 21 
ampere/m
2
Electrical conductivity ()
]LMTI[ 3132 

–1
m
–1

Electric dipole moment (p)
]TIL[ 111
C-m
Electric field (E)
]TILM[ 3111 
V/m
Electrical potential (V)
]TILM[ 3121 
volt
Electric flux ()
]LITM[ 3131 
volt/m
Capacitance (C)
]LMTI[ 2142 
farad (F)
Permittivity ()
]LMTI[ 3142 
C
2
/N-m
2
Permeability ()
]TILM[ 3211 
Newton/A
2
Magnetic dipole moment (M)
]LI[ 21
N-m/T
Magnetic flux ()
]TILM[ 2121 
Weber (Wb)
Magnetic field (B)
]TIM[ 211 
tesla
Inductance (L)
]TILM[ 2221 
henry
Resistance (R)
]TILM[ 3221 
ohm ()
* * * *

2.1
CHAPTER – 2
1. As shown in the figure,
The angle between A

and B

= 110° – 20° = 90°
|A|

= 3 and |B|

= 4m
Resultant R =  cosAB2BA 22
= 5 m
Let  be the angle between R

and A

 = 







90cos43
90sin4
tan 1
= tan
–1
(4/3) = 53°
 Resultant vector makes angle (53° + 20°) = 73° with x-axis.
2. Angle between A

and B

is  = 60° – 30° =30°
| A

| and |B

| = 10 unit
R = 2 2
10 10 2.10.10.cos30   = 19.3
 be the angle between R

and A

 = tan
–1 110sin30 1
tan
10 10cos30 2 3
   
       
= tan
–1
(0.26795) = 15°
 Resultant makes 15° + 30° = 45° angle with x-axis.

3. x component of A

= 100 cos 45° = 2/100 unit
x component of B

= 100 cos 135° = 2/100
x component of C

= 100 cos 315° = 2/100
Resultant x component = 2/100 – 2/100 + 2/100 = 2/100
y component of A

= 100 sin 45° = 2/100 unit
y component of B

= 100 sin 135° = 2/100
y component of C

= 100 sin 315° = – 2/100
Resultant y component = 2/100 + 2/100 – 2/100 = 2/100
Resultant = 100
Tan  =
componentx
componenty
= 1
  = tan
–1
(1) = 45°
The resultant is 100 unit at 45° with x-axis.
4. j3i4a

 , j4i3b


a) 22
34|a| 

= 5
b) 169|b| 

= 5
c) 27|j7i7||ba| 

d) ˆ ˆ ˆ ˆa b ( 3 4)i ( 4 3)j i j        

2 2
| a b | 1 ( 1) 2    

.
x
y

R

B

A

20
x
y

B

A

30°
60°
315°
45°
135°

Chapter-2
2.2
5. x component of OA = 2cos30° = 3
x component of BC = 1.5 cos 120° = –0.75
x component of DE = 1 cos 270° = 0
y component of OA = 2 sin 30° = 1
y component of BC = 1.5 sin 120° = 1.3
y component of DE = 1 sin 270° = –1
Rx = x component of resultant = 075.03  = 0.98 m
Ry = resultant y component = 1 + 1.3 – 1 = 1.3 m
So, R = Resultant = 1.6 m
If it makes and angle  with positive x-axis
Tan  =
componentx
componenty
= 1.32
  = tan
–1
1.32

6. |a|

= 3m |b|

= 4
a) If R = 1 unit   cos.4.3.243 22
= 1
 = 180°
b)  cos.4.3.243 22
= 5
 = 90°
c)  cos.4.3.243 22
= 7
 = 0°
Angle between them is 0°.

7. Kˆ4Jˆ5.0iˆ2AD  = ˆ ˆ6i 0.5j
AD = 22
DEAE  = 6.02 KM
Tan  = DE / AE = 1/12
= tan
–1
(1/12)
The displacement of the car is 6.02 km along the distance tan
–1
(1/12) with positive x-axis.
8. In ABC, tan = x/2 and in DCE, tan = (2 – x)/4 tan  = (x/2) = (2 – x)/4 = 4x
 4 – 2x = 4x
 6x = 4  x = 2/3 ft
a) In ABC, AC = 22
BCAB  =
2
10
3
ft
b) In CDE, DE = 1 – (2/3) = 4/3 ft
CD = 4 ft. So, CE = 22
DECD  =
4
10
3
ft
c) In AGE, AE = 22
GEAG  = 2 2 ft.
9. Here the displacement vector kˆ3jˆ4iˆ7r 

a) magnitude of displacement = 74 ft
b) the components of the displacement vector are 7 ft, 4 ft and 3 ft.
2m
D
A
E
B
x
O
y
1m
1.5m
90°
30° 60°
6m
EA
D
B
0.5 km

C
2m
4m
0.5 km
2–x
G
A
D
B BC = 2 ft
AF = 2 ft
DE = 2x
x
C
E
F
r
z
Y

Chapter-2
2.3
10. a

is a vector of magnitude 4.5 unit due north.
a) 3| a

| = 3  4.5 = 13.5
3a

is along north having magnitude 13.5 units.
b) –4| a

| = –4  1.5 = –6 unit
–4 a

is a vector of magnitude 6 unit due south.
11. | a

| = 2 m, | b

| = 3 m
angle between them  = 60°
a)  60cos|b||a|ba

= 2  3  1/2 = 3 m
2
b)  60sin|b||a||ba|

= 2  3  3/ 2 = 3 3 m
2
.
12. We know that according to polygon law of vector addition, the resultant
of these six vectors is zero.
Here A = B = C = D = E = F (magnitude)
So, Rx = A cos + A cos /3 + A cos 2/3 + A cos 3/3 + A cos 4/4 +
A cos 5/5 = 0
[As resultant is zero. X component of resultant Rx = 0]
= cos  + cos /3 + cos 2/3 + cos 3/3 + cos 4/3 + cos 5/3 = 0
Note : Similarly it can be proved that,
sin  + sin /3 + sin 2/3 + sin 3/3 + sin 4/3 + sin 5/3 = 0
13. a 2i 3 j 4k; b 3i 4 j 5k     
      
 cosabba


ab
ba
cos 1


 
 1 1
2 2 2 2 2 2
2 3 3 4 4 5 38
cos cos
14502 3 4 3 4 5
       
  
    
14. 0)BA(A 

(claim)
As, nˆsinABBA 

AB sin  nˆ is a vector which is perpendicular to the plane containing A

and B

, this implies that it is
also perpendicular to A

. As dot product of two perpendicular vector is zero.
Thus 0)BA(A 

.
15. ˆ ˆ Â 2i 3j 4k  

, ˆ ˆ ˆB 4i 3j 2k  

ˆ ˆ î j k
A B 2 3 4
4 3 2
 
 
 ˆ ˆ ˆ ˆ ˆ î(6 12) j(4 16) k(6 12) 6i 12j 6k         .
16. Given that A

, B

and C

are mutually perpendicular
A

× B

is a vector which direction is perpendicular to the plane containing A

and B

.
Also C

is perpendicular to A

and B

 Angle between C

and A

× B

is 0° or 180° (fig.1)
So, C

× ( A

× B

) = 0
The converse is not true.
For example, if two of the vector are parallel, (fig.2), then also
C

× ( A

× B

) = 0
So, they need not be mutually perpendicular.
A1
60° = /3
A2
A3
A4A5
A6
(A B)
 
A

B

C

A

B

C


Chapter-2
2.4
17. The particle moves on the straight line PP’ at speed v.
From the figure,
nˆsinv)OP(vOP  = v(OP) sin  ˆn = v(OQ) ˆn 
It can be seen from the figure, OQ = OP sin  = OP’ sin ’
So, whatever may be the position of the particle, the magnitude and
direction of vOP

 remain constant.
 vOP

 is independent of the position P.
18. Give 0)Bv(qEqF 

 )Bv(E


So, the direction of Bv

 should be opposite to the direction of E

. Hence,
v

should be in the positive yz-plane.
Again, E = vB sin   v =
sinB
E
For v to be minimum,  = 90° and so vmin = F/B
So, the particle must be projected at a minimum speed of E/B along +ve z-axis ( = 90°) as shown in the
figure, so that the force is zero.
19. For example, as shown in the figure,
A B
 
B

along west
B C

A

along south
C

along north
A B
 
= 0  A B B C  
  
B C

= 0 But B C

20. The graph y = 2x
2
should be drawn by the student on a graph paper for exact
results.
To find slope at any point, draw a tangent at the point and extend the line to meet
x-axis. Then find tan  as shown in the figure.
It can be checked that,
Slope = tan  = )x2(
dx
d
dx
dy 2
 = 4x
Where x = the x-coordinate of the point where the slope is to be measured.
21. y = sinx
So, y + y = sin (x + x)
y = sin (x + x) – sin x
= sin
3 100 3
   
  
 
= 0.0157.
22. Given that, i = RC/t
0ei 
 Rate of change of current = RC/t
0
RC/i
0 e
dt
d
iei
dt
d
dt
di 
 = t / RC0i
e
RC


When a) t = 0,
RC
i
dt
di 

b) when t = RC,
RCe
i
dt
di 

c) when t = 10 RC, 10
0
RCe
i
dt
di 

V

Q

O

P P
 x
E

B

V

y
C

B

A
 B


x
y=2x2 y
x
y = sinx
y

Chapter-2
2.5
23. Equation i = RC/t
0ei 
i0 = 2A, R = 6  10
–5
, C = 0.0500  10
–6
F = 5  10
–7
F
a) i =
3 7
0.3 0.3
6 0 5 10 0.3 2
2 e 2 e amp
e

   
   
     
    .
b) t / RC0idi
e
dt RC

 when t = 0.3 sec  ( 0.3 / 0.3)di 2 20
e Amp/ sec
dt 0.30 3e
 
  
c) At t = 0.31 sec, i = ( 0.3 / 0.3) 5.8
2e Amp
3e

 .
24. y = 3x
2
+ 6x + 7
 Area bounded by the curve, x axis with coordinates with x = 5 and x = 10 is
given by,
Area = 
y
0
dy =
10
2
5
(3x 6x 7)dx  = 
10 103 2
10
5
5 5
x x
3 5 7x
3 3
 
  
 
= 1135 sq.units.
25. Area = 
y
0
dy = 0
0
sinxdx [cosx]


  = 2
26. The given function is y = e
–x
When x = 0, y = e
–0
= 1
x increases, y value deceases and only at x = , y = 0.
So, the required area can be found out by integrating the function from 0 to .
So, Area = x x
0
0
e dx [e ] 1

  
   .
27. bxa
length
mass

a) S.I. unit of ‘a’ = kg/m and SI unit of ‘b’ = kg/m
2
(from principle of
homogeneity of dimensions)
b) Let us consider a small element of length ‘dx’ at a distance x from the
origin as shown in the figure.
 dm = mass of the element =  dx = (a + bx) dx
So, mass of the rod = m = dx)bxa(dm
L
0
  =
L2 2
0
bx bL
ax aL
2 2
 
   
 

28.
dp
dt
= (10 N) + (2 N/S)t
momentum is zero at t = 0
 momentum at t = 10 sec will be
dp = [(10 N) + 2Ns t]dt
p 10 10
0 0 0
dp 10dt (2tdt)    = 
102
10
0
0
t
10t 2
2

 

= 200 kg m/s.
5
y = 3x2
+ 6x + 7
10
x
y
y = sinx
y
x
y
x
x =1
O
y
x

Chapter-2
2.6
29. The change in a function of y and the independent variable x are related as 2
x
dx
dy
 .
 dy = x
2
dx
Taking integration of both sides,
2
dy x dx   y =
3
x
c
3

 y as a function of x is represented by y =
3
x
c
3
 .
30. The number significant digits
a) 1001 No.of significant digits = 4
b) 100.1 No.of significant digits = 4
c) 100.10 No.of significant digits = 5
d) 0.001001 No.of significant digits = 4
31. The metre scale is graduated at every millimeter.
1 m = 100 mm
The minimum no.of significant digit may be 1 (e.g. for measurements like 5 mm, 7 mm etc) and the
maximum no.of significant digits may be 4 (e.g.1000 mm)
So, the no.of significant digits may be 1, 2, 3 or 4.
32. a) In the value 3472, after the digit 4, 7 is present. Its value is greater than 5.
So, the next two digits are neglected and the value of 4 is increased by 1.
 value becomes 3500
b) value = 84
c) 2.6
d) value is 28.
33. Given that, for the cylinder
Length = l = 4.54 cm, radius = r = 1.75 cm
Volume = r
2
l =   (4.54)  (1.75)
2
Since, the minimum no.of significant digits on a particular term is 3, the result should have
3 significant digits and others rounded off.
So, volume V = r
2
l = (3.14)  (1.75)  (1.75)  (4.54) = 43.6577 cm
3
Since, it is to be rounded off to 3 significant digits, V = 43.7 cm
3
.
34. We know that,
Average thickness =
2.17 2.17 2.18
3
 
= 2.1733 mm
Rounding off to 3 significant digits, average thickness = 2.17 mm.
Actual effective length = (90.0 + 2.13) cm
But, in the measurement 90.0 cm, the no. of significant digits is only 2.
So, the addition must be done by considering only 2 significant digits of each
measurement.
So, effective length = 90.0 + 2.1 = 92.1 cm.
* * * *
r
l
90cm
2.13cm

3.1
CHAPTER – 3
1. a) Distance travelled = 50 + 40 + 20 = 110 m
b) AF = AB – BF = AB – DC = 50 – 20 = 30 M
His displacement is AD
AD = m504030DFAF 2222

In AED tan  = DE/AE = 30/40 = 3/4
  = tan–1
(3/4)
His displacement from his house to the field is 50 m,
tan–1
(3/4) north to east.
2. O  Starting point origin.
i) Distance travelled = 20 + 20 + 20 = 60 m
ii) Displacement is only OB = 20 m in the negative direction.
Displacement  Distance between final and initial position.
3. a) Vave of plane (Distance/Time) = 260/0.5 = 520 km/hr.
b) Vave of bus = 320/8 = 40 km/hr.
c) plane goes in straight path
velocity = aveV

= 260/0.5 = 520 km/hr.
d) Straight path distance between plane to Ranchi is equal to the displacement of bus.
 Velocity = aveV

= 260/8 = 32.5 km/hr.
4. a) Total distance covered 12416 – 12352 = 64 km in 2 hours.
Speed = 64/2 = 32 km/h
b) As he returns to his house, the displacement is zero.
Velocity = (displacement/time) = 0 (zero).
5. Initial velocity u = 0 ( starts from rest)
Final velocity v = 18 km/hr = 5 sec
(i.e. max velocity)
Time interval t = 2 sec.
 Acceleration = aave =
2
5
t
uv


= 2.5 m/s2
.
6. In the interval 8 sec the velocity changes from 0 to 20 m/s.
Average acceleration = 20/8 = 2.5 m/s2






time
velocityinchange
Distance travelled S = ut + 1/2 at2
 0 + 1/2(2.5)82
= 80 m.
7. In 1st
10 sec S1 = ut + 1/2 at2
 0 + (1/2 × 5 × 102
) = 250 ft.
At 10 sec v = u + at = 0 + 5 × 10 = 50 ft/sec.
 From 10 to 20 sec (t = 20 – 10 = 10 sec) it moves with uniform
velocity 50 ft/sec,
A
E
S
N
W

40 m
40 m
50 m
20 m
30 m
B C
D
E
A  Initial point
(starting point)
A
X
O
Y
B
(20 m, 0)(–20 m, 0)
Initial velocity
u = 0
20
84
10
Time in sec
2010 30
S (in ft)
0
t (sec)
250
750
1000

Chapter-3
3.2
Distance S2 = 50 × 10 = 500 ft
Between 20 sec to 30 sec acceleration is constant i.e. –5 ft/s2
. At 20 sec velocity is 50 ft/sec.
t = 30 – 20 = 10 s
S3 = ut + 1/2 at2
= 50 × 10 + (1/2)(–5)(10)2
= 250 m
Total distance travelled is 30 sec = S1 + S2 + S3 = 250 + 500 + 250 = 1000 ft.
8. a) Initial velocity u = 2 m/s.
final velocity v = 8 m/s
time = 10 sec,
acceleration =
10
28
ta
uv 


= 0.6 m/s2
b) v2
– u2
= 2aS
 Distance S =
a2
uv 22

=
6.02
28 22


= 50 m.
c) Displacement is same as distance travelled.
Displacement = 50 m.
9. a) Displacement in 0 to 10 sec is 1000 m.
time = 10 sec.
Vave = s/t = 100/10 = 10 m/s.
b) At 2 sec it is moving with uniform velocity 50/2.5 = 20 m/s.
at 2 sec. Vinst = 20 m/s.
At 5 sec it is at rest.
Vinst = zero.
At 8 sec it is moving with uniform velocity 20 m/s
Vinst = 20 m/s
At 12 sec velocity is negative as it move towards initial position. Vinst = – 20 m/s.
10. Distance in first 40 sec is,  OAB + BCD
=
2
1
× 5 × 20 +
2
1
× 5 × 20 = 100 m.
Average velocity is 0 as the displacement is zero.
11. Consider the point B, at t = 12 sec
At t = 0 ; s = 20 m
and t = 12 sec s = 20 m
So for time interval 0 to 12 sec
Change in displacement is zero.
So, average velocity = displacement/ time = 0
 The time is 12 sec.
12. At position B instantaneous velocity has direction along BC . For
average velocity between A and B.
Vave = displacement / time = )t/AB( t = time
10
t
5
2
4
6
8
t
10
(slope of the graph at t = 2 sec)
2.5
50
100
0 t
5 7.5 15
40
t (sec)
20
5 m/s
O
A
B
C
D
20 B
10
10 12 20
4 B
2
C
42
x
6
y

Chapter-3
3.3
We can see that AB is along BC i.e. they are in same direction.
The point is B (5m, 3m).
13. u = 4 m/s, a = 1.2 m/s2
, t = 5 sec
Distance = s = 2
at
2
1
ut 
= 4(5) + 1/2 (1.2)52
= 35 m.
14. Initial velocity u = 43.2 km/hr = 12 m/s
u = 12 m/s, v = 0
a = –6 m/s2
(deceleration)
Distance S =
)6(2
uv 22


= 12 m

Chapter-3
3.4
15. Initial velocity u = 0
Acceleration a = 2 m/s2
. Let final velocity be v (before applying breaks)
t = 30 sec
v = u + at  0 + 2 × 30 = 60 m/s
a) S1 = 2
at
2
1
ut  = 900 m
when breaks are applied u = 60 m/s
v = 0, t = 60 sec (1 min)
Declaration a = (v – u)/t = = (0 – 60)/60 = –1 m/s2
.
S2 =
a2
uv 22


= 1800 m
Total S = S1 + S2 = 1800 + 900 = 2700 m = 2.7 km.
b) The maximum speed attained by train v = 60 m/s
c) Half the maximum speed = 60/2= 30 m/s
Distance S =
a2
uv 22

=
22
030 22


= 225 m from starting point
When it accelerates the distance travelled is 900 m. Then again declarates and attain 30
m/s.
 u = 60 m/s, v = 30 m/s, a = –1 m/s2
Distance =
a2
uv 22

=
)1(2
6030 22


= 1350 m
Position is 900 + 1350 = 2250 = 2.25 km from starting point.
16. u = 16 m/s (initial), v = 0, s = 0.4 m.
Deceleration a =
s2
uv 22

= –320 m/s2
.
Time = t =
320
160
a
uv




= 0.05 sec.
17. u = 350 m/s, s = 5 cm = 0.05 m, v = 0
Deceleration = a =
s2
uv 22

=
05.02
)350(0 2


= –12.2 × 105
m/s2
.
Deceleration is 12.2 × 105
m/s2
.
18. u = 0, v = 18 km/hr = 5 m/s, t = 5 sec
a =
5
05
t
uv 


= 1 m/s2
.
s = 2
at
2
1
ut  = 12.5 m
a) Average velocity Vave = (12.5)/5 = 2.5 m/s.
b) Distance travelled is 12.5 m.
19. In reaction time the body moves with the speed 54 km/hr = 15 m/sec (constant speed)
Distance travelled in this time is S1 = 15 × 0.2 = 3 m.
When brakes are applied,
u = 15 m/s, v = 0, a = –6 m/s2
(deceleration)

Chapter-3
3.5
S2 =
)6(2
150
a2
uv 222




= 18.75 m
Total distance s = s1 + s2 = 3 + 18.75 = 21.75 = 22 m.

Chapter-3
3.6
20.
Driver X
Reaction time 0.25
Driver Y
Reaction time 0.35
A (deceleration on hard
braking = 6 m/s2
)
Speed = 54 km/h
Braking distance a= 19 m
Total stopping distance b =
22 m
Speed = 72 km/h
Braking distance c = 33 m
Total stopping distance d = 39
m.
B (deceleration on hard
braking = 7.5 m/s2
)
Speed = 54 km/h
Braking distance e = 15 m
Total stopping distance f = 18
m
Speed = 72 km/h
Braking distance g = 27 m
Total stopping distance h = 33
m.
a =
)6(2
150 22


= 19 m
So, b = 0.2 × 15 + 19 = 33 m
Similarly other can be calculated.
Braking distance : Distance travelled when brakes are applied.
Total stopping distance = Braking distance + distance travelled in reaction time.
21. VP = 90 km/h = 25 m/s.
VC = 72 km/h = 20 m/s.
In 10 sec culprit reaches at point B from A.
Distance converted by culprit S = vt = 20 × 10 = 200 m.
At time t = 10 sec the police jeep is 200 m behind the
culprit.
Time = s/v = 200 / 5 = 40 s. (Relative velocity is considered).
In 40 s the police jeep will move from A to a distance S, where
S = vt = 25 × 40 = 1000 m = 1.0 km away.
 The jeep will catch up with the bike, 1 km far from the turning.
22. v1 = 60 km/hr = 16.6 m/s.
v2 = 42 km/h = 11.6 m/s.
Relative velocity between the cars = (16.6 – 11.6) = 5 m/s.
Distance to be travelled by first car is 5 + t = 10 m.
Time = t = s/v = 0/5 = 2 sec to cross the 2nd
car.
In 2 sec the 1st
car moved = 16.6 × 2 = 33.2 m
H also covered its own length 5 m.
 Total road distance used for the overtake = 33.2 + 5 = 38 m.
23. u = 50 m/s, g = –10 m/s2
when moving upward, v = 0 (at highest point).
a) S =
)10(2
500
a2
uv 222




= 125 m
maximum height reached = 125 m
b) t = (v – u)/a = (0 – 50)/–10 = 5 sec
c) s = 125/2 = 62.5 m, u = 50 m/s, a = –10 m/s2
,
culprit
A
t=0
Police
t = 10 sec
B
P
C
Before crossing
V1 
5 m
V2 
5 m
After crossing
V1 
V2 
10 m

Chapter-3
3.7
v2
– u2
= 2as
 v = )5.62)(10(250)as2u( 22
 = 35 m/s.
24. Initially the ball is going upward
u = –7 m/s, s = 60 m, a = g = 10 m/s2
s = 2
at
2
1
ut   60 = –7t + 1/2 10t2
 5t2
– 7t – 60 = 0
t =
52
)60(5.4497


=
10
34.357 
taking positive sign t =
10
34.357 
= 4.2 sec ( t  –ve)
Therefore, the ball will take 4.2 sec to reach the ground.
25. u = 28 m/s, v = 0, a = –g = –9.8 m/s2
a) S =
)8.9(2
280
a2
uv 2222



= 40 m
b) time t =
8.9
280
a
uv




= 2.85
t = 2.85 – 1 = 1.85
v = u + at = 28 – (9.8) (1.85) = 9.87 m/s.
 The velocity is 9.87 m/s.
c) No it will not change. As after one second velocity becomes zero for any initial velocity and
deceleration is g = 9.8 m/s2
remains same. Fro initial velocity more than 28 m/s max height
increases.
26. For every ball, u = 0, a = g = 9.8 m/s2
 4th
ball move for 2 sec, 5th
ball 1 sec and 3rd
ball 3 sec when 6th
ball is being dropped.
For 3rd
ball t = 3 sec
S3 = 2
at
2
1
ut  = 0 + 1/2 (9.8)32
= 4.9 m below the top.
For 4th
ball, t = 2 sec
S2 = 0 + 1/2 gt2
= 1/2 (9.8)22
= 19.6 m below the top (u = 0)
For 5th
ball, t = 1 sec
S3 = ut + 1/2 at2
= 0 + 1/2 (9.8)t2
= 4.98 m below the top.
27. At point B (i.e. over 1.8 m from ground) the kid should be catched.
For kid initial velocity u = 0
Acceleration = 9.8 m/s2
Distance S = 11.8 – 1.8 = 10 m
S = 2
at
2
1
ut   10 = 0 + 1/2 (9.8)t2
 t2
= 2.04  t = 1.42.
In this time the man has to reach at the bottom of the building.
Velocity s/t = 7/1.42 = 4.9 m/s.
28. Let the true of fall be ‘t’ initial velocity u = 0
6th
5th
3rd
4th
11.8
7m
1.8m
10m

Chapter-3
3.8
Acceleration a = 9.8 m/s2
Distance S = 12/1 m
 S = 2
at
2
1
ut 
 12.1 = 0 + 1/2 (9.8) × t2
 t2
=
9.4
1.12
= 2.46  t = 1.57 sec
For cadet velocity = 6 km/hr = 1.66 m/sec
Distance = vt = 1.57 × 1.66 = 2.6 m.
The cadet, 2.6 m away from tree will receive the berry on his uniform.
29. For last 6 m distance travelled s = 6 m, u = ?
t = 0.2 sec, a = g = 9.8 m/s2
S = 2
at
2
1
ut   6 = u(0.2) + 4.9 × 0.04
 u = 5.8/0.2 = 29 m/s.
For distance x, u = 0, v = 29 m/s, a = g = 9.8 m/s2
S =
8.92
029
a2
uv 2222




= 42.05 m
Total distance = 42.05 + 6 = 48.05 = 48 m.
30. Consider the motion of ball form A to B.
B  just above the sand (just to penetrate)
u = 0, a = 9.8 m/s2
, s = 5 m
S = 2
at
2
1
ut 
 5 = 0 + 1/2 (9.8)t2
 t2
= 5/4.9 = 1.02  t = 1.01.
 velocity at B, v = u + at = 9.8 × 1.01 (u = 0) =9.89 m/s.
From motion of ball in sand
u1 = 9.89 m/s, v1 = 0, a = ?, s = 10 cm = 0.1 m.
a =
1.02
)89.9(0
s2
uv 22
1
2
1




= – 490 m/s2
The retardation in sand is 490 m/s2
.
31. For elevator and coin u = 0
As the elevator descends downward with acceleration a (say)
The coin has to move more distance than 1.8 m to strike the floor. Time taken t = 1 sec.
Sc = 2
ta
2
1
ut  = 0 + 1/2 g(1)2
= 1/2 g
Se = 2
at
2
1
ut  = u + 1/2 a(1)2
= 1/2 a
Total distance covered by coin is given by = 1.8 + 1/2 a = 1/2 g
 1.8 +a/2 = 9.8/2 = 4.9
 a = 6.2 m/s2
= 6.2 × 3.28 = 20.34 ft/s2
.
32. It is a case of projectile fired horizontally from a height.
2.6m
1.66 m/s
t=0.2 sec6m
x m
10cm
A
6 m
B
C
a
6ft=1.8m
1/2a

Chapter-3
3.9
h = 100 m, g = 9.8 m/s2
a) Time taken to reach the ground t = )g/h2(
=
8.9
1002
= 4.51 sec.
b) Horizontal range x = ut = 20 × 4.5 = 90 m.
c) Horizontal velocity remains constant through out the
motion.
At A, V = 20 m/s
A Vy = u + at = 0 + 9.8 × 4.5 = 44.1 m/s.
Resultant velocity Vr = 22
20)1.44(  = 48.42 m/s.
Tan  =
20
1.44
V
V
x
y
 = 2.205
  = tan–1
(2.205) = 60°.
The ball strikes the ground with a velocity 48.42 m/s at an angle 66° with horizontal.
33. u = 40 m/s, a = g= 9.8 m/s2
,  = 60° Angle of projection.
a) Maximum height h =
102
)60(sin40
g2
sinu 2222




= 60 m
b) Horizontal range X = (u2
sin 2) / g = (402
sin 2(60°)) / 10 = 380 m.
A
100m

20m/s
Vx
Vy
Vr

Chapter-3
3.10
34. g = 9.8 m/s2
, 32.2 ft/s2
; 40 yd = 120 ft
horizontal range x = 120 ft, u = 64 ft/s,  = 45°
We know that horizontal range X = u cos t
 t =


 45cos64
120
cosu
x
= 2.65 sec.
y = u sin (t) – 1/2 gt2
= 2
)65.2)(2.32(
2
1
)65.2(2
1
64 
= 7.08 ft which is less than the height of goal post.
In time 2.65, the ball travels horizontal distance 120 ft (40 yd) and vertical height 7.08 ft which
is less than 10 ft. The ball will reach the goal post.
35. The goli move like a projectile.
Here h = 0.196 m
Horizontal distance X = 2 m
Acceleration g = 9.8 m/s2
.
Time to reach the ground i.e.
t =
8.9
196.02
g
h2 
 = 0.2 sec
Horizontal velocity with which it is projected be u.
 x = ut
 u =
2.0
2
t
x
 = 10 m/s.
36. Horizontal range X = 11.7 + 5 = 16.7 ft covered by te bike.
g = 9.8 m/s2
= 32.2 ft/s2
.
y = x tan  – 2
22
u2
secgx 
To find, minimum speed for just crossing, the ditch
y = 0 ( A is on the x axis)
 x tan  = 2
22
u2
secgx 








2sin
gx
cossin2
gx
tanx2
secgx
u
22
2
 u =
2/1
)7.16)(2.32(
(because sin 30° = 1/2)
 u = 32.79 ft/s = 32 ft/s.
37. tan  = 171/228   = tan–1
(171/228)
The motion of projectile (i.e. the packed) is from A. Taken reference axis at A.
  = –37° as u is below x-axis.
u = 15 ft/s, g = 32.2 ft/s2
, y = –171 ft
y = x tan  – 2
22
u2
secgx 
 –171 = –x (0.7536) –
)225(2
)568.1(gx2
 0.1125x2
+ 0.7536 x – 171 = 0
x = 35.78 ft (can be calculated)
10 ft
120 ft
u
0.196m
2 m
y
15° 15°
11.7ft 5ft5ft
x
u
228ft


171ft

Chapter-3
3.11
Horizontal range covered by the packet is 35.78 ft.
So, the packet will fall 228 – 35.78 = 192 ft short of his friend.

Chapter-3
3.12
38. Here u = 15 m/s,  = 60°, g = 9.8 m/s2
Horizontal range X =
8.9
)602sin()15(
g
2sinu 22



= 19.88 m
In first case the wall is 5 m away from projection point, so it is in the horizontal range of
projectile. So the ball will hit the wall. In second case (22 m away) wall is not within the
horizontal range. So the ball would not hit the wall.
39. Total of flight T =
g
sinu2 
Average velocity =
time
ntdisplacemeinchange
From the figure, it can be said AB is horizontal. So there is no effect of vertical component of
the velocity during this displacement.
So because the body moves at a constant speed of ‘u cos ’ in horizontal direction.
The average velocity during this displacement will be u cos  in the horizontal
direction.
40. During the motion of bomb its horizontal velocity u remains constant and is same
as that of aeroplane at every point of its path. Suppose the bomb explode i.e.
reach the ground in time t. Distance travelled in horizontal direction by bomb =
ut = the distance travelled by aeroplane. So bomb explode vertically below the
aeroplane.
Suppose the aeroplane move making angle  with horizontal. For both bomb and
aeroplane, horizontal distance is u cos  t. t is time for bomb to reach the ground.
So in this case also, the bomb will explode vertically below aeroplane.
41. Let the velocity of car be u when the ball is thrown. Initial velocity of car is = Horizontal
velocity of ball.
Distance travelled by ball B Sb = ut (in horizontal direction)
And by car Sc = ut + 1/2 at2
where t  time of flight of ball in air.
 Car has travelled extra distance Sc – Sb = 1/2 at2
.
Ball can be considered as a projectile having  = 90°.
 t =
8.9
8.92
g
sinu2 


= 2 sec.
 Sc – Sb = 1/2 at2
= 2 m
 The ball will drop 2m behind the boy.
42. At minimum velocity it will move just touching point E reaching the ground.
A is origin of reference coordinate.
If u is the minimum speed.
X = 40, Y = –20,  = 0°
 Y = x tan  – g 2
22
u2
secx 
(because g = 10 m/s2
= 1000
cm/s2
)
 –20 = x tan  – 2
2
u2
1401000 
A
H/2 H/2
B
H

u
9.8 m/s
1 m/s2
A
30 cm
C
E
20 cm
20 cm
20 cm 10 cm

Chapter-3
3.13
 u = 200 cm/s = 2 m/s.
 The minimum horizontal velocity is 2 m/s. 
43. a) As seen from the truck the ball moves vertically upward comes back. Time taken = time
taken by truck to cover 58.8 m.
 time =
7.14
8.58
v
s
 = 4 sec. (V = 14.7 m/s of truck)
u = ?, v = 0, g = –9.8 m/s2
(going upward), t = 4/2 = 2 sec.
v = u + at  0 = u – 9.8 × 2  u = 19.6 m/s. (vertical upward velocity).
b) From road it seems to be projectile motion.
Total time of flight = 4 sec
In this time horizontal range covered 58.8 m = x
 X = u cos  t
 u cos  = 14.7 …(1)
Taking vertical component of velocity into consideration.
y =
)8.9(2
)6.19(0 22


= 19.6 m [from (a)]
 y = u sin  t – 1/2 gt2
 19.6 = u sin  (2) – 1/2 (9.8)22
 2u sin  = 19.6 × 2
 u sin  = 19.6 …(ii)


cosu
sinu
= tan  
7.14
6.19
= 1.333
  = tan–1
(1.333) = 53°
Again u cos  = 14.7
 u =
53cosu
7.14
= 24.42 m/s.
The speed of ball is 42.42 m/s at an angle 53° with horizontal as seen from the road.
44.  = 53°, so cos 53° = 3/5
Sec2
 = 25/9 and tan  = 4/3
Suppose the ball lands on nth bench
So, y = (n – 1)1 …(1) [ball starting point 1 m above ground]
Again y = x tan  – 2
22
u2
secgx 
[x = 110 + n – 1 = 110 + y]
 y = (110 + y)(4/3) – 2
2
352
)9/25()y110(10


 2
2
3518
)y110(250
y
3
4
3
440



From the equation, y can be calculated.
 y = 5
 n – 1 = 5  n = 6.
The ball will drop in sixth bench. 
45. When the apple just touches the end B of the boat.
x = 5 m, u = 10 m/s, g = 10 m/s2
,  = ?
53°
y
53°
35 m/s

Chapter-3
3.14
x =
g
2sinu2

 5 =
10
2sin102

 5 = 10 sin 2
 sin 2 = 1/2  sin 30° or sin 150°
  = 15° or 75°
Similarly for end C, x = 6 m
Then 21 = sin–1
(gx/u2
) = sin–1
(0.6) = 182° or 71°.
So, for a successful shot,  may very from 15° to 18° or 71° to 75°.
46. a) Here the boat moves with the resultant velocity R. But the
vertical component 10 m/s takes him to the opposite shore.
Tan  = 2/10 = 1/5
Velocity = 10 m/s
distance = 400 m
Time = 400/10 = 40 sec.
b) The boat will reach at point C.
In ABC, tan  =
5
1
400
BC
AB
BC

 BC = 400/5 = 80 m.
47. a) The vertical component 3 sin  takes him to
opposite side.
Distance = 0.5 km, velocity = 3 sin  km/h
Time = hr
sin3
5.0
Velocity
Distance


= 10/sin min.
b) Here vertical component of velocity i.e. 3 km/hr takes him to opposite side.
Time = 16.0
3
5.0
Velocity
Distance
 hr
 0.16 hr = 60 × 0.16 = 9.6 = 10 minute.
48. Velocity of man mV

= 3 km/hr
BD horizontal distance for resultant velocity R.
X-component of resultant Rx = 5 + 3 cos 
t = 0.5 / 3sin
which is same for horizontal component of velocity.
H = BD = (5 + 3 cos ) (0.5 / 3 sin ) =


sin6
cos35
For H to be min (dH/d) = 0
 0
sin6
cos35
d
d









 –18 (sin2
 + cos2
) – 30 cos  = 0
 –30 cos  = 18  cos  = –18 / 30 = –3/5

10 m/s
5 m 1 m
1/2 m 1/2 m
10m/s
2m/s
400m
C
A
B

3sin 3km/h5km/h
5km/h
E
S
N
W
3km/h
5km/h

5km/h
D
3sin
B
500m
R

5km/h
3km/h

Chapter-3
3.15
Sin  =  2
cos1 = 4/5
 H =
)5/4(6
)5/3(35
sin6
cos35





=
3
2
km.
49. In resultant direction R

the plane reach the point B.
Velocity of wind wV

= 20 m/s
Velocity of aeroplane aV

= 150 m/s
In ACD according to sine formula

15
1
2
1
150
20
30sin
150
20
Asin
30sin
150
Asin
20



 A = sin–1
(1/15)
a) The direction is sin–1
(1/15) east of the line AB.
b) sin–1
(1/15) = 3°48
 30° + 3°48 = 33°48
R = 8433cos20)150(220150 22
 = 167 m/s.
Time =
167
500000
v
s
 = 2994 sec = 49 = 50 min.
50. Velocity of sound v, Velocity of air u, Distance between A and B be x.
In the first case, resultant velocity of sound = v + u
 (v + u) t1 = x
 v + u = x/t1 …(1)
In the second case, resultant velocity of sound = v – u
(v – u) t2 = x
 v – u = x/t2 …(2)
From (1) and (2) 2v = 






2121 t
1
t
1
x
t
x
t
x
 v = 








21 t
1
t
1
2
x
From (i) u = 






2111 t2
x
t2
x
t
x
v
t
x
= 






21 t
1
t
1
2
x
 Velocity of air V = 






21 t
1
t
1
2
x
And velocity of wind u = 






21 t
1
t
1
2
x
51. Velocity of sound v, velocity of air u
Velocity of sound be in direction AC so it can reach B with resultant velocity AD.
Angle between v and u is  > /2.
Resultant )uv(AD 22

Here time taken by light to reach B is neglected. So time lag between seeing and hearing =
time to here the drum sound.
W
S
150m/s
C R

N
30 D
s/m20Vw


20m/s30
20

R
150
x
vy B 
v v
x A
u
B
v
DA x

Chapter-3
3.16
t =
22
uv
x
velocity
ntDisplaceme



)t/x)(t/x(
x
)uv)(uv(
x
21


[from question no. 50]
= 21tt .
52. The particles meet at the centroid O of the triangle. At any instant the particles will form an
equilateral ABC with the same centroid.
Consider the motion of particle A. At any instant its velocity makes angle 30°. This component
is the rate of decrease of the distance AO.
Initially AO =
3
a
2
a
a
3
2
2
2







Therefore, the time taken for AO to become zero.
=
v3
a2
3v3
a2
30cosv
3/a




.
* * * *
CB
A
O

4.1
CHAPTER – 4
1. m = 1 gm = 1/1000 kg
F = 6.67 × 10
–17
N  F = 2
21
r
mGm
 6.67 × 20
–17
= 2
11
r
)1000/1()1000/1(1067.6  
 r
2
= 17
17
17
611
10
10
1064.6
101067.6







= 1
 r = 1 = 1 metre.
So, the separation between the particles is 1 m.
2. A man is standing on the surface of earth
The force acting on the man = mg ………(i)
Assuming that, m = mass of the man = 50 kg
And g = acceleration due to gravity on the surface of earth = 10 m/s
2
W = mg = 50× 10= 500 N = force acting on the man
So, the man is also attracting the earth with a force of 500 N
3. The force of attraction between the two charges
= 2
9
2
21
o r
1
109
r
qq
4
1


The force of attraction is equal to the weight
Mg = 2
9
r
109
 r
2
=
m
109
10m
109 89




[Taking g=10 m/s
2
]
 r =
m
103
m
109 48



mt
For example, Assuming m= 64 kg,
r = 4
4
10
8
3
64
103


= 3750 m
4. mass = 50 kg
r = 20 cm = 0.2 m
04.0
25001067.6
r
mm
GF
11
2
21
G



Coulomb’s force FC = 2
21
o r
qq
4
1

= 9 × 10
9
04.0
q2
Since, FG = Fc =
04.0
q109
04.0
2500107.6 2911


 
 q
2
= 25
109
107.6
04.0
2500107.6
9
911




 
= 18.07 × 10
–18
q = -18
1018.07  = 4.3 × 10
-9
C.
m1 = 1 gm m2 = 1 gm
r

Chapter-4
4.2
5. The limb exerts a normal force 48 N and frictional force of 20 N. Resultant magnitude of
the force,
R = 22
)20()48( 
= 4002304 
= 2704
= 52 N
6. The body builder exerts a force = 150 N.
Compression x = 20 cm = 0.2 m
Total force exerted by the man = f = kx
 kx = 150
 k =
2.0
150
=
2
1500
= 750 N/m
7. Suppose the height is h.
At earth station F = GMm/R
2
M = mass of earth
m = mass of satellite
R = Radius of earth
F= 2
)hR(
GMm

= 2
R2
GMm
 2R
2
= (R + h)
2
 R
2
– h
2
– 2Rh = 0
 h
2
+ 2Rh – R
2
= 0
H =
2
R4R4R2 22





 
=
2
R22R2 
= –R ± R2 = R  12 
= 6400 × (0.414)
= 2649.6 = 2650 km
8. Two charged particle placed at a sehortion 2m. exert a force of 20m.
F1 = 20 N. r1 = 20 cm
F2 = ? r2 = 25 cm
Since, F = 2
21
o r
qq
4
1

, F  2
r
1
2
1
2
2
2
1
r
r
F
F
  F2 = F1 ×
2
2
1
r
r






= 20 ×
2
25
20






= 20 ×
25
16
=
5
64
= 12.8 N = 13 N.
9. The force between the earth and the moon, F= G 2
cm
r
mm
F =
 28
242211
108.3
1061036.71067.6

 
=
  162
35
108.3
1036.767.6


= 20.3 × 10
19
=2.03 × 10
20
N = 2 ×10
20
N
10. Charge on proton = 1.6 × 10
–19
 Felectrical = 2
21
o r
qq
4
1


=
 
2
3829
r
106.1109 

mass of proton = 1.732 × 10
–27
kg
48N
x
FF

Chapter-4
4.3
Fgravity = 2
21
r
mm
G =
 
2
5411
r
10732.11067.6 

 
 
2
5411
2
3829
g
e
r
10732.11067.6
r
106.1109
F
F




 =
 
  652
292
10732.167.6
106.19



= 1.24 × 10
36
11. The average separation between proton and electron of Hydrogen atom is r= 5.3 10
–11
m.
a) Coulomb’s force = F = 9 × 10
9
× 2
21
r
qq
=
 
 211
2199
103.5
100.1109




= 8.2 × 10
–8
N.
b) When the average distance between proton and electron becomes 4 times that of its ground state
Coulomb’s force F =
 2
21
o r4
qq
4
1


=
 
  222
2199
103.516
106.1109




=
 
 
7
2
2
10
3.516
6.19 



= 0.0512 × 10
–7
= 5.1 × 10
–9
N.
12. The geostationary orbit of earth is at a distance of about 36000km.
We know that, g = GM / (R+h)
2
At h = 36000 km. g = GM / (36000+6400)
2
 0227.0
106106
256
4240042400
64006400
g
`g






 g = 0.0227 × 9.8 = 0.223
[ taking g = 9.8 m/s
2
at the surface of the earth]
A 120 kg equipment placed in a geostationary satellite will have weight
Mg` = 0.233 × 120 = 26.79 = 27 N
* * * *

5.1
CHAPTER – 5
1. m = 2kg
S = 10m
Let, acceleration = a, Initial velocity u = 0.
S= ut + 1/2 at
2
 10 = ½ a (2
2
)  10 = 2a  a = 5 m/s
2
Force: F = ma = 2 × 5 = 10N (Ans)
2. u = 40 km/hr =
3600
40000
= 11.11 m/s.
m = 2000 kg ; v = 0 ; s = 4m
acceleration ‘a’ =
s2
uv 22

=
 
42
11.110
22


=
8
43.123
 = –15.42 m/s
2
(deceleration)
So, braking force = F = ma = 2000 × 15.42 = 30840 = 3.08 10
4
N (Ans)
3. Initial velocity u = 0 (negligible)
v = 5 × 10
6
m/s.
s = 1cm = 1 × 10
–2
m.
acceleration a =
s2
uv 22

=
 
2
26
1012
0105



= 2
12
102
1025



= 12.5 × 10
14
ms
–2
F = ma = 9.1 × 10
–31
× 12.5 × 10
14
= 113.75 × 10
–17
= 1.1 × 10
–15
N.
4.
g = 10m/s
2
T – 0.3g = 0  T = 0.3g = 0.3 × 10 = 3 N
T1 – (0.2g + T) =0  T1 = 0.2g + T = 0.2 × 10 + 3 = 5N
Tension in the two strings are 5N & 3N respectively.
5.
T + ma – F = 0 T – ma = 0  T = ma …………(i)
 F= T + ma  F= T + T from (i)
 2T = F  T = F / 2
6. m = 50g = 5 × 10
–2
kg
As shown in the figure,
Slope of OA = Tanθ
OD
AD
=
3
15
= 5 m/s
2
So, at t = 2sec acceleration is 5m/s
2
Force = ma = 5 × 10
–2
× 5 = 0.25N along the motion
fig 1
0.2kg
0.3kg
0.2kg
fig 3
0.2g
T1
T
fig 2
0.3g
0.3kg
A
Fig 2
mg
T
R
F
B
Fig 3
mg
ma
R
T
B
Fig 1
S
A
2
v(m/s)
180°–
A B
CD E

4 6
5
15
10

Chapter-5
5.2
At t = 4 sec slope of AB = 0, acceleration = 0 [ tan 0° = 0]
Force = 0
At t = 6 sec, acceleration = slope of BC.
In ∆BEC = tan θ =
EC
BE
=
3
15
= 5.
Slope of BC = tan (180° – θ) = – tan θ = –5 m/s
2
(deceleration)
Force = ma = 5 × 10
–2
5 = 0.25 N. Opposite to the motion.
7. Let, F  contact force between mA & mB.
And, f  force exerted by experimenter.
F + mA a –f = 0 mB a –f =0
 F = f – mA a ……….(i)  F= mB a ……...(ii)
From eqn (i) and eqn (ii)
 f – mA a = mB a  f = mB a + mA a  f = a (mA + m B).
 f =
Bm
F
(mB + mA) = 






B
A
m
m
1F [because a = F/mB]
 The force exerted by the experimenter is 






B
A
m
m
1F
8. r = 1mm = 10
–3
‘m’ = 4mg = 4 × 10
–6
kg
s = 10
–3
m.
v = 0
u = 30 m/s.
So, a =
s2
uv 22

= 3
102
3030



= –4.5 × 10
5
m/s
2
(decelerating)
Taking magnitude only deceleration is 4.5 × 10
5
m/s
2
So, force F = 4 × 10
–6
× 4.5 × 10
5
= 1.8 N
9. x = 20 cm = 0.2m, k = 15 N/m, m = 0.3kg.
Acceleration a =
m
F
=
x
kx
=
 
3.0
2.015
=
3.0
3
 = –10m/s
2
(deceleration)
So, the acceleration is 10 m/s
2
opposite to the direction of motion
10. Let, the block m towards left through displacement x.
F1 = k1 x (compressed)
F2 = k2 x (expanded)
They are in same direction.
Resultant F = F1 + F2  F = k1 x + k2 x  F = x(k1 + k2)
So, a = acceleration =
m
F
=
m
)kx(k 21 
opposite to the displacement.
11. m = 5 kg of block A.
ma = 10 N
 a 10/5 = 2 m/s
2
.
As there is no friction between A & B, when the block A moves, Block B remains at rest in its position.
Fig 1
s
m1 m2f
F
R
mBg
mBa
Fig 3
F
R
mAg
mBa
Fig 2
F1
m
x
K1
F2
K2
10N
A
B
0.2m

Chapter-5
5.3
Initial velocity of A = u = 0.
Distance to cover so that B separate out s = 0.2 m.
Acceleration a = 2 m/s2
 s= ut + ½ at
2
 0.2 = 0 + (½) ×2 × t
2
 t
2
= 0.2  t = 0.44 sec  t= 0.45 sec.
12. a) at any depth let the ropes make angle θ with the vertical
From the free body diagram
F cos θ + F cos θ – mg = 0
 2F cos θ = mg  F =
cos2
mg
As the man moves up. θ increases i.e. cos decreases. Thus F
increases.
b) When the man is at depth h
cos  =
22
h)2/d(
h

Force = 22
2
2
h4d
h4
mg
h
4
d
h
mg


13. From the free body diagram
 R + 0.5 × 2 – w = 0
 R = w – 0.5 × 2
= 0.5 (10 – 2) = 4N.
So, the force exerted by the block A on the block B, is 4N.
14. a) The tension in the string is found out for the different conditions from the free body diagram as
shown below.
T – (W + 0.06 × 1.2) = 0
 T = 0.05 × 9.8 + 0.05 × 1.2
= 0.55 N.
b)  T + 0.05 × 1.2 – 0.05 × 9.8 = 0
 T = 0.05 × 9.8 – 0.05 × 1.2
= 0.43 N.
c) When the elevator makes uniform motion
T – W = 0
 T = W = 0.05 × 9.8
= 0.49 N
d) T + 0.05 × 1.2 – W = 0
 T = W – 0.05 × 1.2
= 0.43 N.
e) T – (W + 0.05 × 1.2) = 0
 T = W + 0.05 × 1.2
= 0.55 N
s
10N
R
w
ma
F

d
Fig-1

d/2
F

Fig-2

mg
F
F

h
d/2
A
mg
2 m/s2
B
A
0.5×2
R
W=mg=0.5×10
Fig-1
2m/s
2 W
0.05×1.2
T
Fig-2
W
0.05×1.2
T
Fig-4
–a
Fig-3
1.2m/s2
W
T
Fig-6
a=0
Fig-5
Uniform
velocity
Fig-7
a=1.2m/s2
W
0.05×1.2
T
Fig-8
W
0.05×1.2
T
Fig-10
–a
Fig-9
1.2m/s2

Chapter-5
5.4
f) When the elevator goes down with uniform velocity acceleration = 0
T – W = 0
 T = W = 0.05 × 9.8
= 0.49 N.
15. When the elevator is accelerating upwards, maximum weight will be recorded.
R – (W + ma ) = 0
 R = W + ma = m(g + a) max.wt.
When decelerating upwards, maximum weight will be recorded.
R + ma – W = 0
R = W – ma = m(g – a)
So, m(g + a) = 72 × 9.9 …(1)
m(g – a) = 60 × 9.9 …(2)
Now, mg + ma = 72 × 9.9  mg – ma = 60 × 9.9
 2mg = 1306.8
 m =
9.92
8.1306

= 66 Kg
So, the true weight of the man is 66 kg.
Again, to find the acceleration, mg + ma = 72 × 9.9
 a = 9.0
11
9.9
66
9.9669.972


m/s
2
.
16. Let the acceleration of the 3 kg mass relative to the elevator is ‘a’ in the downward direction.
As, shown in the free body diagram
T – 1.5 g – 1.5(g/10) – 1.5 a = 0 from figure (1)
and, T – 3g – 3(g/10) + 3a = 0 from figure (2)
 T = 1.5 g + 1.5(g/10) + 1.5a … (i)
And T = 3g + 3(g/10) – 3a … (ii)
Equation (i) × 2  3g + 3(g/10) + 3a = 2T
Equation (ii) × 1  3g + 3(g/10) – 3a = T
Subtracting the above two equations we get, T = 6a
Subtracting T = 6a in equation (ii)
6a = 3g + 3(g/10) – 3a.
 9a =
10
g33
 a = 34.32
10
33)8.9(

a = 3.59  T = 6a = 6 × 3.59 = 21.55
T
1
= 2T = 2 × 21.55 = 43.1 N cut is T1 shown in spring.
Mass =
8.9
1.43
g
wt
 = 4.39 = 4.4 kg
17. Given, m = 2 kg, k = 100 N/m
From the free body diagram, kl – 2g = 0  kl = 2g
 l =
100
6.19
100
8.92
k
g2


 = 0.196 = 0.2 m
Suppose further elongation when 1 kg block is added be x,
Then k(1 + x) = 3g
 kx = 3g – 2g = g = 9.8 N
 x =
100
8.9
= 0.098 = 0.1 m
W
T
Fig-12Fig-11
Uniform
velocity
m
W
R
a
ma
W
R
a
ma
–a
Fig-1
1.5g
T
1.5(g/10)
1.5a
Fig-2
3g
T
3(g/10)
3a
kl
2g

Chapter-5
5.5
18. a = 2 m/s
2
kl – (2g + 2a) = 0
kl = 2g + 2a
= 2 × 9.8 + 2 × 2 = 19.6 + 4
 l =
100
6.23
= 0.236 m = 0.24 m
When 1 kg body is added total mass (2 + 1)kg = 3kg.
elongation be l1
kl1 = 3g + 3a = 3 × 9.8 + 6
 l1 =
100
4.33
= 0.0334 = 0.36
Further elongation = l1 – l = 0.36 – 0.12 m.
19. Let, the air resistance force is F and Buoyant force is B.
Given that
Fa  v, where v  velocity
 Fa = kv, where k  proportionality constant.
When the balloon is moving downward,
B + kv = mg …(i)
 M =
g
kvB 
For the balloon to rise with a constant velocity v, (upward)
let the mass be m
Here, B – (mg + kv) = 0 …(ii)
 B = mg + kv
 m =
g
kwB 
So, amount of mass that should be removed = M – m.
=
g
kv2
g
kvBkvB
g
kvB
g
kvB






=
G
)BMg(2 
= 2{M – (B/g)}
20. When the box is accelerating upward,
U – mg – m(g/6) = 0
 U = mg + mg/6 = m{g + (g/6)} = 7 mg/7 …(i)
 m = 6U/7g.
When it is accelerating downward, let the required mass be M.
U – Mg + Mg/6 = 0
 U =
6
Mg5
6
MgMg6


 M =
g5
U6
Mass to be added = M – m = 






7
1
5
1
g
U6
g7
U6
g5
U6
= 











g
U
35
12
35
2
g
U6
= 






g
1
6
mg7
35
12
from (i)
= 2/5 m.
 The mass to be added is 2m/5.
2a
a
kl
2g
a
2×2
kl
3g
2m/s2
M
Fig-1
v
kV
mg
B
Fig-2
v
kV
mg
B
Fig-1
g/6T
mg/6
mg
V
Fig-2
g/6T
mg/6
mg
V

Chapter-5
5.6
21. Given that, AuF

 and mg act on the particle.
For the particle to move undeflected with constant velocity, net force should be zero.
 mg)Au( 

= 0
 mg)Au( 

Because, )Au(

 is perpendicular to the plane containing u

and A

, u

should be in the xz-plane.
Again, u A sin  = mg
 u =
sinA
mg
u will be minimum, when sin  = 1   = 90°
 umin =
A
mg
along Z-axis.
22.
m1 = 0.3 kg, m2 = 0.6 kg
T – (m1g + m1a) = 0 …(i)  T = m1g + m1a
T + m2a – m2g = 0 …(ii)  T = m2g – m2a
From equation (i) and equation (ii)
m1g + m1a + m2a – m2g = 0, from (i)
 a(m1 + m2) = g(m2 – m1)
 a = 















3.06.0
3.06.0
8.9
mm
mm
f
21
12
= 3.266 ms
–2
.
a) t = 2 sec acceleration = 3.266 ms
–2
Initial velocity u = 0
So, distance travelled by the body is,
S = ut + 1/2 at
2
 0 + ½(3.266) 2
2
= 6.5 m
b) From (i) T = m1(g + a) = 0.3 (9.8 + 3.26) = 3.9 N
c) The force exerted by the clamp on the pully is given by
F – 2T = 0
F = 2T = 2 × 3.9 = 7.8 N.
23. a = 3.26 m/s
2
T = 3.9 N
After 2 sec mass m1 the velocity
V = u + at = 0 + 3.26 × 2 = 6.52 m/s upward.
At this time m2 is moving 6.52 m/s downward.
At time 2 sec, m2 stops for a moment. But m1 is moving upward with velocity 6.52 m/s.
It will continue to move till final velocity (at highest point) because zero.
Here, v = 0 ; u = 6.52
A = –g = –9.8 m/s
2
[moving up ward m1]
V = u + at  0 = 6.52 +(–9.8)t
 t = 6.52/9.8 = 0.66 =2/3 sec.
During this period 2/3 sec, m2 mass also starts moving downward. So the string becomes tight again
after a time of 2/3 sec.
gm

A

y
F

m
yZ
x
m1g
T
m1a
m2g
T
m2a
m1
m2
T
F
T
a
0.3kg m1
m2
0.6kg

Chapter-5
5.7
24. Mass per unit length 3/30 kg/cm = 0.10 kg/cm.
Mass of 10 cm part = m1 = 1 kg
Mass of 20 cm part = m2 = 2 kg.
Let, F = contact force between them.
F – 20 – 10 = 0 …(i)
And, 32 – F – 2a = 0 …(ii)
From eqa (i) and (ii) 3a – 12 = 0  a = 12/ 3 = 4 m/s
2
Contact force F = 20 + 1a = 20 + 1 × 4 = 24 N.
25.
Sin 1 = 4/5 g sin 1 – (a + T) = 0 T – g sin 2 – a = 0
sin 2 = 3/5  g sing 1 = a + T …(i) T = g sin 2 + a …(ii)
 T + a – g sin  
From eqn (i) and (ii), g sin 2 + a + a – g sin 1 = 0
 2a = g sin 1 – g sin 2 = g 






5
3
5
4
= g / 5
 a =
10
g
2
1
5
g

26.
From the above Free body diagram From the above Free body diagram
M1a + F – T = 0  T = m1a + F …(i) m2a + T – m2g =0 ….(ii)
 m2a + m1a + F – m2g = 0 (from (i))
 a(m1 + m2) + m2g/2 – m2g = 0 {because f = m
2
g/2}
 a(m1 + m2) – m2g =0
 a(m1 + m2) = m2g/2  a =
)mm(2
gm
21
2

Acceleration of mass m1 is
)mm(2
gm
21
2

towards right.
27.
From the above free body diagram From the free body diagram
T + m1a – m(m1g + F ) = 0 T – (m2g + F + m2a)=0
20N m1
10m
m2
20m
32N
a
1a 20N
R1
F
1g
2a F
R2
32N
2g
3m
Fig-1

1kg 1kg

4m
5m

1g
a
R2
1a
T
Fig-3

1g
a
R1
1a
T
Fig-2
a
m1a F
R
T
m1g
Fig-2
T
m2a
a
m2g
Fig-3
m2
m1
a
Fig-1
T
Fig-2
m1a
m1g
F
a
m1
a
Fig-1
m2
T
Fig-3
m2g
F
m2a

Chapter-5
5.8
 T = m1g + F – m1a  T = 5g + 1 – 5a …(i)
T = m2g +F + m2a  T = 2g + 1 + 2a …(ii)
From the eqn (i) and eqn (ii)
5g + 1 – 5a = 2g + 1 + 2a  3g – 7a = 0 7a = 3g
 a =
7
g3
=
7
4.29
= 4.2 m/s
2
[ g= 9.8m/s
2
]
a) acceleration of block is 4.2 m/s
2
b) After the string breaks m1 move downward with force F acting down ward.
m1a = F + m1g = (1 + 5g) = 5(g + 0.2) Force = 1N, acceleration = 1/5= 0.2m/s.
So, acceleration =
mass
Force
=
5
0.2)5(g 
= (g + 0.2) m/s
2
28.
Let the block m+1+ moves upward with acceleration a, and the two blocks m2 an m3 have relative
acceleration a2 due to the difference of weight between them. So, the actual acceleration at the blocks
m1, m2 and m3 will be a1.
(a1 – a2) and (a1 + a2) as shown
T = 1g – 1a2 = 0 ...(i) from fig (2)
T/2 – 2g – 2(a1 – a2) = 0 ...(ii) from fig (3)
T/2 – 3g – 3(a1 + a2) = 0 ...(iii) from fig (4)
From eqn (i) and eqn (ii), eliminating T we get, 1g + 1a2 = 4g + 4(a1 + a2)  5a2 – 4a1 = 3g (iv)
From eqn (ii) and eqn (iii), we get 2g + 2(a1 – a2) = 3g – 3(a1 – a2)  5a1 + a2 = (v)
Solving (iv) and (v) a1 =
29
g2
and a2 = g – 5a1 =
29
g10
g  =
29
g19
So, a1 – a2 =
29
g17
29
g19
29
g2

a1 + a2 =
29
g21
29
g19
29
g2
 So, acceleration of m1, m2, m3 ae
29
g19
(up)
29
g17
(doan)
29
g21
(down)
respectively.
Again, for m1, u = 0, s= 20cm=0.2m and a2 = g
29
19
[g = 10m/s
2
]
S = ut + ½ at
2
= 2
gt
29
19
2
1
2.0   t = 0.25sec.
29.
(a1+a2)
m1
a1
Fig-1
m2
m3
m2
Fig-3
T/2
2g
2(a1–a2)
m1
Fig-2
T
a
lg
la1
m3
Fig-4
T/2
3g
3(a1+a2)
5a
5g
F=1N
a2=0
m1
a1
Fig-1
m2
m3
Fig-3
T/2
2g
2a1
Fig-2
T
m1g
Fig-4
T/2
3g
3a1

Chapter-5
5.9
m1 should be at rest.
T – m1g = 0 T/2 – 2g – 2a1 = 0 T/2 – 3g – 3a1 =0
 T = m1g …(i) T – 4g – 4a1 = 0 …(ii)  T = 6g – 6a1 …(iii)
From eqn (ii) & (iii) we get
3T – 12g = 12g – 2T  T = 24g/5= 408g.
Putting yhe value of T eqn (i) we get, m1 = 4.8kg.
30.
T + 1a = 1g ...(i) T – 1a =0  T = 1a (ii)
From eqn (i) and (ii), we get
1a + 1a = 1g  2a = g  a =
2
g
=
2
10
= 5m/s
2
From (ii) T = 1a = 5N.
31.
Ma – 2T = 0 T + Ma – Mg = 0
Ma = 2T  T = Ma /2.  Ma/2 + ma = Mg. (because T = Ma/2)
 3 Ma = 2 Mg  a = 2g/3
a) acceleration of mass M is 2g/3.
b) Tension T =
2
Ma
=
2
M
=
3
g2
=
3
Mg
c) Let, R
1
= resultant of tensions = force exerted by the clamp on the pulley
R
1
= T2TT 22

 R =
3
Mg2
3
Mg
2T2 
Again, Tan =
T
T
= 1   = 45°.
So, it is
3
Mg2
at an angle of 45° with horizontal.
32.
B
1a
Fig-2
T
1g
BFig-1
1kg
1kg
1g
Fig-3
T
R
1a
2mg
Fig-2
2m(a/2)
R
2Ta
Fig-1
2m
M
B Aa/2
Fig-3
T
mg
ma
Fig-3
2T
2mg
2mamg
Fig-2
2ma
T
M
a
Fig-1
A
30° 2M
T
T
R

45°
T
T
R
45°

Chapter-5
5.10
2Ma + Mg sin – T = 0 2T + 2Ma – 2Mg = 0
 T = 2Ma + Mg sin  …(i)  2(2Ma + Mg sin ) + 2Ma – 2Mg = 0 [From (i)]
 4Ma + 2Mgsin + 2 Ma – 2Mg =0
 6Ma + 2Mg sin30° – 2Mg = 0
 6Ma = Mg a =g/6.
Acceleration of mass M is 2a = s × g/6 = g/3 up the plane. 
33.
As the block ‘m’ does not slinover M, ct will have same acceleration as that of M
From the freebody diagrams.
T + Ma – Mg = 0 ...(i) (From FBD – 1)
T – Ma – R sin  = 0 ...(ii) (From FBD -2)
R sin  – ma = 0 ...(iii) (From FBD -3)
R cos – mg =0 ...(iv) (From FBD -4)
Eliminating T, R and a from the above equation, we get M =
1cot
mM


34. a) 5a + T – 5g = 0  T = 5g – 5a ...(i) (From FBD-1)
Again (1/2) – 4g – 8a = 0  T = 8g – 16a ...(ii) (from FBD-2)
From equn (i) and (ii), we get
5g – 5a = 8g + 16a  21a = –3g  a = – 1/7g
So, acceleration of 5 kg mass is g/7 upward and that of 4 kg mass
is 2a = 2g/7 (downward).
b)
4a – t/2 = 0  8a – T = 0  T = 8a … (ii) [From FBD -4]
Again, T + 5a – 5g = 0  8a + 5a – 5g = 0
 13a – 5g = 0  a = 5g/13 downward. (from FBD -3)
Acceleration of mass (A) kg is 2a = 10/13 (g) & 5kg (B) is 5g/13.
c)
T + 1a – 1g = 0  T = 1g – 1a …(i) [From FBD – 5]
Again,
2
T
– 2g – 4a = 0  T – 4g – 8a = 0 …(ii) [From FBD -6]
 1g – 1a – 4g – 8a = 0 [From (i)]
Ma
FBD-1
T
Mg

FBD-2

Mg
T
R
FBD-3

ma
mg
MFBD-4
a
a
M a
5kg
FBD-3
2a
4kg
2a
8a
FBD-2
T/2
4g
2a
5a
FBD-1
T
5g
a
a
B 5kg
2kg
2a
5a
FBD-3
T
5g
4a
R
T/2
2g
FBD-4
a  2a
2kg
1kg
T/2
T C 
T/2
B 
1a
FBD-5
T
1g
4a
FBD-6
T
2g

Chapter-5
5.11
 a = –(g/3) downward.
Acceleration of mass 1kg(b) is g/3 (up)
Acceleration of mass 2kg(A) is 2g/3 (downward).
35. m1 = 100g = 0.1kg
m2 = 500g = 0.5kg
m3 = 50g = 0.05kg.
T + 0.5a – 0.5g = 0 ...(i)
T1 – 0.5a – 0.05g = a ...(ii)
T1 + 0.1a – T + 0.05g = 0 ...(iii)
From equn (ii) T1 = 0.05g + 0.05a ...(iv)
From equn (i) T1 = 0.5g – 0.5a ...(v)
Equn (iii) becomes T1 + 0.1a – T + 0.05g = 0
 0.05g + 0.05a + 0.1a – 0.5g + 0.5a + 0.05g = 0 [From (iv)
and (v)]
 0.65a = 0.4g  a =
65.0
4.0
=
65
40
g =
13
8
g downward
Acceleration of 500gm block is 8g/13g downward.
36. m = 15 kg of monkey. a = 1 m/s
2
.
 T – [15g + 15(1)] = 0  T = 15 (10 + 1)  T = 15 × 11  T = 165 N.
The monkey should apply 165N force to the rope.
Initial velocity u = 0 ; acceleration a = 1m/s
2
; s = 5m.
 s = ut + ½ at
2
5 = 0 + (1/2)1 t
2
 t
2
= 5 × 2  t = 10 sec.
Time required is 10 sec.
37. Suppose the monkey accelerates upward with acceleration ’a’ & the block, accelerate downward with
acceleration a1. Let Force exerted by monkey is equal to ‘T’
From the free body diagram of monkey
 T – mg – ma = 0 ...(i)
 T = mg + ma.
Again, from the FBD of the block,
T = ma1 – mg = 0.
 mg + ma + ma1 – mg = 0 [From (i)]  ma = –ma1  a = a1.
Acceleration ‘–a’ downward i.e. ‘a’ upward.
The block & the monkey move in the same direction with equal acceleration.
If initially they are rest (no force is exertied by monkey) no motion of monkey of block occurs as they
have same weight (same mass). Their separation will not change as time passes.
38. Suppose A move upward with acceleration a, such that in the tail of A maximum tension 30N produced.
T – 5g – 30 – 5a = 0 ...(i) 30 – 2g – 2a = 0 ...(ii)
 T = 50 + 30 +(5 × 5) = 105 N (max)  30 – 20 – 2a = 0  a = 5 m/s
2
So, A can apply a maximum force of 105 N in the rope to carry the monkey B with it.
a 
0.5a
FBD-1
T
0.5g
a 
0.5a
FBD-2
T
0.5g 0.1a
R

FBD-3
T
T
0.1g
a 
100g
m3 50g
30° 500g
m2 a a 
15g
a 
15a 
ma
mg
a 
T
ma1
mg
A
B
T1 = 30N
a 
5a
Fig-2
T
5g
T1 = 30N
a 
2a
Fig-3
2g

Chapter-5
5.12
For minimum force there is no acceleration of monkey ‘A’ and B.  a = 0
Now equation (ii) is T1 – 2g = 0  T1 = 20 N (wt. of monkey B)
Equation (i) is T – 5g – 20 = 0 [As T1 = 20 N]
 T = 5g + 20 = 50 + 20 = 70 N.
 The monkey A should apply force between 70 N and 105 N to carry the monkey B with it.
39. (i) Given, Mass of man = 60 kg.
Let R = apparent weight of man in this case.
Now, R + T – 60g = 0 [From FBD of man]
 T = 60g – R ...(i)
T – R – 30g = 0 ...(ii) [ From FBD of box]
 60g – R – R – 30g = 0 [ From (i)]
 R = 15g The weight shown by the machine is 15kg.
(ii) To get his correct weight suppose the applied force is ‘T’ and so, acclerates upward with ‘a’.
In this case, given that correct weight = R = 60g, where g = acc
n
due to gravity
From the FBD of the man From the FBD of the box
T
1
+ R – 60g – 60a = 0 T
1
– R – 30g – 30a = 0
T
1
– 60a = 0 [R = 60g] T
1
– 60g – 30g – 30a = 0
T
1
= 60a ...(i)  T
1
– 30a = 90g = 900
 T
1
= 30a – 900 ...(ii)
From eqn (i) and eqn (ii) we get T
1
= 2T
1
– 1800  T
1
= 1800N.
 So, he should exert 1800 N force on the rope to get correct reading.
40. The driving force on the block which n the body to move sown the plane is F = mg sin ,
So, acceleration = g sin 
Initial velocity of block u = 0.
s = ℓ, a = g sin 
Now, S = ut + ½ at
2
 ℓ = 0 + ½ (g sin ) t
2
 g
2
=
sing
2


 t =
sing
2


Time taken is
sing
2



41. Suppose pendulum makes  angle with the vertical. Let, m = mass of the pendulum.
T cos  – mg = 0 ma – T sin  =0
 T cos  = mg  ma = T sin 
 T =
cos
mg
...(i)  t =
sin
ma
...(ii)
R
60a
60g
a 
T1
60g
T
R 30g
R
T
a
30g
R
T
30a
A

v
R

mg

ma
a  
mg
T

Chapter-5
5.13
From (i) & (ii)


 sin
ma
cos
mg
 tan  =
g
a
  =
g
a
tan 1
The angle is Tan
–1
(a/g) with vertical.
(ii) m  mass of block.
Suppose the angle of incline is ‘’
From the diagram
ma cos  – mg sin  = 0  ma cos  = mg sin  
g
a
cos
sin



 tan  = a/g   = tan
–1
(a/g).
42. Because, the elevator is moving downward with an acceleration 12 m/s
2
(>g), the bodygets separated.
So, body moves with acceleration g = 10 m/s
2
[freely falling body] and the elevator move with
acceleration 12 m/s
2
Now, the block has acceleration = g = 10 m/s
2
u = 0
t = 0.2 sec
So, the distance travelled by the block is given by.
 s = ut + ½ at
2
= 0 + (½) 10 (0.2)
2
= 5 × 0.04 = 0.2 m = 20 cm.
The displacement of body is 20 cm during first 0.2 sec.
* * * *

a
R

mg
ma
mg
ma

12 m/s2
10 m/s
2

6.1
CHAPTER 6
1. Let m = mass of the block
From the freebody diagram,
R – mg = 0  R = mg ...(1)
Again ma –  R = 0  ma =  R =  mg (from (1))
 a = g  4 = g   = 4/g = 4/10 = 0.4
The co-efficient of kinetic friction between the block and the plane is 0.4
2. Due to friction the body will decelerate
Let the deceleration be ‘a’
R – mg = 0  R = mg ...(1)
ma –  R = 0  ma =  R =  mg (from (1))
 a = g = 0.1 × 10 = 1m/s
2.
Initial velocity u = 10 m/s
Final velocity v = 0 m/s
a = –1m/s
2
(deceleration)
S =
a2
uv 22

=
)1(2
100 2


=
2
100
= 50m
It will travel 50m before coming to rest.
3. Body is kept on the horizontal table.
If no force is applied, no frictional force will be there
f  frictional force
F  Applied force
From grap it can be seen that when applied force is zero,
frictional force is zero.
4. From the free body diagram,
R – mg cos  = 0  R = mg cos  ..(1)
For the block
U = 0, s = 8m, t = 2sec.
s = ut + ½ at
2
 8 = 0 + ½ a 2
2
 a = 4m/s
2
Again, R + ma – mg sin  = 0
  mg cos  + ma – mg sin  = 0 [from (1)]
 m(g cos  + a – g sin ) = 0
  × 10 × cos 30° = g sin 30° – a
 × 10 × )3/3( = 10 × (1/2) – 4
 )3/5(  =1   = 1/ )3/5( = 0.11
 Co-efficient of kinetic friction between the two is 0.11.
4 – 4a – R + 4g sin 30° = 0 …(1)
R – 4g cos 30° = 0 ...(2)
 R = 4g cos 30°
Putting the values of R is & in equn. (1)
4 – 4a – 0.11 × 4g cos 30° + 4g sin 30° = 0
 4 – 4a – 0.11 × 4 × 10 × ( 2/3 ) + 4 × 10 × (1/2) = 0
 4 – 4a – 3.81 + 20 = 0  a  5 m/s
2
For the block u =0, t = 2sec, a = 5m/s
2
Distance s = ut + ½ at
2
 s = 0 + (1/2) 5 × 2
2
= 10m
The block will move 10m.
R
mg
ma R
a
R
mg
ma R
avelocity
F
p
o
mg
R
30°
R

R
mg
ma
4kg
30°
4N
R

R
mg
ma
avelocity

Chapter 6
6.2
6. To make the block move up the incline, the force should be equal and opposite to
the net force acting down the incline =  R + 2 g sin 30°
= 0.2 × (9.8) 3 + 2 I 9.8 × (1/2) [from (1)]
= 3.39 + 9.8 = 13N
With this minimum force the body move up the incline with a constant velocity as net
force on it is zero.
b) Net force acting down the incline is given by,
F = 2 g sin 30° – R
= 2 × 9.8 × (1/2) – 3.39 = 6.41N
Due to F = 6.41N the body will move down the incline with acceleration.
No external force is required.
 Force required is zero.
g = 10m/s
2
, m = 2kg,  = 30°,  = 0.2
R – mg cos  - F sin  = 0
 R = mg cos  + F sin  ...(1)
And mg sin  + R – F cos  = 0
 mg sin  + (mg cos  + F sin ) – F cos  = 0
 mg sin  +  mg cos  +  F sin  – F cos  = 0
 F =
)cossin(
)cosmgsinmg(


 F =
)2/3()2/1(2.0
)2/3(1022.0)2/1(102


=
76.0
464.13
= 17.7N  17.5N
8. m  mass of child
R – mg cos 45° = 0
 R = mg cos 45° = mg /v
2
...(1)
Net force acting on the boy due to which it slides down is mg sin 45° - R
= mg sin 45° -  mg cos 45°
= m × 10 (1/ 2 ) – 0.6 × m × 10 × (1/ 2 )
= m [(5/ 2 ) – 0.6 × (5 / 2 )]
= m(2 2 )
acceleration =
mass
Force
=
m
)22(m
= 22 m/s
2

9. Suppose, the body is accelerating down with acceleration ‘a’.
R – mg cos  = 0
 R = mg cos  ...(1)
ma + mg sin  –  R = 0
 a =
m
)cos(sinmg 
= g (sin  –  cos )
For the first half mt. u = 0, s = 0.5m, t = 0.5 sec.
So, v = u + at = 0 + (0.5)4 = 2 m/s
S = ut + ½ at
2
 0.5 = 0 + ½ a (0/5)
2
 a = 4m/s
2
...(2)
For the next half metre
u` = 2m/s, a = 4m/s
2
, s= 0.5.
 0.5 = 2t + (1/2) 4 t
2
 2 t
2
+ 2 t – 0.5 =0
(body moving us)
R
R

F
mg
(body moving down)
R
R

mg
R
45°
R
mg
F
30°
R
R
30°
mg
mg
ma
R
R

Chapter 6
6.3
 4 t
2
+ 4 t – 1 = 0
 =
42
16164


=
8
656.1
= 0.207sec
Time taken to cover next half meter is 0.21sec.
10. f  applied force
Fi  contact force
F  frictional force
R  normal reaction
 = tan  = F/R
When F = R, F is the limiting friction (max friction). When applied force increase, force of friction
increase upto limiting friction (R)
Before reaching limiting friction
F < R
 tan  =
R
R
R
F 
  tan      tan
–1
 
T + 0.5a – 0.5 g = 0 ...(1)
R + 1a + T1 – T = 0 ...(2)
R + 1a – T1 = 0
R + 1a = T1 ...(3)
From (2) & (3)  R + a = T – T1
 T – T1 = T1
 T = 2T1
Equation (2) becomes R + a + T1 – 2T1 = 0
 R + a – T1 = 0
 T1 = R + a = 0.2g + a ...(4)
Equation (1) becomes 2T1 + 0/5a – 0.5g = 0
 T1 =
2
a5.0g5.0 
= 0.25g – 0.25a ...(5)
From (4) & (5) 0.2g + a = 0.25g – 0.25a
 a =
25.1
05.0
× 10 = 0.04 I 10 = 0.4m/s
2
a) Accln of 1kg blocks each is 0.4m/s
2
b) Tension T1 = 0.2g + a + 0.4 = 2.4N
c) T = 0.5g – 0.5a = 0.5 × 10 – 0.5 × 0.4 = 4.8N
1 R + 1 – 16 = 0
 1 (2g) + (–15) = 0
 1 = 15/20 = 0.75
2 R1 + 4 × 0.5 + 16 – 4g sin 30° = 0
2 (20 3 ) + 2 + 16 – 20 = 0
 2 =
320
2
=
32.17
1
= 0.057  0.06
Co-efficient of friction 1 = 0.75 & 2 = 0.06 
F f

Fi
R
Limiting
Friction
R
T1
R
A 1g
1a

R
R
1g
1a

0.5g
0.5g
a
BA
=0.2
0.5kg
1kg 1kg
=0.2
2×0.5
16N
R1
4g
16N=T
2R
4×0.5
30°
a
1
0.5 m/s2
2kg
4kg
2

Chapter 6
6.4
13.
T + 15a – 15g = 0 T – (T1 + 5a+ R)= 0 T1 – 5g – 5a = 0
 T = 15g – 15 a ...(i)  T – (5g + 5a + 5a + R) = 0 T1=5g + 5a …(iii)
 T = 5g + 10a + R …(ii)
From (i) & (ii) 15g – 15a = 5g + 10a + 0.2 (5g)
 25a = 90  a = 3.6m/s
2
Equation (ii)  T = 5 × 10 + 10 × 3.6 + 0.2 × 5 × 10
 96N in the left string
Equation (iii) T1 = 5g + 5a = 5 × 10 + 5 × 3.6 =68N in the right string.
14. s = 5m,  = 4/3, g = 10m/s
2
u = 36km/h = 10m/s, v = 0,
a =
s2
uv 22

=
52
100 2


= –10m/s
2
From the freebody diagrams,
R – mg cos  = 0 ; g = 10m/s
2
 R = mg cos ….(i) ;  = 4/3.
Again, ma + mg sin  -  R = 0
 ma + mg sin  –  mg cos  = 0
a + g sin  – mg cos  = 0
 10 + 10 sin  - (4/3) × 10 cos  = 0
 30 + 30 sin  – 40 cos  =0
 3 + 3 sin  – 4 cos  = 0
 4 cos  - 3 sin  = 3
 4  2
sin1 = 3 + 3 sin 
 16 (1 – sin
2
) = 9 + 9 sin
2
+ 18 sin 
sin  =
252
)7)(25(41818 2


=
50
3218 
=
50
14
= 0.28 [Taking +ve sign only]
  = sin
–1
(0.28) = 16°
Maximum incline is  = 16°
15. to reach in minimum time, he has to move with maximum possible acceleration.
Let, the maximum acceleration is ‘a’
 ma – R = 0  ma =  mg
 a =  g = 0.9 × 10 = 9m/s
2
a) Initial velocity u = 0, t = ?
a = 9m/s
2
, s = 50m
s = ut + ½ at
2
 50 = 0 + (1/2) 9 t
2
 t =
9
100
=
3
10
sec.
b) After overing 50m, velocity of the athelete is
V = u + at = 0 + 9 × (10/3) = 30m/s
He has to stop in minimum time. So deceleration ia –a = –9m/s
2
(max)
T1
5g
5a
T
15g
15a
A
B
R
5g
T1
T
r=5g
R
C
B
A
a
15kg 15kg
a
a
R
ma
mg
R
a

velocity
 the max.
angle
R
a
R
mg
ma
R
a
R
mg
ma

Chapter 6
6.5













)onDecelerati(s/m9ga
)forcefrictional(maxRma
maR
2

u
1
= 30m/s, v
1
= 0
t =
a
uv 11

=
a
300


=
a
30


=
3
10
sec.
16. Hardest brake means maximum force of friction is developed between car’s type & road.
Max frictional force = R
R – mg cos  =0
 R = mg cos  ...(i)
and R + ma – mg sin ) = 0 …(ii)
 mg cos  + ma – mg sin  = 0
 g cos  + a – 10 × (1/2) = 0
a = 5 – {1 – (2 3 )} × 10 ( 2/3 ) = 2.5 m/s
2
When, hardest brake is applied the car move with acceleration 2.5m/s
2
S = 12.8m, u = 6m/s
S0, velocity at the end of incline
V = as2u2
 = )8.12)(5.2(262
 = 6436  = 10m/s = 36km/h
Hence how hard the driver applies the brakes, that car reaches the bottom with least velocity 36km/h.
17. Let, , a maximum acceleration produced in car.
 ma = R [For more acceleration, the tyres will slip]
 ma =  mg  a = g = 1 × 10 = 10m/s
2
For crossing the bridge in minimum time, it has to travel with maximum
acceleration
u = 0, s = 500m, a = 10m/s
2
s = ut + ½ at
2
 500 = 0 + (1/2) 10 t
2
 t = 10 sec.
If acceleration is less than 10m/s
2
, time will be more than 10sec. So one can’t drive through the bridge
in less than 10sec.
R = 4g cos 30° = 4 × 10 × 2/3 = 20 3 ...(i)
2 R + 4a – P – 4g sin 30° = 0  0.3 (40) cos 30° + 4a – P – 40 sin 20° = 0 …(ii)
P + 2a + 1 R1 – 2g sin 30° = 0 …(iii)
R1 = 2g cos 30° = 2 × 10 × 2/3 = 10 3 ...(iv)
Equn. (ii) 6 3 + 4a – P – 20 = 0
Equn (iv) P + 2a + 2 3 – 10 = 0
From Equn (ii) & (iv) 6 3 + 6a – 30 + 2 3 = 0
 6a = 30 – 8 3 = 30 – 13.85 = 16.15
 a =
6
15.16
= 2.69 = 2.7m/s
2
b) can be solved. In this case, the 4 kg block will travel with more acceleration because, coefficient of
friction is less than that of 2kg. So, they will move separately. Drawing the free body diagram of 2kg
mass only, it can be found that, a = 2.4m/s
2
.
ma
R
a
mg
R
R
a
mg
R
4kg
30°
2kg
a
a
R
4g
R
P
2a
R
2g
P
1 R1

Chapter 6
6.6
R1= M1 g cos  ...(i)
R2= M2 g cos  ...(ii)
T + M1g sin  – m1 a –  R1 = 0 ...(iii)
T – M2 – M2 a +  R2 = 0 ...(iv)
Equn (iii)  T + M1g sin  – M1 a –  M1g cos  = 0
Equn (iv)  T – M2 g sin  + M2 a +  M2g cos  = 0 ...(v)
Equn (iv) & (v)  g sin  (M1 + M2) – a(M1 + M2) – g cos  (M1 + M2) = 0
 a (M1 + M2) = g sin  (M1 + M2) –  g cos  (M1 + M2)
 a = g(sin  –  cos )
 The blocks (system has acceleration g(sin  –  cos )
The force exerted by the rod on one of the blocks is tension.
Tension T = – M1g sin  + M1a +  M1g sin 
 T = – M1g sin  + M1(g sin  –  g cos ) +  M1g cos 
 T = 0
20. Let ‘p’ be the force applied to at an angle 
R + P sin  – mg = 0
 R = – P sin  + mg ...(i)
R – p cos  ...(ii)
Equn. (i) is (mg – P sin ) – P cos  = 0
  mg =   sin  – P cos   =


cossin
mg
Applied force P should be minimum, when  sin  + cos  is maximum.
Again,  sin  + cos  is maximum when its derivative is zero.
d/d ( sin  + cos ) = 0
  cos  – sin  = 0   = tan
–1

So, P =


cossin
mg
=






cos
cos
cos
sin
cos/mg
=


tan1
secmg
=


2
tan1
secmg
=


sec
mg
=


2
tan1(
mg
=
2
1
mg


Minimum force is
2
1
mg


at an angle  = tan
–1
.
21. Let, the max force exerted by the man is T.
R + T – Mg = 0
 R = Mg – T ...(i)
R1 – R – mg = 0
 R1 = R + mg ...(ii)
And T –  R1 = 0
T
R1
a
M1p
R1
T
R2
M2a
M2g
R2
a

M1
M2
R

P
mg
R
T
R1
mR1
R
mg
R
T
mg

Chapter 6
6.7
 T –  (R + mg) = 0 [From equn. (ii)]
 T –  R –  mg = 0
 T –  (Mg + T) –  mg = 0 [from (i)]
 T (1 + ) = Mg +  mg
 T =


1
g)mM(
Maximum force exerted by man is


1
g)mM(

22.
R1 – 2g = 0
 R1 = 2 × 10 = 20 4a1 –  R1 = 0
2a + 0.2 R1 – 12 = 0  4a1 =  R1 = 0.2 (20)
 2a + 0.2(20) = 12  4a1 = 4
 2a = 12 – 4 = 8  a1 = 1m/s
2
 a = 4m/s
2
2kg block has acceleration 4m/s
2
& that of 4 kg is 1m/s
2
(ii) R1 = 2g = 20 4a + 0.2 × 2 × 10 – 12 = 0
Ma –  R1 = 0  4a + 4 = 12
 2a = 0.2 (20) = 4  4a = 8
 a = 2m/s
2
 a = 2 m/s
2
23.
a) When the 10N force applied on 2kg block, it experiences maximum frictional force
R1 =  × 2kg = (0.2) × 20 = 4N from the 3kg block.
So, the 2kg block experiences a net force of 10 – 4 = 6N
So, a1 = 6/2 = 3 m/s
2
But for the 3kg block, (fig-3) the frictional force from 2kg block (4N) becomes the driving force and the
maximum frictional force between 3kg and 7 kg block is
2R2 = (0.3) × 5kg = 15N
So, the 3kg block cannot move relative to the 7kg block. The 3kg block and 7kg block both will have
same acceleration (a2 = a3) which will be due to the 4N force because there is no friction from the floor.
a2 = a3 = 4/10 = 0.4m/s
2
12N
2kg
4kg
aR1
2a
12N
0.2R1
2g
R1
aR1
2a
4g
2g
4a
12N
2kg
4kg
aR1
2a
R1
2g
R1
R1
4g
2g
4a
12
1 = 0.2
C
10N
7 kg
3 kg
2 kg
B
A
1 = 0.3
1 = 0.5
R1
10N
2g
R1=4N
R2=5g
3g
15N
10N

Chapter 6
6.8
b) When the 10N force is applied to the 3kg block, it can experience maximum frictional force of 15 + 4
= 19N from the 2kg block & 7kg block.
So, it can not move with respect to them.
As the floor is frictionless, all the three bodies will move together
 a1 = a2 = a3 = 10/12 = (5/6)m/s
2
c) Similarly, it can be proved that when the 10N force is applied to the 7kg block, all the three blocks will
move together.
Again a1 = a2 = a3 = (5/6)m/s
2
24. Both upper block & lower block will have acceleration 2m/s
2
R1 = mg ...(i) T – R1 = 0
F – R1 – T = 0  F – mg –T = 0 ...(ii)  T = mg
 F =  mg +  mg = 2  mg [putting T =  mg]
b) 2F – T –  mg – ma = 0 …(i) T – Ma –  mg = 0 [ R1 = mg]
 T = Ma +  mg
Putting value of T in (i)
2f – Ma– mg –  mg – ma = 0
 2(2mg) – 2  mg = a(M + m) [Putting F = 2 mg]
 4 mg – 2  mg = a (M + m)  a =
mM
mg2


Both blocks move with this acceleration ‘a’ in opposite direction.
25.
R1 + ma – mg =0
 R1 = m(g–a) = mg – ma ...(i)
T –  R1 = 0  T = m (mg – ma) ...(ii)
Again, F – T –  R1 =0
C
10N
7 kg
3 kg
2 kg
B
A
3kg
10N
15N
4N
R=5g
3g
2g
mg
R1
T
F
R1
mg
R
TR1
R1
M
m
ma
a
mg
2F T R1
R1
ma
a
mg
T
R1
R1
R
F
ma
a
mg
T R1
R1
R1
ma
mg
T
R1
R2
M
ma
F
T = mR1
= m (mg–ma)

Chapter 6
6.9
 F – {(mg –ma)} – u(mg – ma) = 0
 F –  mg +  ma –  mg +  ma = 0
 F = 2  mg – 2 ma  F = 2 m(g–a)
b) Acceleration of the block be a1
R1 = mg – ma ...(i) T – R1 – M a1 = 0
2F – T – R1 – ma1 =0  T = R1 + M a1
 2F – t – mg + a – ma1 = 0 ...(ii) T =  (mg – ma) + Ma1
 T =  mg –  ma + M a1
Subtracting values of F & T, we get
2(2m(g – a)) – 2(mg – ma + Ma1) – mg +  ma –  a1 = 0
 4 mg – 4  ma – 2  mg + 2 ma = ma1 + M a1  a1 =
mM
)ag(m2


Both blocks move with this acceleration but in opposite directions.
26. R1 + QE – mg = 0
R 1 = mg – QE ...(i)
F – T – R1 = 0
 F – T (mg – QE) = 0
 F – T –  mg + QE = 0 …(2)
T -  R1 = 0
 T =  R1 =  (mg – QE) =  mg – QE
Now equation (ii) is F – mg +  QE –  mg +  QE = 0
 F – 2  mg + 2 QE = 0
 F = 2mg – 2 QE
 F= 2(mg – QE)
Maximum horizontal force that can be applied is 2(mg – QE).
27. Because the block slips on the table, maximum frictional force acts on it.
R = mg
 F –  R = 0  F = R =  mg
But the table is at rest. So, frictional force at the legs of the table is not  R1. Let be
f, so form the free body diagram.
o –  R = 0  o = R =  mg.
Total frictional force on table by floor is  mg.
28. Let the acceleration of block M is ‘a’ towards right. So, the block ‘m’ must go down with an acceleration ‘2a’.
As the block ‘m’ is in contact with the block ‘M’, it will also have acceleration ‘a’ towards right. So, it will
experience two inertia forces as shown in the free body diagram-1.
From free body diagram -1
R1
a1
ma
mg
T
R1
R2
ma12F
ma
a
m
T R1
R1
ma1
a1
F=QE
E
F
M
m
F
R1
QE
m
T
R1
R2
T
mg
R1
R1
R2
R
FR
m

R
R
mg
m
a 2a
M
R1
R1
T1
mg
(FBD-1)
ma
R1
R2
T
Mg
(FBD-2)
Ma
R1
R2

Chapter 6
6.10
R1 – ma = 0  R1 = ma ...(i)
Again, 2ma + T – mg + 1R1 = 0
 T = mg – (2 – 1)ma …(ii)
From free body diagram-2
T + 1R1 + mg – R2 =0
 R2 = T + 1 ma + Mg [Putting the value of R1 from (i)]
= (mg – 2ma – 1 ma) + 1 ma + Mg [Putting the value of T from (ii)]
R2 = Mg + mg – 2ma …(iii)
Again, form the free body diagram -2
T + T – R – Ma –2R2 = 0
 2T – MA – mA – 2 (Mg + mg – 2ma) = 0 [Putting the values of R1 and R2 from (i) and (iii)]
 2T = (M + m) + 2(Mg + mg – 2ma) ...(iv)
From equation (ii) and (iv)
2T = 2 mg – 2(2 + 1)mg = (M + m)a + 2(Mg + mg – 2ma)
 2mg – 2(M + m)g = a (M + m – 22m + 4m + 21m)
 a =
)](25[mM
g)]mM(m2[
21
2



29. Net force = *(202 + (15)2 – (0.5) × 40 = 25 – 20 = 5N
 tan  = 20/15 = 4/3   = tan
–1
(4/3) = 53°
So, the block will move at an angle 53 ° with an 15N force
30. a) Mass of man = 50kg. g = 10 m/s
2
Frictional force developed between hands, legs & back side with the wall the wt of
man. So he remains in equilibrium.
He gives equal force on both the walls so gets equal reaction R from both the walls. If
he applies unequal forces R should be different he can’t rest between the walls.
Frictional force 2R balance his wt.
R + R = 40g  2 R = 40 × 10 R =
8.02
1040


= 250N
b) The normal force is 250 N. 
31. Let a1 and a2 be the accelerations of ma and M respectively.
Here, a1 > a2 so that m moves on M
Suppose, after time ‘t’ m separate from M.
In this time, m covers vt + ½ a1t
2
and SM = vt + ½ a2 t
2
For ‘m’ to m to ‘m’ separate from M. vt + ½ a1 t
2
= vt + ½ a2 t
2
+l ...(1)
Again from free body diagram
Ma1 + /2 R = 0
 ma1 = – (/2) mg = – (/2)m × 10  a1= –5
Again,
Ma2 +  (M + m)g – (/2)mg = 0
 2Ma2 + 2 (M + m)g –  mg = 0
 2 M a2 =  mg – 2Mg – 2 mg
 a2
M2
Mg2mg 
Putting values of a1 & a2 in equation (1) we can find that
T = 





 g)mM(
ml4
    
velocity
a1
ℓ
M
a2
R
R
2

a1
mg
M+mg
Ma2
<
mg
2

a1
(M+m)g
R
R
40g
R

7.1
SOLUTIONS TO CONCEPTS circular motion;;
CHAPTER 7
1. Distance between Earth & Moon
r = 3.85 × 10
5
km = 3.85 × 10
8
m
T = 27.3 days = 24 × 3600 × (27.3) sec = 2.36 × 10
6
sec
v =
T
r2
= 6
8
1036.2
1085.314.32


= 1025.42m/sec
a =
r
v2
= 8
2
1085.3
)42.1025(

= 0.00273m/sec
2
= 2.73 × 10
–3
m/sec
2
2. Diameter of earth = 12800km
Radius R = 6400km = 64 × 10
5
m
V =
T
R2
=
360024
106414.32 5


m/sec = 465.185
a =
R
V2
= 5
2
1064
)5185.46(

= 0.0338m/sec
2
3. V = 2t, r = 1cm
a) Radial acceleration at t = 1 sec.
a =
r
v2
=
1
22
= 4cm/sec
2
b) Tangential acceleration at t = 1sec.
a =
dt
dv
= )t2(
dt
d
= 2cm/sec
2
c) Magnitude of acceleration at t = 1sec
a = 22
24  = 20 cm/sec
2
4. Given that m = 150kg,
v= 36km/hr = 10m/sec, r = 30m
Horizontal force needed is
r
mv2
=
30
)10(150 2

=
30
100150
= 500N
5. in the diagram
R cos  = mg ..(i)
R sin  =
r
mv2
..(ii)
Dividing equation (i) with equation (ii)
Tan  =
rmg
mv2
=
rg
v2
v = 36km/hr = 10m/sec, r = 30m
Tan  =
rg
v2
=
1030
100

= (1/3)
  = tan
–1
(1/3)
6. Radius of Park = r = 10m
speed of vehicle = 18km/hr = 5 m/sec
Angle of banking tan =
rg
v2
  = tan
–1
rg
v2
= tan
–1
100
25
= tan
–1
(1/4)
R
mv2
/R
mg

Chapter 7
7.2
7. The road is horizontal (no banking)
R
mv2
= N
and N = mg
So
R
mv2
=  mg v = 5m/sec, R = 10m

10
25
= g   =
100
25
= 0.25
8. Angle of banking =  = 30°
Radius = r = 50m
tan  =
rg
v2
 tan 30° =
rg
v2

3
1
=
rg
v2
 v
2
=
3
rg
=
3
1050
 v =
3
500
= 17m/sec.
9. Electron revolves around the proton in a circle having proton at the centre.
Centripetal force is provided by coulomb attraction.
r = 5.3 t 10
–11
m m = mass of electron = 9.1 × 10
–3
kg.
charge of electron = 1.6 × 10–19
c.
r
mv2
= 2
2
r
q
k  v
2
=
rm
kq2
= 3111
389
101.9103.5
106.16.1109




= 13
10
23.48
04.23

 v
2
= 0.477 × 10
13
= 4.7 × 10
12
 v = 12
107.4  = 2.2 × 10
6
m/sec
10. At the highest point of a vertical circle
R
mv2
= mg
 v
2
= Rg  v = Rg
11. A celling fan has a diameter = 120cm.
Radius = r = 60cm = 0/6m
Mass of particle on the outer end of a blade is 1g.
n = 1500 rev/min = 25 rev/sec
 = 2 n = 2  ×25 = 157.14
Force of the particle on the blade = Mr
2
= (0.001) × 0.6 × (157.14) = 14.8N
The fan runs at a full speed in circular path. This exerts the force on the particle (inertia). The particle
also exerts a force of 14.8N on the blade along its surface.
12. A mosquito is sitting on an L.P. record disc & rotating on a turn table at
3
1
33 rpm.
n =
3
1
33 rpm =
603
100

rps
 = 2  n = 2  ×
180
100
=
9
10
rad/sec
r = 10cm =0.1m, g = 10m/sec2
mg  mr
2
  =
g
r 2


10
9
10
1.0
2





 

  
81
2


R
mg
g mv2
/R

Chapter 7
7.3
13. A pendulum is suspended from the ceiling of a car taking a turn
r = 10m, v = 36km/hr = 10 m/sec, g = 10m/sec
2
From the figure T sin  =
r
mv2
..(i)
T cos  = mg ..(ii)



cos
sin
=
rmg
mv2
 tan  =
rg
v2
  = tan
–1








rg
v2
= tan
–1
1010
100
 = tan
–1
(1)   = 45°
14. At the lowest pt.
T = mg +
r
mv2
Here m = 100g = 1/10 kg, r = 1m, v = 1.4 m/sec
T = mg +
r
mv2
=
10
)4.1(
8.9
10
1 2
 = 0.98 + 0.196 = 1.176 = 1.2 N
15. Bob has a velocity 1.4m/sec, when the string makes an angle of 0.2 radian.
m = 100g = 0.1kg, r = 1m, v = 1.4m/sec.
From the diagram,
T – mg cos  =
R
mv2
 T =
R
mv2
+ mg cos 
 T =







 


2
18.9)1.0(
1
)4.1(1.0 22
 T = 0.196 + 9.8 ×









2
)2(.
1
2
( cos  =
2
1
2

 for small )
 T = 0.196 + (0.98) × (0.98) = 0.196 + 0.964 = 1.156N  1.16 N
16. At the extreme position, velocity of the pendulum is zero.
So there is no centrifugal force.
So T = mg cos o
17. a) Net force on the spring balance.
R = mg – m
2
r
So, fraction less than the true weight (3mg) is
=
mg
)rmmg(mg 2

=
g
2

=
10
106400
360024
2 32









= 3.5 × 10
–3
b) When the balance reading is half the true weight,
mg
)rmmg(mg 2

= 1/2

2
r = g/2  
r2
g
 3
1064002
10

rad/sec
 Duration of the day is
T =

2
=
8.9
1064002
2
3

 sec =
49
1064
2
6

 sec =
36007
80002


hr = 2hr
mv2
/R
mg

mg
mv2
/r
T
mg sin 
mg cos 
T
mg sin 
mg cos

T
m
2
/R
mg
R

Chapter 7
7.4
18. Given, v = 36km/hr = 10m/s, r = 20m,  = 0.4
The road is banked with an angle,
 = tan
–1








rg
v2
= tan
–1






1020
100
= tan
–1 





2
1
or tan  = 0.5
When the car travels at max. speed so that it slips upward, R1
acts downward as shown in Fig.1
So, R1 – mg cos  –
r
mv
2
1
sin  = 0 ..(i)
And R1 + mg sin  –
r
mv
2
1
cos  = 0 ..(ii)
Solving the equation we get,
V1 =


tan1
tan
rg =
2.1
1.0
1020  = 4.082 m/s = 14.7 km/hr
So, the possible speeds are between 14.7 km/hr and 54km/hr.
19. R = radius of the bridge
L = total length of the over bridge
a) At the highest pt.
mg =
R
mv2
 v
2
= Rg  v = Rg
b) Given, v = Rg
2
1
suppose it loses contact at B. So, at B, mg cos  =
R
mv2
 v
2
= Rg cos 

2
2
Rv








= Rg cos  
2
Rg
= Rg cos   cos  = 1/2   = 60° = /3
 =
r

 ℓ = r =
3
R
So, it will lose contact at distance
3
R
from highest point
c) Let the uniform speed on the bridge be v.
The chances of losing contact is maximum at the end of the bridge for which  =
R2
L
.
So,
R
mv2
= mg cos   v = 





R2
L
cosgR 
20. Since the motion is nonuniform, the acceleration has both radial & tangential
component
ar =
r
v
2
at =
dt
dv
=a
Resultant magnitude = 2
2
2
a
r
v









Now N = m 2
2
2
a
r
v









  mg = m 2
2
2
a
r
v









 
2
g
2
= 2
4
a
2r
v








 v
4
= (
2
g
2
– a
2
) r
2
 v = [(
2
g
2
– a
2
) r
2
]
1/4

R1

mv1
2
/r
mg
R1

mv2
2
/r
R2

mg
R2

2
mv2
/R
mg
2= L/R

mv2
/R
mg
2
2= L/R

mv2
/R
2
2= L/R
mv2
/R mg
m dv/dt
m
N
mv2
/R
m

Chapter 7
7.5
21. a) When the ruler makes uniform circular motion in the horizontal
plane, (fig–a)
 mg = m

L



L
g

b) When the ruler makes uniformly accelerated circular motion,(fig–b)
 mg = 222
2 )mL()Lm(   2
4
+ 
2
= 2
22
L
g
 2 =
4/1
2
2
L
g













 
(When viewed from top)
22. Radius of the curves = 100m
Weight = 100kg
Velocity = 18km/hr = 5m/sec
a) at B mg –
R
mv2
= N  N = (100 × 10) –
100
25100
= 1000 – 25 = 975N
At d, N = mg +
R
mv2
= 1000 + 25 = 1025 N
b) At B & D the cycle has no tendency to slide. So at B & D, frictional force is zero.
At ‘C’, mg sin  = F  F = 1000 ×
2
1
= 707N
c) (i) Before ‘C’ mg cos  – N =
R
mv2
 N = mg cos  –
R
mv2
= 707 – 25 = 683N
(ii) N – mg cos  =
R
mv2
 N =
R
mv2
+ mg cos  = 25 + 707 = 732N
d) To find out the minimum desired coeff. of friction, we have to consider a point just before C. (where
N is minimum)
Now,  N = mg sin    × 682 = 707
So,  = 1.037
23. d = 3m  R = 1.5m
R = distance from the centre to one of the kids
N = 20 rev per min = 20/60 = 1/3 rev per sec
= 2r = 2/3
m = 15kg
 Frictional force F = mr
2
= 15 × (1.5) ×
9
)2( 2

= 5 × (0.5) × 4
2
= 10
2
 Frictional force on one of the kids is 10
2

24. If the bowl rotates at maximum angular speed, the block tends to slip upwards. So, the frictional force
acts downward.
Here, r = R sin 
From FBD –1
R1 – mg cos  – m
2
(R sin ) sin  = 0 ..(i) [because r = R sin ]
and R1 mg sin  – m1
2
(R sin ) cos  = 0 ..(ii)
Substituting the value of R1 from Eq (i) in Eq(ii), it can be found out that
1 =
2/1
)sin(cossinR
)cos(sing








Again, for minimum speed, the frictional force
R2 acts upward. From FBD–2, it can be proved
that,
1
2
L mg
mg

RL
(Fig–a)
(Fig–b)
m2
2
L mg
mL
R1

m1
2
r
R1
(FBD – 1)
R2

m2
2
r
R2
(FBD – 2)
 E
D
CB
A
mv2
/R
mg
N
B
15kg
mg
m2
rF
15kg

Chapter 7
7.6
2 =
2/1
)sin(cossinR
)cos(sing








 the range of speed is between 1 and 2
25. Particle is projected with speed ‘u’ at an angle . At the highest pt. the vertical component of velocity is
‘0’
So, at that point, velocity = u cos 
centripetal force = m u
2
cos
2 




 
r
At highest pt.
mg =
r
mv2
 r =
g
cosu 22

26. Let ‘u’ the velocity at the pt where it makes an angle /2 with horizontal. The
horizontal component remains unchanged
So, v cos /2 =  cos   v =





 

2
cos
cosu
...(i)
From figure
mg cos (/2) =
r
mv2
 r =
 2/cosg
v2

putting the value of ‘v’ from equn(i)
r =
)2/(cosg
cosu
3
22



27. A block of mass ‘m’ moves on a horizontal circle against the wall of a cylindrical room of radius ‘R’
Friction coefficient between wall & the block is .
a) Normal reaction by the wall on the block is =
R
mv2
b)  Frictional force by wall =
R
mv2

c)
R
mv2

= ma  a = –
R
v2

(Deceleration)
d) Now,
dt
dv
=
ds
dv
v = –
R
v2

 ds = –
v
dvR

 s =


R
In V + c
At s = 0, v = v0
Therefore, c =

R
In V0
so, s =
0v
v
In
R

 
0v
v
= e
–s/R
For, one rotation s = 2R, so v = v0e
–2
28. The cabin rotates with angular velocity  & radius R
 The particle experiences a force mR
2
.
The component of mR
2
along the groove provides the required force to the particle to move along AB.
 mR
2
cos  = ma  a = R
2
cos 
length of groove = L
L = ut + ½ at
2
 L = ½ R
2
cos  t
2
 t
2
=
 cosR
L2
2
=  t =
 cosR
L2
1 2

/
mg
mv2
/

 cos

mgcos/2
R
m

mv2
/R
mv2
/R
R
A
B
mv2
/R
mv2
/r
mg
u sin 

u cos 

Chapter 7
7.7
29. v = Velocity of car = 36km/hr = 10 m/s
r = Radius of circular path = 50m
m = mass of small body = 100g = 0.1kg.
 = Friction coefficient between plate & body = 0.58
a) The normal contact force exerted by the plate on the block
N =
r
mv2
=
50
1001.0 
= 0.2N
b) The plate is turned so the angle between the normal to the plate & the radius of the road slowly
increases
N =
r
mv2
cos  ..(i)
 N =
r
mv2
sin  ..(ii)
Putting value of N from (i)

r
mv2
cos  =
r
mv2
sin    = tan    = tan
–1
 = tan
–1
(0.58) = 30° 
30. Let the bigger mass accelerates towards right with ‘a’.
From the free body diagrams,
T – ma – m
R = 0 …(i)
T + 2ma – 2m
2
R = 0 …(ii)
Eq (i) – Eq (ii)  3ma = m
2
R
 a =
3
Rm 2

Substituting the value of a in Equation (i), we get T = 4/3 m
2
R.
* * * *
R
a
m
2m
ma T
m2
R
2ma T
2m2
R

8.1
CHAPTER – 8
1. M = mc + mb = 90kg
u = 6 km/h = 1.666 m/sec
v = 12 km/h = 3.333 m/sec
Increase in K.E. = ½ Mv2
– ½ Mu2
= ½ 90 × (3.333)2
– ½ × 90 × (1.66)2
= 494.5 – 124.6 = 374.8  375 J
2. mb = 2 kg.
u = 10 m/sec
a = 3 m/aec2
t = 5 sec
v = u + at = 10 + 3 I 5 = 25 m/sec.
F.K.E = ½ mv2
= ½ × 2 × 625 = 625 J.
3. F = 100 N
S = 4m,  = 0°
 = S.F

100 × 4 = 400 J
4. m = 5 kg
 = 30°
S = 10 m
F = mg
So, work done by the force of gravity
 = mgh = 5 × 9.8 × 5 = 245 J
5. F= 2.50N, S = 2.5m, m =15g = 0.015kg.
So, w = F × S  a =
m
F
=
015.0
5.2
=
3
500
m/s2
=F × S cos 0° (acting along the same line)
= 2.5 × 2.5 = 6.25J
Let the velocity of the body at b = U. Applying work-energy principle ½ mv2
– 0 = 6.25
 V =
015.0
225.6 
= 28.86 m/sec.
So, time taken to travel from A to B.
 t =
a
uv 
=
500
386.28 
 Average power =
t
W
=
3)86.28(
50025.6


= 36.1
6. Given
jˆ3iˆ2r1 

jˆ2iˆ3r2 
So, displacement vector is given by,
21 rrr

  jˆiˆ)jˆ3iˆ2()jˆ2iˆ3(r 

u=1.66 m/s
90kg 90kg
v=3.33 m/s
u=10 m/s
2 kg a

= 3m/s2
F 4m
R
100 N
mg
30°
5 
mg
F 
5 log
10m
30°
v
A  B 

Chapter 8
8.2
So, work done = sF

 = 5 × 1 + 5(-1) = 0
7. mb = 2kg, s = 40m, a = 0.5m/sec2
So, force applied by the man on the box
F = mba = 2 × (0.5) = 1 N
 = FS = 1 × 40 = 40 J
8. Given that F= a + bx
Where a and b are constants.
So, work done by this force during this force during the displacement x = 0 and x = d is given
by
W =  
d
0
d
0
dx)bxa(dxF = ax + (bx2
/2) = [a + ½ bd] d
9. mb = 250g = .250 kg
 = 37°, S = 1m.
Frictional force f = R
mg sin  =  R ..(1)
mg cos  ..(2)
so, work done against R = RS cos 0° = mg sin  S = 0.250 × 9.8 × 0.60 × 1 = 1.5 J
10. a =
)mM(2
F

(given)
a) from fig (1)
ma = k R1 and R1 = mg
  =
1R
ma
=
g)mM(2
F

b) Frictional force acting on the smaller block f = R =
)mM(2
Fm
mg
g)mM(2
F




c) Work done w = fs s = d
w = d
)mM(2
mF


=
)mM(2
mFd


11. Weight = 2000 N, S = 20m,  = 0.2
a) R + Psin - 2000 = 0 ..(1)
P cos - 0.2 R =0 ..(2)
From (1) and (2) P cos – 0.2 (2000 – P sin)=0
P =
 sin2.0cos
400
..(3)
So, work done by the person, W = PS cos =


sin2.0cos
cos8000
=
 sin2.01
8000
=
 tan5
40000
b) For minimum magnitude of force from equn(1)
d/d (cos + 0.2 sin) = 0  tan  = 0.2
putting the value in equn (3)
W =
 tan5
40000
=
)2.5(
40000
= 7690 J
12. w = 100 N,  = 37°, s = 2m
R
mbg
mba F
R R
1 m
mg
37°
M

F
m

R1
 kR1ma 
mg
R1
R2
fma 
mg
R2

R
P
0.2R
2000 N

Chapter 8
8.3
Force F= mg sin 37° = 100 × 0.60 = 60 N
So, work done, when the force is parallel to incline.
w = Fs cos  = 60 × 2 × cos  = 120 J
In ∆ABC AB= 2m
CB = 37°
so, h = C = 1m
work done when the force in horizontal direction
W = mgh = 100 × 1.2 = 120 J
13. m = 500 kg, s = 25m, u = 72km/h= 20 m/s, v = 0
(-a) =
S2
uv 22

 a =
50
400
= 8m/sec2
Frictional force f = ma = 500 × 8 = 4000 N
14. m = 500 kg, u = 0, v = 72 km/h = 20m/s
a =
s2
uv 22

=
50
400
= 8m/sec2
force needed to accelerate the car F = ma = 500 × 8 = 4000 N
15. Given, v = a x (uniformly accelerated motion)
displacement s = d – 0 = d
putting x = 0, v1= 0
putting x = d, v2= a d
a =
s2
uv
2
2
2
2 
=
d2
da2
=
2
a2
force f = ma =
2
ma2
work done w = FS cos  = d
2
ma2
 =
2
dma2

16. a) m = 2kg,  = 37°, F = 20 N
F = (2g sin ) + ma  a = (20 – 20 sin)/s = 4m/sec2
S = ut + ½ at2
(u = 0, t = 1s, a = 1.66)
= 2m
So, work, done w = Fs = 20 × 2 = 40 J
b) If W = 40 J
S =
F
W
=
20
40
h = 2 sin 37° = 1.2 m
So, work done W = –mgh = – 20 × 1.2 = –24 J
c) v = u + at = 4 × 10 = 40 m/sec
So, K.E. = ½ mv2
= ½ × 2 × 16 = 16 J
17. m = 2kg,  = 37°, F = 20 N, a = 10 m/sec2
a) t = 1sec
So, s= ut + ½ at2
= 5m
37°
A
A
A
B
v=0

v=20 m/s
m=500 kg
–a

25m

a

R
mg

maf

500 kg
a

25m

R
mg

F
F
ma

ma
2g cos
20N

R

ma 2gsin
20N
ma
mg cos
20N

R

mg sin
R
h
37°
5m
C 
A
B
37°

Chapter 8
8.4
Work done by the applied force w = FS cos 0° = 20 × 5 = 100 J
b) BC (h) = 5 sin 37° = 3m
So, work done by the weight W = mgh = 2 × 10 × 3 = 60 J
c) So, frictional force f = mg sin
work done by the frictional forces w = fs cos0° = (mg sin) s = 20 × 0.60 × 5 = 60 J
18. Given, m = 25o g = 0.250kg,
u = 40 cm/sec = 0.4m/sec
 = 0.1, v=0
Here,  R = ma {where, a = deceleration}
a =
m
R
=
m
mg
= g = 0.1 × 9.8 = 0.98 m/sec2
S =
a2
uv 22

= 0.082m = 8.2 cm
Again, work done against friction is given by
– w =  RS cos 
= 0.1 × 2.5 × 0.082 × 1 ( = 0°) = 0.02 J
 W = – 0.02 J
19. h = 50m, m = 1.8 × 105
kg/hr, P = 100 watt,
P.E. = mgh = 1.8 × 105
× 9.8 × 50 = 882 × 105
J/hr
Because, half the potential energy is converted into electricity,
Electrical energy ½ P.E. = 441 × 105
J/hr
So, power in watt (J/sec) is given by =
3600
10441 5

 number of 100 W lamps, that can be lit
1003600
10441 5


= 122.5 122
20. m = 6kg, h = 2m
P.E. at a height ‘2m’ = mgh = 6 × (9.8) × 2 = 117.6 J
P.E. at floor = 0
Loss in P.E. = 117.6 – 0 = 117. 6 J  118 J
21. h = 40m, u = 50 m/sec
Let the speed be ‘v’ when it strikes the ground.
Applying law of conservation of energy
mgh + ½ mu2
= ½ mv2
 10 × 40 + (1/2) × 2500 = ½ v2
 v2
= 3300  v = 57.4 m/sec 58 m/sec
22. t = 1 min 57.56 sec = 11.56 sec, p= 400 W, s =200 m
p =
t
w
, Work w = pt = 460 × 117.56 J
Again, W = FS =
200
56.117460
= 270.3 N  270 N
23. S = 100 m, t = 10.54 sec, m = 50 kg
The motion can be assumed to be uniform because the time taken for acceleration is
minimum.

Chapter 8
8.5
a) Speed v = S/t = 9.487 e/s
So, K.E. = ½ mv2
= 2250 J
b) Weight = mg = 490 J
given R = mg /10 = 49 J
so, work done against resistance WF = – RS = – 49 × 100 = – 4900 J
c) To maintain her uniform speed, she has to exert 4900 j of energy to over come friction
P =
t
W
= 4900 / 10.54 = 465 W
24. h = 10 m
flow rate = (m/t) = 30 kg/min = 0.5 kg/sec
power P =
t
mgh
= (0.5) × 9.8 × 10 = 49 W
So, horse power (h.p) P/746 = 49/746 = 6.6 × 10–2
hp
25. m = 200g = 0.2kg, h = 150cm = 1.5m, v = 3m/sec, t = 1 sec
Total work done = ½ mv2
+ mgh = (1/2) × (0.2) ×9 + (0.2) × (9.8) × (1.5) = 3.84 J
h.p. used =
746
84.3
= 5.14 × 10–3
26. m = 200 kg, s = 12m, t = 1 min = 60 sec
So, work W = F cos  = mgs cos0° [ = 0°, for minimum work]
= 2000 × 10 × 12 = 240000 J
So, power p =
t
W
=
60
240000
= 4000 watt
h.p =
746
4000
= 5.3 hp.
27. The specification given by the company are
U = 0, m = 95 kg, Pm = 3.5 hp
Vm = 60 km/h = 50/3 m/sec tm = 5 sec
So, the maximum acceleration that can be produced is given by,
a =
5
0)3/50( 
=
3
10
So, the driving force is given by
F = ma = 95 ×
3
10
=
3
950
N
So, the velocity that can be attained by maximum h.p. white supplying
3
950
will be
v =
F
p
 v =
950
57465.3 
= 8.2 m/sec.
Because, the scooter can reach a maximum of 8.s m/sec while producing a force of 950/3 N,
the specifications given are some what over claimed.
28. Given m = 30kg, v = 40 cm/sec = 0.4 m/sec s = 2m
From the free body diagram, the force given by the chain is,
F = (ma – mg) = m(a – g) [where a = acceleration of the block]
a =
2s
u2)(v2
=
4.0
16.0
= 0.04 m/sec2
mg

F

ma

mg

F


Chapter 8
8.6
So, work done W = Fs cos  = m(a –g) s cos 
 W = 30 (0.04 – 9.8) × 2  W = –585.5  W = –586 J.
So, W = – 586 J
29. Given, T = 19 N
From the freebody diagrams,
T – 2 mg + 2 ma = 0 …(i)
T – mg – ma = 0 …(ii)
From, Equation (i) & (ii) T = 4ma  a =
m4
T
 A =
m4
16
=
m
4
m/s2
.
Now, S = ut + ½ at2
 S = 1
m
4
2
1
  S =
m
2
m [ because u=0]
Net mass = 2m – m = m
Decrease in P.E. = mgh P.E. = m × g ×
m
2
 P.E. = 9.8 × 2 P.E. = 19.6 J
30. Given, m1 = 3 kg, m2 = 2kg, t = during 4th
second
From the freebody diagram
T – 3g + 3a = 0 ..(i)
T – 2g – 2a = 0 ..(ii)
Equation (i) & (ii), we get 3g – 3a = 2g + 2a  a =
5
g
m/sec2
Distance travelled in 4th
sec is given by
S4th = )1n2(
2
a
 = )142(
s
5
g







=
10
g7
=
10
8.97
m
Net mass ‘m’ = m1 – m2 = 3 – 2 = 1kg
So, decrease in P.E. = mgh = 1 × 9.8 ×
10
7
× 9.8 = 67.2 = 67 J
31. m1 = 4kg, m2 = 1kg, V2 = 0.3m/sec V1 = 2 × (0.3) = 0.6 m/sec
(v1 = 2x2 m this system)
h = 1m = height descent by 1kg block
s = 2 × 1 = 2m distance travelled by 4kg block
u = 0
Applying change in K.E. = work done (for the system)
[(1/2)m1v1
2
+ (1/2) m2vm
2
] –0 = (–R)S + m2g [R = 4g = 40 N]
 ½ × 4 × (0.36) × ½ ×1 × (0.09) = –  × 40 × 2 + 1 × 40 × 1
 0.72 + 0.045 = – 80 + 10
  =
80
235.9
= 0.12
32. Given, m = 100g = 0.1kg, v = 5m/sec, r = 10cm
Work done by the block = total energy at A – total energy at B
(1/2 mv2
+ mgh) – 0
 W = ½ mv2
+ mgh – 0 = ½ × (0.1) × 25 + (0.1) × 10 × (0.2) [h = 2r = 0.2m]
2ma

2mg

T

ma

mg

T

2m

m
a
 a

3kg
3a

3g

T

3kg

2kg
a
 a

2a

2g

T

2kg
4 kg

1kg

10cm
A
B

Chapter 8
8.7
 W = 1.25 – 0.2  W = 1.45 J
So, the work done by the tube on the body is
Wt = –1.45 J
33. m = 1400kg, v = 54km/h = 15m/sec, h = 10m
Work done = (total K.E.) – total P.E.
= 0 + ½ mv2
– mgh = ½ × 1400 × (15)2
– 1400 × 9.8 × 10 = 157500 – 137200 = 20300 J
So, work done against friction, Wt = 20300 J
34. m = 200g = 0.2kg,s = 10m, h = 3.2m, g = 10 m/sec2
a) Work done W = mgh = 0.2 × 10 × 3.2 = 6.4 J
b) Work done to slide the block up the incline
w = (mg sin ) = (0.2) × 10 ×
10
2.3
× 10 = 6.4 J
c) Let, the velocity be v when falls on the ground vertically,
½ mv2
– 0 = 6.4J  v = 8 m/s
d) Let V be the velocity when reaches the ground by liding
½ mV2
– 0 = 6.4 J  V = 8m/sec
35. ℓ = 10m, h = 8m, mg = 200N
f = 200 ×
10
3
= 60N
a) Work done by the ladder on the boy is zero when the boy is going up
because the work is done by the boy himself.
b) Work done against frictional force, W = RS = f ℓ = (–60) × 10 = – 600 J
c) Work done by the forces inside the boy is
Wb = (mg sin) × 10 = 200 ×
10
8
× 10 = 1600 J
36. H = 1m, h = 0.5m
Applying law of conservation of Energy for point A & B
mgH = ½ mv2
+ mgh  g = (1/2) v2
+ 0.5g  v2
2(g – 0.59) = g  v = g = 3.1 m/s
After point B the body exhibits projectile motion for which
 = 0°, v = – 0.5
So, –0.5 = (u sin) t - (1/2) gt2
 0.5 = 4.9 t2
 t = 0.31 sec.
So, x = (4 cos ) t = 3.1 × 3.1 = 1m.
So, the particle will hit the ground at a horizontal distance in from B.
37. mg = 10N, = 0.2, H = 1m, u = v = 0
change in P.E. = work done.
Increase in K.E.
 w = mgh = 10 × 1 = 10 J
Again, on the horizontal surface the fictional force
F = R = mg = 0.2 × 10 = 2 N
So, the K.E. is used to overcome friction
 S =
F
W
=
N2
J10
= 5m
15m/s
A

B

10m
h

10m 3.2m



mg sin mg cos
F

R

F

mg

8m
10m



mg
R=f

R

x

H=1m

A
B
h=0.5m
1m

A

B

Chapter 8
8.8
38. Let ‘dx’ be the length of an element at a distance × from the table
mass of ‘dx’ length = (m/ℓ) dx
Work done to put dx part back on the table
W = (m/ℓ) dx g(x)
So, total work done to put ℓ/3 part back on the table
W = 
3/1
0
dxgx)/m(   w = (m/ℓ) g
3
0
2
2
x









=


18
mg 2
=
18
mg
39. Let, x length of chain is on the table at a particular instant.
So, work done by frictional force on a small element ‘dx’
dWf = Rx = gxdx
L
M






 [where dx = dx
L
M
]
Total work don by friction,
Wf = 
0
3/L2
dxgx
L
M
 Wf =
0
3/L2
2
2
x
g
L
m








 =









18
L4
L
M 2
= 2Mg L/9
40. Given, m = 1kg, H = 1m, h = 0.8m
Here, work done by friction = change in P.E. [as the body comes to rest]
 Wf = mgh – mgH
= mg (h – H)
= 1 × 10 (0.8 – 1) = – 2J
41. m = 5kg, x = 10cm = 0.1m, v = 2m/sec,
h =? G = 10m/sec2
S0, k =
x
mg
=
1.0
50
= 500 N/m
Total energy just after the blow E = ½ mv2
+ ½ kx2
…(i)
Total energy a a height h = ½ k (h – x)2
+ mgh …(ii)
½ mv2
+ ½ kx2
= ½ k (h – x)2
+ mgh
On, solving we can get,
H = 0.2 m = 20 cm
42. m = 250 g = 0.250 kg,
k = 100 N/m, m = 10 cm = 0.1m
g = 10 m/sec2
½ kx2
= mgh  h = 







mg
kx
2
1 2
=
1025.02
)1.0(100 2


= 0.2 m = 20 cm
43. m = 2kg, s1 = 4.8m, x = 20cm = 0.2m, s2 = 1m,
sin 37° = 0.60 = 3/5,  = 37°, cos 37° = .79 = 0.8 = 4/5
g = 10m/sec2
Applying work – Energy principle for downward motion of the body
dx

x

2l/3
dx
 x
2l/3
H=1m

A
B
h=0.8m
h
B
A 
0.1m
m

0.2 
4.8m
37°
R 

Chapter 8
8.9
0 – 0 = mg sin 37° × 5 – R × 5 – ½ kx2
 20 × (0.60) × 1 –  × 20 × (0.80) × 1 + ½ k (0.2)2
= 0
 60 – 80 - 0.02k = 0  80 + 0.02k = 60 …(i)
Similarly, for the upward motion of the body the equation is
0 – 0 = (–mg sin 37°) × 1 –  R × 1 + ½ k (0.2)2
 –20 × (0.60) × 1 –  ×20 × (0.80) × 1 + ½ k (0.2)2
= 0
 –12 – 16 + 0.02 K = 0 ..(ii)
Adding equation (i) & equation (ii), we get 96  = 48
  = 0.5
Now putting the value of  in equation (i) K = 1000N/m
44. Let the velocity of the body at A be v
So, the velocity of the body at B is v/2
Energy at point A = Energy at point B
So, ½ mvA
2
= ½ mvB
2
+ ½ kx2+
 ½ kx2
= ½ mvA
2
- ½ mvB
2
 kx2
= m (vA
2+ -
vB
2
)  kx2
= 








4
v
vm
2
2
 k = 2
2
x3
mv3
45. Mass of the body = m
Let the elongation be x
So, ½ kx2
= mgx
 x = 2mg / k
46. The body is displaced x towards right
Let the velocity of the body be v at its mean position
½ mv2
= ½ k1x2
+ ½ k2x2
 mv2
= x2
(k1 + k2)  v2
= m
)kk(x 21
2

 v =
m
kk
x 21 
47. Let the compression be x
According to law of conservation of energy
½ mv2
=½ kx2
 x2
= mv2
/ k  x = v )k/m(
b) No. It will be in the opposite direction and magnitude will be less due to loss in spring.
48. m = 100g = 0.1kg, x = 5cm = 0.05m, k = 100N/m
when the body leaves the spring, let the velocity be v
½ mv2
= ½ kx2
 v = m/kx = 0.05 ×
1.0
100
= 1.58m/sec
For the projectile motion,  = 0°, Y = –2
Now, y = (u sin )t – ½ gt2
 –2 = (-1/2) × 9.8 × t2
 t = 0.63 sec.
So, x = (u cos ) t  1.58 × 0.63 = 1m
x 
v B  A 
k1 
m 
k2 
x 
k

x



v 
k 
A
l

v 
m 
B 
l


Chapter 8
8.10
49. Let the velocity of the body at A is ‘V’ for minimum velocity given at A velocity of the body at
point B is zero.
Applying law of conservation of energy at A & B
½ mv2
= mg (2ℓ)  v = )g4(  = g2
50. m = 320g = 0.32kg
k = 40N/m
h = 40cm = 0.4m
g = 10 m/s2
From the free body diagram,
kx cos  = mg
(when the block breaks off R = 0)
 cos  = mg/kx
So,
x40
2.3
x4.0
4.0



 16x = 3.2x + 1.28  x = 0.1 m
S0, s = AB = 3.0)4.0()5.0(h)xh( 2222
 m
Let the velocity of the body at B be v
Charge in K.E. = work done (for the system)
(1/2 mv2
+ ½ mv2
) = –1/2 kx2
+ mgs
 (0.32) × v2
= –(1/2) × 40 × (0.1)2
+ 0.32 × 10 × (0.3)  v = 1.5 m/s.
51.  = 37° ; l = h = natural length
Let the velocity when the spring is vertical be ‘v’.
Cos 37° = BC/AC = 0.8 = 4/5
Ac = (h + x) = 5h/4 (because BC = h)
So, x = (5h/4) – h = h/4
Applying work energy principle ½ kx2
= ½ mv2
 v = x )m/k( =
m
k
4
h
52. The minimum velocity required to cross the height point c =
gl2
Let the rod released from a height h.
Total energy at A = total energy at B
mgh = 1/2 mv2
; mgh = 1/2 m (2gl)
[Because v = required velocity at B such that the block
makes a complete circle. [Refer Q – 49]
So, h = l.
53. a) Let the velocity at B be v2
1/2 mv1
2
= 1/2 mv2
2
+ mgl
 1/2 m (10 gl) = 1/2 mv2
2
+ mgl
v2
2
= 8 gl
So, the tension in the string at horizontal position
kx sin
S
A
0.4
m
m
B 
A
m
m

S
B
(0.4 +x)
mg
kx cos
C
h

37°
B  A
C
h

l
B 
A
B 
l
Av1
l
60°
B
D
C
v2
v3
mv
2
/R
mg
mg
mv3
2
/ l
mg

Chapter 8
8.11
T =
l
gl8m
R
mv2
 = 8 mg
b) Let the velocity at C be V3
1/2 mv1
2
= 1/2 mv3
2
+ mg (2l)
 1/2 m (log l) = 1/2 mv3
2
+ 2mgl
 v3
2
= 6 mgl
So, the tension in the string is given by
Tc = mg
l
mv2
 =
l
glm6
mg = 5 mg
c) Let the velocity at point D be v4
Again, 1/2 mv1
2
= 1/2 mv4
2
+ mgh
1/2 × m × (10 gl) = 1.2 mv4
2
+ mgl (1 + cos 60°)
 v4
2
= 7 gl
So, the tension in the string is
TD = (mv2
/l) – mg cos 60°
= m(7 gl)/l – l – 0.5 mg  7 mg – 0.5 mg = 6.5 mg.
54. From the figure, cos  = AC/AB
 AC = AB cos   (0.5) × (0.8) = 0.4.
So, CD = (0.5) – (0.4) = (0.1) m
Energy at D = energy at B
1/2 mv2
= mg (CD)
v2
= 2 × 10 × (0.1) = 2
So, the tension is given by,
T = mg
r
mv2
 = (0.1) 





10
5.0
2
= 1.4 N.
55. Given, N = mg
As shown in the figure, mv2
/ R = mg
 v2
= gR …(1)
Total energy at point A = energy at P
1/2 kx2
=
2
mgR2mgR 
[because v2
= gR]
 x2
= 3mgR/k  x = k/)mgR3( .
56. V = gl3
1/2 mv2
– 1/2 mu2
= –mgh
v2
= u2
– 2g(l + lcos)
 v2
= 3gl – 2gl (1 + cos ) …(1)
Again,
mv2
/l = mg cos 
v2
= lg cos 
From equation (1) and (2), we get
3gl – 2gl – 2gl cos  = gl cos 
A
0.1 kg
37°
0.5m
CB
0.5m
D
T
mv2
/ r
mg
R
A
P 
m
k
N mv
2
/R
m 
v
u=3 gl
 l
l
mg 
mv2
/R 
T=0

Chapter 8
8.12
3 cos  = 1  cos  = 1/3
 = cos–1
(1/3)
So, angle rotated before the string becomes slack
= 180° – cos–1
(1/3) = cos–1
(–1/3)
57. l = 1.5 m; u = 57 m/sec.
a) mg cos  = mv2
/ l
v2
= lg cos  …(1)
change in K.E. = work done
1/2 mv2
– 1/2 mu2
= mgh
 v2
– 57 = –2 × 1.5 g (1 + cos )…(2)
 v2
= 57 – 3g(1 + cos )
Putting the value of v from equation (1)
15 cos  = 57 – 3g (1 + cos )  15 cos  = 57 – 30 – 30 cos 
 45 cos  = 27  cos  = 3/5.
  = cos–1
(3/5) = 53°
b) v = )cos1(g357  from equation (2)
= 9 = 3 m/sec.
c) As the string becomes slack at point B, the particle will start making projectile motion.
H = OE + DC = 1.5 cos  +
g2
sinu 22

= (1.5) × (3/5) +
102
)8.0(9 2


= 1.2 m.
58.
a) When the bob has an initial height less than the peg and then released from rest (figure 1),
let body travels from A to B.
Since, Total energy at A = Total energy at B
 (K.E)A = (PE)A = (KE)B + (PE)B
 (PE)A = (PE)B [because, (KE)A = (KE)B = 0]
So, the maximum height reached by the bob is equal to initial height.
b) When the pendulum is released with  = 90° and x = L/2, (figure 2) the path of the particle
is shown in the figure 2.
At point C, the string will become slack and so the particle will start making projectile
motion. (Refer Q.No. 56)
(1/2)mvc
2
– 0 = mg (L/2) (1 – cos )
D 
C 
E  B
A
v
1.5 m
 l
mg 
mv2
/ l 
T=0

C 
x
O 
B
Fig-1
A
P L
m 

Fig-2
x=L/2
C 
D
B
A
P 
F E 
mv2
/(L/2)
mg
TC=0

Fig-3


Chapter 8
8.13
because, distance between A nd C in the vertical direction is L/2 (1 – cos )
 vc
2
= gL(1 – cos ) ..(1)
Again, form the freebody diagram (fig – 3)
2/L
mv
2
c
mg cos  {because Tc = 0}
So, VC
2
=
2
gL
cos  ..(2)
From Eqn.(1) and equn (2),
gL (1 – cos ) =
2
gL
cos 
 1 – cos  = 1/2 cos 
 3/2 cos  = 1  cos  = 2/3 ..(3)
To find highest position C, before the string becomes slack
BF =  cos
2
L
2
L
=
3
2
2
L
2
L
 = 






3
1
2
1
L
So, BF = (5L/6)
c) If the particle has to complete a vertical circle, at the point C.
)xL(
mv 2
c

mg
 vc
2
= g (L – x) ..(1)
Again, applying energy principle between A and C,
1/2 mvc
2
– 0 = mg (OC) 
 1/2 vc
2
= mg [L – 2(L – x)] = mg (2x – L)
 vc
2
= 2g(2x – L) ..(2)
From equn. (1) and equn (2)
g(L – x) = 2g (2x – L)
 L – x = 4x – 2L
 5x = 3L

L
x
=
5
3
= 0.6
So, the rates (x/L) should be 0.6
59. Let the velocity be v when the body leaves the surface.
R
mv2
= mg cos  [Because normal reaction]
v2
= Rg cos  ..(1)
Again, form work-energy principle,
Change in K.E. = work done
 1/2 mv2
– 0 = mg(R – R cos)
 v2
= 2gR (1 – cos ) ..(2)
From (1) and (2)
Rg cos  = 2gR (1 – cos )
O 
OP=x
C 
B
A
P 
L–x
mv2
/(L–x)
mg
A 
B 

mg sing
mg 
mg cos
mv2
/R

Chapter 8
8.14
3gR cos  = 2 gR
Cos  = 2/3
 = cos –1
(2/3)
60. a) When the particle is released from rest (fig–1), the centrifugal force is zero.
N force is zero = mg cos 
= mg cos 30° =
2
mg3
b) When the particle leaves contact with the surface (fig–2), N = 0.
So,
R
mv2
mg cos 
 v2
= Rg cos ..(1)
Again, ½ mv2
= mgR (cos 30° – cos )
 v2
= 2Rg 







 cos
2
3
..(2)
From equn. (1) and equn. (2)
Rg cos  = 3 Rg – 2Rg cos 
 3 cos  = 3
 cos  =
3
1
  = cos–1






3
1
So, the distance travelled by the particle before leaving contact,
ℓ = R( - /6) [because 30° = /6]
putting the value of , we get ℓ = 0.43R
61. a) Radius =R
horizontal speed = v
From the free body diagram, (fig–1)
N = Normal force = mg -
R
mv2
b) When the particle is given maximum velocity so that the centrifugal force balances the
weight, the particle does not slip on the sphere.
R
mv2
= mg  v = gR
c) If the body is given velocity v1
v1 = 2/gR
v1
2
– gR / 4
Let the velocity be v2 when it leaves contact with the surface, (fig–2)
So,
R
mv2
= mg cos 
v2
2
= Rg cos  ..(1)
Again, 1/2 mv2
2
– 1/2 mv1
2
= mgR (1 – cos )
 v2
2
= v1
2
+ 2gR (1 – cos ) ..(2)
From equn. (1) and equn (2)
mg 
N

Fig-1
mv2
/R
30°

Fig-2
mg 
mg 
mv2
/R
N
Fig-1
v1

Fig-2
mg 
mv2
2
/R

Chapter 8
8.15
Rg cos  = (Rg/4) + 2gR (1 – cos )
 cos  = (1/4) + 2 – 2 cos 
 3 cos  = 9/4
 cos  = 3/4
  = cos–1
(3/4)
62. a) Net force on the particle between A & B, F = mg sin 
work done to reach B, W = FS = mg sin  ℓ
Again, work done to reach B to C = mgh = mg R (1 – cos )
So, Total workdone = mg[ℓsin  + R(1 – cos )]
Now, change in K.E. = work done
 1/2 mvO
2
= mg [ ℓ sin  + R (1 – cos )
 vo = )sin)cos1((2  Rg
b) When the block is projected at a speed 2vo.
Let the velocity at C will be Vc.
Applying energy principle,
1/2 mvc
2
- 1/2 m (2vo)2
= –mg [ℓsin  + R(1 – cos )]
 vc
2
= 4vo – 2g [ℓsin  + R(1 – cos]
4.2g [ℓsin  + R(1 – cos )] – 2g [ℓsin  R(1 – cos )
= 6g [ℓsin  + R(1 – cos )]
So, force acting on the body,
 N =
R
mv
2
c
= 6mg [(ℓ/R) sin  + 1 – cos ]
c) Let the block loose contact after making an angle 
R
mv2
= mg cos   v2
= Rg cos  ..(1)
Again, 1/2 mv2
= mg (R – R cos )  v2
= 2gR (1 – cos ) ..(2)……..(?)
From (1) and (2) cos  = 2/3   = cos –1
(2/3)
63. Let us consider a small element which makes angle ‘d’ at the centre.
 dm = (m/ℓ)Rd 
a) Gravitational potential energy of ‘dm’ with respect to centre of the sphere
= (dm)g R cos 
= (mg/ℓ) Rcos  d
So, Total G.P.E. = 
r/
0
2
mgR


cos  d  [  = (ℓ/R)](angle subtended by the chain at the
centre)……….
=

gmR2
[sin ] (ℓ/R) =

mRg
sin (ℓ/R)
b) When the chain is released from rest and slides down through an angle , the K.E. of the
chain is given
K.E. = Change in potential energy.
v1
 mg 
mv2
/R
l
A 
B 

C
R

R

l

Chapter 8
8.16
=

gmR2
sin(ℓ/R) -m  
2
gR
cos  d 
=

gmR2
[ sin (ℓ/R) + sin  - sin { +(ℓ/R)}]
c) Since, K.E. = 1/2 mv2
=

gmR2
[ sin (ℓ/R) + sin  - sin { +(ℓ/R)}]
Taking derivative of both sides with respect to ‘t’
 × 2v ×
dt
dv
=

gR2
[cos  ×
dt
d
– cos ( +ℓ/R)
dt
d
]
 (R
dt
d
)
dt
dv
=

gR2
×
dt
d
[cos  – cos ( +(ℓ/R))]
When the chain starts sliding down,  = 0.
So,
dt
dv
=

Rg
[1 – cos (ℓ/R)]
64. Let the sphere move towards left with an acceleration ‘a
Let m = mass of the particle
The particle ‘m’ will also experience the inertia due to acceleration ‘a’ as it is on the sphere. It
will also experience the tangential inertia force (m (dv/dt)) and centrifugal force (mv2
/R).
dt
dv
m = ma cos  + mg sin   mv
dt
dv
= ma cos  




 
dt
d
R + mg sin 





 
dt
d
R
Because, v =
dt
d
R

 vd v = a R cos  d + gR sin  d
Integrating both sides we get,
2
v2
 a R sin  – gR cos  + C
Given that, at  = 0, v = 0 , So, C = gR
So,
2
v2
= a R sin  – g R cos  + g R
 v2
= 2R (a sin  + g – g cos )  v =[2R (a sin  + g – g cos )]1/2
* * * * *
R
a

mv
2
/R
ma
mdv/dt
N
mg

9.1
CHAPTER 9
1. m1 = 1kg, m2 = 2kg, m3 = 3kg,
x1 = 0, x2 = 1, x3=1/2
y1 = 0, y2 = 0, y3 = 2/3
The position of centre of mass is
C.M = 









321
332211
321
332211
mmm
ymymym
,
mmm
xmxmxm
=












321
))2/3(3()02()01(
,
321
)2/13()12()01(
=








12
33
,
12
7
from the point B.
2. Let  be the origin of the system
In the above figure
m1 = 1gm, x1 = – (0.96×10
–10
)sin 52° y1 = 0
m2 = 1gm, x2 = – (0.96×10
–10
)sin 52° y2 = 0
x3 = 0 y3= (0.96 × 10
–10
) cos 52°
The position of centre of mass










321
332211
321
332211
mmm
ymymym
,
mmm
xmxmxm
=







 

 
18
y1600
,
1611
01652sin)1096.0(52sin)1096.0( 3
1010
=   o10
52cos1096.09/8,0 

3. Let ‘O’ (0,0) be the origin of the system.
Each brick is mass ‘M’ & length ‘L’.
Each brick is displaced w.r.t. one in contact by ‘L/10’
The X coordinate of the centre of mass
m7
2
L
m
10
L
2
L
m
10
L
10
L3
2
L
m
10
L3
2
L
m
10
L2
2
L
m
10
L
2
L
m
2
L
m
Xcm











































=
7
2
L
10
L
2
L
5
L
2
L
10
L3
2
L
5
L
2
L
10
L
2
L
2
L

=
7
5
L2
10
L5
2
L7

=
710
L4L5L35


=
70
L44
= L
35
11
4. Let the centre of the bigger disc be the origin.
2R = Radius of bigger disc
R = Radius of smaller disc
m1 = R
2
× T × 
m2 = (2R)
2
I T × 
where T = Thickness of the two discs
 = Density of the two discs
 The position of the centre of mass
(0, 0)
CB
1m
(1, 1)








2
3
,
2
1
1m
A
1m
X
Y
0.96×10–10
m
m2H
(0, 0)
m1
Q
104°
52° 52°
H
m3
0.96×10–10
m
XO
L
L/10
m1
(0, 0)
O
m2
R
(R, 0)

Chapter 9
9.2










21
2211
21
2211
mm
ymym
,
mm
xmxm
x1 = R y1 = 0
x2 = 0 y2 = 0










21
22
2
mm
0
,
T)R2(TR
0RTR











0,
TR5
RTR
2
2
 





0,
5
R

At R/5 from the centre of bigger disc towards the centre of smaller disc.
5. Let ‘0’ be the origin of the system.
R = radius of the smaller disc
2R = radius of the bigger disc
The smaller disc is cut out from the bigger disc
As from the figure
m1 = R
2
T x1 = R y1 = 0
m2 = (2R)
2
T x2 = 0 y2 = 0
The position of C.M. =












21
22
2
mm
00
,
RT)R2(TR
0RTR
=










0,
TR3
RTR
2
2
= 





 0,
3
R
C.M. is at R/3 from the centre of bigger disc away from centre of the hole. 
6. Let m be the mass per unit area.
 Mass of the square plate = M1 = d
2
m
Mass of the circular disc = M2 = m
4
d2

Let the centre of the circular disc be the origin of the system.
 Position of centre of mass
=












21
22
22
MM
00
,
m)4/d(md
0m)4/d(mdd
=

















 

0,
4
1md
md
2
3
= 






0,
4
d4
The new centre of mass is 





 4
d4
right of the centre of circular disc.
7. m1 = 1kg. 1v

= –1.5 cos 37 iˆ – 1.55 sin 37 jˆ = – 1.2 iˆ – 0.9 jˆ
m2 = 1.2kg. 2v

= 0.4 jˆ
m3 = 1.5kg 3v

= – 0.8 iˆ + 0.6 jˆ
m4 = 0.5kg 4v

= 3 iˆ
m5 = 1kg 5v

= 1.6 iˆ – 1.2 jˆ
So, cv

=
54321
5544332211
mmmmm
vmvmvmvmvm



=
2.5
)jˆ2.1iˆ6.1(1)iˆ3(5.0)jˆ6.0iˆ8.0(5.1)jˆ4.0(2.1)jˆ9.0iˆ2.1(1 
=
2.5
jˆ2.1iˆ6.1iˆ5.1jˆ90.iˆ2.1jˆ8.4jˆ9.0iˆ2.1 
=
2.5
jˆ72.0
2.5
iˆ7.0

m1
(0, 0)
O
m2
R
(R, 0)
(x1, y1)
M1dM1
d/2 d/2
d/2O (0, 0)
(d, 0)
(x2, y2)
37°
1.5m/s
1kg 0.4m/s
1.2kg
37°
1m/s
1.5kg
37°
2m/s
1kg
05kg 3m/s

Chapter 9
9.3
8. Two masses m1 & m2 are placed on the X-axis
m1 = 10 kg, m2 = 20kg.
The first mass is displaced by a distance of 2 cm

21
2211
cm
mm
xmxm
X


 =
30
x20210 2
 0 =
30
x2020 2
 20 + 20x2 = 0
 20 = – 20x2  x2 = –1.
 The 2
nd
mass should be displaced by a distance 1cm towards left so as to kept the position of centre
of mass unchanged.
9. Two masses m1 & m2 are kept in a vertical line
m1 = 10kg, m2 = 30kg
The first block is raised through a height of 7 cm.
The centre of mass is raised by 1 cm.
 1 =
21
2211
mm
ymym


=
40
y30710 2
 1 =
40
y3070 2
 70 +30y2 = 40  30y2 = – 30  y2 = –1.
The 30 kg body should be displaced 1cm downward inorder to raise the centre of mass through 1 cm.
10. As the hall is gravity free, after the ice melts, it would tend to acquire a
spherical shape. But, there is no external force acting on the system. So, the
centre of mass of the system would not move.
11. The centre of mass of the blate will be on the symmetrical axis.

2
R
2
R
3
R4
2
R
3
R4
2
R
y 2
1
2
2
1
2
12
2
2
cm
















 













 

=
3
2
1
2
2
1
3
2
)RR(2/
R)3/2(R)3/2(


=
)RR)(RR(
)RRRR)(RR(
3
4
1212
21
2
1
2
212



=
21
21
2
1
2
2
RR
)RRRR(
3
4



above the centre.
12. m1 = 60kg, m2 = 40kg , m3 = 50kg,
Let A be the origin of the system.
Initially Mr. Verma & Mr. Mathur are at extreme position of the boat.
 The centre of mass will be at a distance
=
150
450240060 
=
150
280
= 1.87m from ‘A’
When they come to the mid point of the boat the CM lies at 2m from ‘A’.
 The shift in CM = 2 – 1.87 = 0.13m towards right.
But as there is no external force in longitudinal direction their CM would not shift.
So, the boat moves 0.13m or 13 cm towards right.
13. Let the bob fall at A,. The mass of bob = m.
The mass of cart = M.
Initially their centre of mass will be at
mM
0MLm


= L
mM
m







Distance from P
When, the bob falls in the slot the CM is at a distance ‘O’ from P.
mM
L
R1R2
60kg
A
20kg40kg
B
M
m
A
P

Chapter 9
9.4
Shift in CM = 0 –
mM
mL

= –
mM
mL

towards left
=
mM
mL

towards right.
But there is no external force in horizontal direction.
So the cart displaces a distance
mM
mL

towards right.
14. Initially the monkey & balloon are at rest.
So the CM is at ‘P’
When the monkey descends through a distance ‘L’
The CM will shift
to =
mM
0MLm


=
mM
mL

from P
So, the balloon descends through a distance
mM
mL

15. Let the mass of the to particles be m1 & m2 respectively
m1 = 1kg, m2 = 4kg
According to question
½ m1v1
2
= ½ m2v2
2
 2
1
2
2
2
1
v
v
m
m
 
2
1
1
2
m
m
v
v
 
1
2
2
1
m
m
v
v

Now,
2
1
1
2
2
1
22
11
m
m
m
m
m
m
vm
vm
 =
4
1
= 1/2

22
11
vm
vm
= 1 : 2
16. As uranium 238 nucleus emits a -particle with a speed of 1.4 × 10
7
m/sec. Let v2 be the speed of the
residual nucleus thorium 234.
 m1v1 = m2v2
 4 × 1.4 × 10
7
= 234 × v2
 v2 =
234
104.14 7

= 2.4 × 10
5
m/sec.
17. m1v1 = m2v2
 50 × 1.8 = 6 × 10
24
× v2
 v2 = 24
106
8.150


= 1.5 × 10
–23
m/sec
so, the earth will recoil at a speed of 1.5 × 10
–23
m/sec.
18. Mass of proton = 1.67 × 10
–27
Let ‘Vp’ be the velocity of proton
Given momentum of electron = 1.4 × 10
–26
kg m/sec
Given momentum of antineutrino = 6.4 × 10
–27
kg m/sec
a) The electron & the antineutrino are ejected in the same direction. As the total momentum is
conserved the proton should be ejected in the opposite direction.
1.67 × 10
–27
× Vp = 1.4 × 10
–26
+ 6.4 × 10
–27
= 20.4 × 10
–27
 Vp = (20.4 /1.67) = 12.2 m/sec in the opposite direction.
b) The electron & antineutrino are ejected 
r
to each other.
Total momentum of electron and antineutrino,
= 2722
10)4.6()14( 
 kg m/s = 15.4 × 10
–27
kg m/s
Since, 1.67 × 10
–27
Vp = 15.4 × 10
–27
kg m/s
So Vp = 9.2 m/s
pea
p
a
e
L
M
mg

Chapter 9
9.5
19. Mass of man = M, Initial velocity = 0
Mass of bad = m
Let the throws the bag towards left with a velocity v towards left. So,
there is no external force in the horizontal direction.
The momentum will be conserved. Let he goes right with a velocity
mv = MV  V =
M
mv
 v =
m
MV
..(i)
Let the total time he will take to reach ground = g/H2 = t1
Let the total time he will take to reach the height h = t2 = g/)hH(2 
Then the time of his flying = t1 – t2 = g/H2 – g/)hH(2  =  hHHg/2 
Within this time he reaches the ground in the pond covering a horizontal distance x
 x = V × t  V = x / t
 v =
t
x
m
M
=
 hHH2
g
m
M


As there is no external force in horizontal direction, the x-coordinate of CM will remain at that position.
 0 =
mM
xm)x(M 1


 x1 = x
m
M

 The bag will reach the bottom at a distance (M/m) x towards left of the line it falls.
20. Mass = 50g = 0.05kg
v = 2 cos 45° iˆ – 2 sin 45° jˆ
v1 = – 2 cos 45° iˆ – 2 sin 45° jˆ
a) change in momentum = m v

– m 1v

= 0.05 (2 cos 45° iˆ – 2 sin 45° jˆ ) – 0.05 (– 2 cos 45° iˆ – 2 sin 45° jˆ )
= 0.1 cos 45° iˆ – 0.1 sin 45° jˆ +0.1 cos 45° iˆ + 0.1 sin 45° jˆ
= 0.2 cos 45° iˆ
 magnitude =
2
2
2.0






=
2
2.0
= 0.14 kg m/s
c) The change in magnitude of the momentum of the ball
– iP

– fP

= 2 × 0.5 – 2 × 0.5 = 0.
21. incidenceP

= (h/) cos  iˆ – (h/) sin  jˆ
PReflected = – (h/) cos  iˆ – (h/) sin  jˆ
The change in momentum will be only in the x-axis direction. i.e.
P = (h/) cos  – ((h/) cos ) = (2h/) cos  
22. As the block is exploded only due to its internal energy. So net
external force during this process is 0. So the centre mass will not change.
Let the body while exploded was at the origin of the co-ordinate system.
If the two bodies of equal mass is moving at a speed of 10m/s in + x & +y axis
direction respectively,
o22
90cos10.2101010  = 10 2 m/s 45° w.r.t. + x axis
If the centre mass is at rest, then the third mass which have equal mass with other
two, will move in the opposite direction (i.e. 135° w.r.t. + x- axis) of the resultant at the same velocity.
23. Since the spaceship is removed from any material object & totally isolated from surrounding, the
missions by astronauts couldn’t slip away from the spaceship. So the total mass of the spaceship
remain unchanged and also its velocity.
v
45°
45°
P1 – h/cos 

x
y1
x1
PR – h/cos 
PR – h/
P1 – h/sin = PR
x
y
Hard ground
h
h
pound

Chapter 9
9.6
24. d = 1cm, v = 20 m/s, u = 0,  = 900 kg/m
3
= 0.9gm/cm
3
volume = (4/3)r
3
= (4/3)  (0.5)
3
= 0.5238cm
3
 mass = v = 0.5238 × 0.9 = 0.4714258gm
 mass of 2000 hailstone = 2000 × 0.4714 = 947.857
 Rate of change in momentum per unit area = 947.857 × 2000 = 19N/m
3
 Total force exerted = 19 × 100 = 1900 N.
25. A ball of mass m is dropped onto a floor from a certain height let ‘h’.
 v1 = gh2 , v1 = 0, v2 = gh2 & v2 = 0
 Rate of change of velocity :-
F =
t
gh22m 
 v = gh2 , s = h, v = 0
 v = u + at
 gh2 = g t  t =
g
h2
 Total time
t
h2
2
 F =
g
h2
2
gh22m
= mg
26. A railroad car of mass M is at rest on frictionless rails when a man of mass m starts moving on the car
towards the engine. The car recoils with a speed v backward on the rails.
Let the mass is moving with a velocity x w.r.t. the engine.
The velocity of the mass w.r.t earth is (x – v) towards right
Vcm = 0 (Initially at rest)
 0 = –Mv + m(x – v)
 Mv = m(x – v)  mx = Mv + mv  x = v
m
mM





 
 x = v
m
M
1 






27. A gun is mounted on a railroad car. The mass of the car, the gun, the shells and the operator is 50m
where m is the mass of one shell. The muzzle velocity of the shells is 200m/s.
Initial, Vcm = 0.
 0 = 49 m × V + m × 200  V =
49
200
m/s

49
200
m/s towards left.
When another shell is fired, then the velocity of the car, with respect to the platform is,
 V` =
49
200
m/s towards left.
When another shell is fired, then the velocity of the car, with respect to the platform is,
 v` =
48
200
m/s towards left
 Velocity of the car w.r.t the earth is 






48
200
49
200
m/s towards left.
28. Two persons each of mass m are standing at the two extremes of a railroad car of mass m resting on a
smooth track.
Case – I
Let the velocity of the railroad car w.r.t the earth is V after the jump of the left man.
 0 = – mu + (M + m) V

Chapter 9
9.7
 V =
mM
mu

towards right
Case – II
When the man on the right jumps, the velocity of it w.r.t the car is u.
 0 = mu – Mv’
 v =
M
mu
(V is the change is velocity of the platform when platform itself is taken as reference assuming the car
to be at rest)
 So, net velocity towards left (i.e. the velocity of the car w.r.t. the earth)
=
mM
mv
M
mv

 =
)mM(M
MmuvmmMu 2


=
)mM(M
vm2

29. A small block of mass m which is started with a velocity V on the horizontal part of the bigger block of
mass M placed on a horizontal floor.
Since the small body of mass m is started with a velocity V in the horizontal direction, so the total initial
momentum at the initial position in the horizontal direction will remain same as the total final momentum
at the point A on the bigger block in the horizontal direction.
From L.C.K. m:
mv + M×O = (m + M) v  v =
mM
mv

30. Mass of the boggli = 200kg, VB = 10 km/hour.
 Mass of the boy = 2.5kg & VBoy = 4km/hour.
If we take the boy & boggle as a system then total momentum before the process of sitting will remain
constant after the process of sitting.
 mb Vb = mboyVboy = (mb + mboy) v
 200 × 10 + 25 × 4 = (200 +25) × v
 v =
225
2100
=
3
28
= 9.3 m/sec
31. Mass of the ball = m1 = 0.5kg, velocity of the ball = 5m/s
Mass of the another ball m2 = 1kg
Let it’s velocity = v m/s
Using law of conservation of momentum,
0.5 × 5 + 1 × v = 0  v = – 2.5
 Velocity of second ball is 2.5 m/s opposite to the direction of motion of 1
st
ball.
32. Mass of the man = m1 = 60kg
Speed of the man = v1 = 10m/s
Mass of the skater = m2 = 40kg
let its velocity = v
 60 × 10 + 0 = 100 × v  v = 6m/s
loss in K.E.= (1/2)60 ×(10)
2
– (1/2)× 100 × 36 = 1200 J
33. Using law of conservation of momentum.
m1u1 + m2u2 = m1v(t) + m2v
Where v = speed of 2
nd
particle during collision.
 m1u1 + m2u2 = m1u1 + m1 + (t/∆t)(v1 – u1) + m2v
 v)uv(
t
t
2m
m
m
um
11
1
2
22 


 v = )uv(
t
t
m
m
u 1
2
1
2 


34. Mass of the bullet = m and speed = v
Mass of the ball = M
m = frictional mass from the ball.
U1st
‘–ve’ ‘+ve’
U2nd
vA
m

Chapter 9
9.8
mv + 0 = (m+ m) v + (M – m) v1
where v = final velocity of the bullet + frictional mass
 v =
mm
V)mM(mv 1


35. Mass of 1
st
ball = m and speed = v
Mass of 2
nd
ball = m
Let final velocities of 1
st
and 2
nd
ball are v1 and v2 respectively
m(v1 + v2) = mv.
 v1 + v2 = v …(1)
Also
v1 – v2 = ev …(2)
Given that final K.E. = ¾ Initial K.E.
 ½ mv1
2
+ ½ mv2
2
= ¾ × ½ mv
2
 v1
2
+ v2
2
= ¾ v
2

    2
2
21
2
21
v
4
3
2
vvvv



  2
22
v
4
3
2
ve1


 1 + e
2
=
2
3
 e
2
=
2
1
 e =
2
1
36. Mass of block = 2kg and speed = 2m/s
Mass of 2
nd
block = 2kg.
Let final velocity of 2
nd
block = v
using law of conservation of momentum.
2 × 2 = (2 + 2) v  v = 1m/s
 Loss in K.E. in inelastic collision
= (1/2) × 2 × (2)
2
v – (1/2) (2 + 2) ×(1)
2
= 4 – 2 = 2 J
b) Actual loss =
2
lossMaximum
= 1J
(1/2) × 2 × 22
– (1/2) 2 × v1
2
+ (1/2) × 2× v2
2
= 1
 4 – (v1
2
+ v2
2
) = 1
 1
2
4)e1(
4
2



2(1 + e
2
) =3  1 + e
2
=
2
3
 e
2
=
2
1
 e =
2
1
37. Final K.E. = 0.2J
Initial K.E. = ½ mV1
2
+ 0 = ½ × 0.1 u
2
= 0.05 u
2
mv1 = mv2 = mu
Where v1 and v2 are final velocities of 1
st
and 2
nd
block respectively.
 v1 + v2 = u …(1)
(v1 – v2) + ℓ (a1 – u2) = 0  ℓa = v2 – v1 ..(2)
u2 = 0, u1= u.
Adding Eq.(1) and Eq.(2)
2v2 = (1 + ℓ)u  v2 = (u/2)(1 + ℓ)
 v1 = 
2
u
2
u
u 
v1 =
2
u
(1 – ℓ)
Given (1/2)mv1
2
+(1/2)mv2
2
= 0.2
 v1
2
+ v2
2
= 4
u1
u2 = 0100 g100 g

Chapter 9
9.9
 4)1(
4
u
)1(
4
u 2
2
2
2
   )1(
2
u 2
2
 = 4  u
2
= 2
1
8

For maximum value of u, denominator should be minimum,
 ℓ = 0.
 u
2
= 8  u = 22 m/s
For minimum value of u, denominator should be maximum,
 ℓ = 1
u
2
= 4  u = 2 m/s
38. Two friends A & B (each 40kg) are sitting on a frictionless platform some distance d apart A rolls a ball
of mass 4kg on the platform towards B, which B catches. Then B rolls the ball towards A and A catches
it. The ball keeps on moving back & forth between A and B. The ball has a fixed velocity 5m/s.
a) Case – I :– Total momentum of the man A & the ball will remain constant
 0 = 4 × 5 – 40 × v  v = 0.5 m/s towards left
b) Case – II :– When B catches the ball, the momentum between the B & the
ball will remain constant.
 4 × 5 = 44v  v = (20/44) m/s
Case – III :– When B throws the ball, then applying L.C.L.M
 44 × (20/44) = – 4 × 5 + 40 × v  v = 1m/s (towards right)
Case – IV :– When a Catches the ball, the applying L.C.L.M.
 –4 × 5 + (–0.5)× 40 = – 44v  v =
11
10
m/s towards left.
c) Case – V :– When A throws the ball, then applying L.C.L.M.
 44 × (10/11) = 4 × 5 – 40 × V  V = 60/40 = 3/2 m/s towards left.
Case – VI :– When B receives the ball, then applying L.C.L.M
 40 × 1 + 4 × 5 = 44 × v  v = 60/44 m/s towards right.
Case – VII :– When B throws the ball, then applying L.C.L.M.
 44 × (66/44) = – 4 × 5 + 40 × V  V = 80/40 = 2 m/s towards right.
Case – VIII :– When A catches the ball, then applying L.C.L.M
 – 4 × 5 – 40 × (3/2) = – 44 v  v = (80/44) = (20/11) m/s towards left.
Similarly after 5 round trips
The velocity of A will be (50/11) & velocity of B will be 5 m/s.
d) Since after 6 round trip, the velocity of A is 60/11 i.e.
> 5m/s. So, it can’t catch the ball. So it can only roll the ball six.
e) Let the ball & the body A at the initial position be at origin.
 XC =
44040
d4004040


=
11
10
d
39. u = gh2 = velocity on the ground when ball approaches the ground.
 u = 28.92 
v = velocity of ball when it separates from the ground.
0uv 



 vu

   ℓ =
28.92
5.18.92


=
4
3
=
2
3
40. K.E. of Nucleus = (1/2)mv
2
= (1/2) m
2
mc
E






= 2
2
mc2
E
Energy limited by Gamma photon = E.
Decrease in internal energy = 2
2
mc2
E
E 
B
o
sm/s
d
A
(core -1) ‘+ve’‘–ve’
BA
d
M
V
linear momentum = E/c

Chapter 9
9.10
41. Mass of each block MA and MB = 2kg.
Initial velocity of the 1
st
block, (V) = 1m/s
VA = 1 m/s, VB = 0m/s
Spring constant of the spring = 100 N/m.
The block A strikes the spring with a velocity 1m/s/
After the collision, it’s velocity decreases continuously and at a instant the whole system (Block A + the
compound spring + Block B) move together with a common velocity.
Let that velocity be V.
Using conservation of energy, (1/2) MAVA
2
+ (1/2)MBVB
2
= (1/2)MAv
2
+ (1/2)MBv
2
+ (1/2)kx
2
.
(1/2) × 2(1)
2
+ 0 = (1/2) × 2× v
2
+ (1/2) × 2 × v
2
+ (1/2) x
2
× 100
(Where x = max. compression of spring)
 1 = 2v
2
+ 50x
2
…(1)
As there is no external force in the horizontal direction, the momentum should be conserved.
 MAVA + MBVB = (MA + MB)V.
 2 × 1 = 4 × v
 V = (1/2) m/s. …(2)
Putting in eq.(1)
1 = 2 × (1/4) + 50x+2+
 (1/2) = 50x
2
 x
2
= 1/100m
2
 x = (1/10)m = 0.1m = 10cm.
42. Mass of bullet m = 0.02kg.
Initial velocity of bullet V1 = 500m/s
Mass of block, M = 10kg.
Initial velocity of block u2 = 0.
Final velocity of bullet = 100 m/s = v.
Let the final velocity of block when the bullet emerges out, if block = v.
mv1 + Mu2 = mv + Mv
 0.02 × 500 = 0.02 × 100 + 10 × v
 v = 0.8m/s
After moving a distance 0.2 m it stops.
 change in K.E. = Work done
 0 – (1/2) × 10× (0.8)
2
= – × 10 × 10 × 0.2   =0.16
43. The projected velocity = u.
The angle of projection = .
When the projectile hits the ground for the 1
st
time, the velocity would be the same i.e. u.
Here the component of velocity parallel to ground, u cos should remain constant. But the vertical
component of the projectile undergoes a change after the collision.
 e =
v
sinu 
 v = eu sin .
Now for the 2nd
projectile motion,
U = velocity of projection = 22
)sineu()cosu( 
and Angle of projection =  = tan
–1








cosa
sineu
= tan
–1
(e tan )
or tan  = e tan  …(2)
Because, y = x tan  – 2
22
u2
secgx 
…(3)
Here, y = 0, tan  = e tan , sec
2
 = 1 + e
2
tan
2

And u
2
= u
2
cos
2
 + e
2
sin
2

Putting the above values in the equation (3),
500 m/s

2 m/s
B
2kg
A
2kg
u sin 
u

u
u cos 

Chapter 9
9.11
x e tan  =
)sine(cosu2
)tane1(gx
2222
222


 x =
)tane1(g
)sine(costaneu2
22
2222


 x =
g
costaneu2 22

=
g
2sineu2

 So, from the starting point O, it will fall at a distance
=
g
2sineu
g
2sinu 22



= )e1(
g
2sinu2



44. Angle inclination of the plane = 
M the body falls through a height of h,
The striking velocity of the projectile with the indined plane v = gh2
Now, the projectile makes on angle (90° – 2)
Velocity of projection = u = gh2
Let AB = L.
So, x = ℓ cos , y = – ℓ sin 
From equation of trajectory,
y = x tan  – 2
22
u2
secgx 
– ℓ sin  = ℓ cos  . tan (90° – 2) –
gh22
)290(seccosg o222

 
 – ℓ sin  = ℓ cos  . cot 2 –
gh4
2eccoscosg 222

So,
h4
2eccoscos 22

= sin  + cos  cot 2
 ℓ =
 2eccoscos
h4
22
(sin  + cos  cot 2) = 










2sin
2cos
cossin
cos
2sinh4
2
2
= 









2sin
2coscos2sinsin
cos
cossin4h4
2
22
= 16 h sin
2
 ×


cossin2
cos
= 8h sin 
45. h = 5m,  = 45°, e = (3/4)
Here the velocity with which it would strike = v = 5g2  = 10m/sec
After collision, let it make an angle  with horizontal. The horizontal component of velocity 10 cos 45°
will remain unchanged and the velocity in the perpendicular direction to the plane after wllisine.
 Vy = e × 10 sin 45°
= (3/4) × 10 ×
2
1
= (3.75) 2 m/sec
Vx = 10 cos 45° = 5 2 m/sec
So, u =
2
y
2
x VV  = 125.2850  = 125.78 = 8.83 m/sec
Angle of reflection from the wall  = tan
–1








25
275.3
= tan
–1






4
3
= 37°
 Angle of projection  = 90 – ( + ) = 90 – (45° + 37°) = 8°
Let the distance where it falls = L
 x = L cos , y = – L sin 
Angle of projection () = –8°

A
ℓ





A
ℓ





Chapter 9
9.12
Using equation of trajectory, y = x tan  – 2
22
u2
secgx 
 – ℓ sin  = ℓ cos  × tan 8° – 2
22
u
8seccos
2
g 


 – sin 45° = cos 45° – tan 8° – )(
)83.8(
8sec45cos10
2
2


Solving the above equation we get,
ℓ = 18.5 m.
46. Mass of block
Block of the particle = m = 120gm = 0.12kg.
In the equilibrium condition, the spring is stretched by a distance x = 1.00 cm = 0.01m.
 0.2 × g = K. x.
 2 = K × 0.01  K = 200 N/m.
The velocity with which the particle m will strike M is given by u
= 45.0102  = 9 = 3 m/sec.
So, after the collision, the velocity of the particle and the block is
V =
32.0
312.0 
=
8
9
m/sec.
Let the spring be stretched through an extra deflection of .
0 –(1/2) × 0.32 × (81/64) = 0.32 × 10 ×  – ( 1/2 × 200 × ( + 0.1)
2
– (1/2) × 200 × (0.01)
2
Solving the above equation we get
 = 0.045 = 4.5cm
47. Mass of bullet = 25g = 0.025kg.
Mass of pendulum = 5kg.
The vertical displacement h = 10cm = 0.1m
Let it strike the pendulum with a velocity u.
Let the final velocity be v.
 mu = (M + m)v.
 v = u
)mM(
m

= u
025.5
025.0
 =
201
u
Using conservation of energy.
0 – (1/2) (M + m). V
2
= – (M + m) g × h  2
2
)201(
u
= 2 × 10 × 0.1 = 2
 u = 201 × 2 = 280 m/sec.
48. Mass of bullet = M = 20gm = 0.02kg.
Mass of wooden block M = 500gm = 0.5kg
Velocity of the bullet with which it strikes u = 300 m/sec.
Let the bullet emerges out with velocity V and the velocity of block = V
As per law of conservation of momentum.
mu = Mv+ mv ….(1)
Again applying work – energy principle for the block after the collision,
0 – (1/2) M × V
2
= – Mgh (where h = 0.2m)
V
2
= 2gh
V = gh2 = 2.020  = 2m/sec
Substituting the value of V in the equation (1), we get
0.02 × 300 = 0.5 × 2 + 0.2 × v
 V =
02.0
1.6
= 250m/sec.
x
m
M

Chapter 9
9.13
49. Mass of the two blocks are m1 , m2.
Initially the spring is stretched by x0
Spring constant K.
For the blocks to come to rest again,
Let the distance travelled by m1 & m2
Be x1 and x2 towards right and left respectively.
As o external forc acts in horizontal direction,
m1x1 = m2x2 …(1)
Again, the energy would be conserved in the spring.
 (1/2) k × x
2
= (1/2) k (x1 + x2 – x0)
2
 xo = x1 + x2 – x0
 x1 + x2 = 2x0 …(2)
 x1 = 2x0 – x2 similarly x1 = 0
21
2
x
mm
m2







 m1(2x0 – x2) = m2x2  2m1x0 – m1x2 = m2x2  x2 = 0
21
1
x
mm
m2







50. a)  Velocity of centre of mass =
21
102
mm
0mvm


=
21
02
mm
vm

b) The spring will attain maximum elongation when both velocity of two blocks will attain the velocity of
centre of mass.
d) x maximum elongation of spring.
Change of kinetic energy = Potential stored in spring.
 (1/2) m2 v0
2
– (1/2) (m1 + m2) (
2
21
02
mm
vm







= (1/2) kx
2
 m2 v0
2








21
2
mm
m
1 = kx
2
 x =
2/1
21
21
mm
mm







× v0
51. If both the blocks are pulled by some force, they suddenly move with some acceleration and
instantaneously stop at same position where the elongation of spring is maximum.
 Let x1, x2  extension by block m1 and m2
Total work done = Fx1 + Fx2 …(1)
 Increase the potential energy of spring = (1/2) K (x1+ x2)
2
…(2)
Equating (1) and (2)
F(x1 + x2) = (1/2) K (x1+ x2)
2
 (x1+ x2) =
K
F2
Since the net external force on the two blocks is zero thus same force act on opposite direction.
 m1x1 = m2x2 …(3)
And (x1+ x2) =
K
F2
x2 = 1
m
m
2
1

Substituting
2
1
m
m
× 1 + x1 =
K
F2

K
F2
m
m
1x
2
1
1 





  x1 =
21
2
mm
m
K
F2

Similarly x2 =
21
1
mm
m
K
F2

x x2
m2m1
x1
v0
K
m2m1
FK
m2m1
F

Chapter 9
9.14
52. Acceleration of mass m1 =
21
21
mm
FF


Similarly Acceleration of mass m2 =
21
12
mm
FF


Due to F1 and F2 block of mass m1 and m2 will experience different acceleration and experience an
inertia force.
 Net force on m1 = F1 – m1 a
= F1 – m1 ×
21
21
mm
FF


=
21
12111211
mm
mFFmFmFm


=
21
2112
mm
FmFm


Similarly Net force on m2 = F2 – m2 a
= F2 – m2 ×
21
12
mm
FF


=
21
21222221
mm
mFFmFmFm


=
21
2221
mm
FmFm


 If m1 displaces by a distance x1 and x2 by m2 the maximum extension of the spring is x1 + m2.
 Work done by the blocks = energy stored in the spring.,

21
2112
mm
FmFm


× x1 +
21
2112
mm
FmFm


× x2 = (1/2) K (x1+ x2)
2
 x1+ x2 =
K
2
21
2112
mm
FmFm


53. Mass of the man (Mm) is 50 kg.
Mass of the pillow (Mp) is 5 kg.
When the pillow is pushed by the man, the pillow will go down while the man goes up. It becomes the
external force on the system which is zero.
 acceleration of centre of mass is zero
 velocity of centre of mass is constant
As the initial velocity of the system is zero.
 Mm × Vm = Mp × Vp …(1)
Given the velocity of pillow is 80 ft/s.
Which is relative velocity of pillow w.r.t. man.
m/pV

= mp VV

 = Vp – (–Vm) = Vp + Vm  Vp = m/pV – Vm
Putting in equation (1)
Mm × Vm = Mp ( m/pV – Vm)
 50 × Vm = 5 × (8 – Vm)
 10 × Vm = 8 – Vm  Vm =
11
8
= 0.727m/s
 Absolute velocity of pillow = 8 – 0.727 = 7.2 ft/sec.
 Time taken to reach the floor =
v
S
=
2.7
8
= 1.1 sec.
As the mass of wall >>> then pillow
The velocity of block before the collision = velocity after the collision.
 Times of ascent = 1.11 sec.
 Total time taken = 1.11 + 1.11 = 2.22 sec.
54. Let the velocity of A = u1.
Let the final velocity when reaching at B becomes collision = v1.
 (1/2) mv1
2
– (1/2)mu1
2
= mgh
 v1
2
– u1
2
= 2 gh  v1 =
2
1ugh2  …(1)
When the block B reached at the upper man’s head, the velocity of B is just zero.
For B, block
 (1/2) × 2m × 0
2
– (1/2) × 2m × v
2
= mgh  v = gh2
F2
K
m2m1
F1
pillow
B
A
h h

Chapter 9
9.15
 Before collision velocity of uA = v1, uB = 0.
After collision velocity of vA = v (say) vB = gh2
Since it is an elastic collision the momentum and K.E. should be coserved.
 m × v1 + 2m × 0 = m × v + 2m × gh2
 v1 – v = 2 gh2
Also, (1/2) × m × v1
2
+ (1/2) I 2m × 0
2
= (1/2) × m × v
2
+ (1/2) ×2m ×  2
gh2
 v1
2
– v
2
= 2 × gh2 × gh2 …(2)
Dividing (1) by (2)
)vv(
)vv)(vv(
1
11


=
gh22
gh2gh22


 v1 + v = gh2 …(3)
Adding (1) and (3)
2v1 = 3 gh2  v1 = 





2
3
gh2
But v1 = 2
ugh2  = 





2
3
gh2
 2gh + u
2
= gh2
4
9

 u = 2.5 gh2
So the block will travel with a velocity greater than 2.5 gh2 so awake the man by B.
55. Mass of block = 490 gm.
Mass of bullet = 10 gm.
Since the bullet embedded inside the block, it is an plastic collision.
Initial velocity of bullet v1 = 50 7 m/s.
Velocity of the block is v2 = 0.
Let Final velocity of both = v.
 10 × 10
–3
× 50 × 7 + 10
–3
× 190 I 0 = (490 + 10) × 10
–3
× VA
 VA = 7 m/s.
When the block losses the contact at ‘D’ the component mg will act on it.
 
r
Vm
2
B
= mg sin   (VB)
2
= gr sin  …(1)
Puttin work energy principle
(1/2) m × (VB)
2
– (1/2) × m × (VA)
2
= – mg (0.2 + 0.2 sin )
 (1/2) × gr sin  – (1/2) ×  2
7 = – mg (0.2 + 0.2 sin )
 3.5 – (1/2) × 9.8 × 0.2 × sin  = 9.8 × 0.2 (1 + sin )
 3.5 – 0.98 sin  = 1.96 + 1.96 sin 
 sin  = (1/2)   = 30°
 Angle of projection = 90° - 30° = 60°.
 time of reaching the ground =
g
h2
=
8.9
)30sin2.02.0(2 
= 0.247 sec.
 Distance travelled in horizontal direction.
s = V cos  × t = tsingr  = 247.0)2/1(28.9  = 0.196m
 Total distance = (0.2 – 0.2 cos 30°) + 0.196 = 0.22m.
D 90°- 
490gm

10gm

C
BO
MVB
2
/r

Chapter 9
9.16
56. Let the velocity of m reaching at lower end = V1
From work energy principle.
 (1/2) × m × V1
2
– (1/2) × m × 0
2
= mg ℓ
 v1 = g2 .
Similarly velocity of heavy block will be v2 = gh2 .
 v1 = V2 = u(say)
Let the final velocity of m and 2m v1 and v2 respectively.
According to law of conservation of momentum.
m × x1 + 2m × V2 = mv1 + 2mv2
 m × u – 2 m u = mv1 + 2mv2
 v1 + 2v2 = – u …(1)
Again, v1 – v2 = – (V1 – V2)
 v1 – v2 = – [u – (–v)] = – 2V …(2)
Subtracting.
3v2 = u  v2 =
3
u
=
3
g2 
Substituting in (2)
v1 – v2 = - 2u  v1 = – 2u + v2 = –2u +
3
u
= -
3
5
u = – g2
3
5
 = –
3
g50 
b) Putting the work energy principle
(1/2) × 2m × 0
2
– (1/2) × 2m × (v2)
2
= – 2m × g × h
[ h  height gone by heavy ball]
 (1/2)
9
g2
= ℓ × h  h =
9

Similarly, (1/2) × m × 0
2
– (1/2) × m × v1
2
= m × g × h2
[ height reached by small ball]
 (1/2) ×
9
g50 
= g × h2  h2 =
9
25
Someh2 is more than 2ℓ, the velocity at height point will not be zero. And the ‘m’ will rise by a distance 2ℓ.
57. Let us consider a small element at a distance ‘x’ from the floor of length ‘dy’ .
So, dm =
L
M
dx
So, the velocity with which the element will strike the floor is, v = gx2
 So, the momentum transferred to the floor is,
M = (dm)v = gx2dx
L
M
 [because the element comes to rest]
So, the force exerted on the floor change in momentum is given by,
F1 =
dt
dM
= gx2
dt
dx
L
M

Because, v =
dt
dx
= gx2 (for the chain element)
F1 = gx2gx2
L
M
 = gx2
L
M
 =
L
Mgx2
Again, the force exerted due to ‘x’ length of the chain on the floor due to its own weight is given by,
W = g)x(
L
M
 =
L
Mgx
So, the total forced exerted is given by,
F = F1 + W =
L
Mgx
L
Mgx2
 =
L
Mgx3
m m
(Initial position)
L
M
x
dx

Chapter 9
9.17
58. V1 = 10 m/s V2 = 0
V1, v2  velocity of ACB after collision.
a) If the edlision is perfectly elastic.
mV1 + mV2 = mv1 + mv2
 10 + 0 = v1 + v2
 v1 + v2 = 10 …(1)
Again, v1 – v2 = – (u1 – v2) = – (10 – 0) = –10 …(2)
Subtracting (2) from (1)
2v2 = 20  v2 = 10 m/s.
The deacceleration of B = g
Putting work energy principle
 (1/2) × m × 0
2
– (1/2) × m × v2
2
= – m × a × h
 – (1/2) × 10
2
= -  g × h  h =
101.02
100

= 50m
b) If the collision perfectly in elastic.
m × u1 + m × u2 = (m + m) × v
 m × 10 + m × 0 = 2m × v  v =
2
10
= 5 m/s.
The two blocks will move together sticking to each other.
 Putting work energy principle.
(1/2) × 2m × 0
2
– (1/2) × 2m × v
2
= 2m ×  g × s

2101.0
52

= s  s = 12.5 m.
59. Let velocity of 2kg block on reaching the 4kg block before collision =u1.
Given, V2 = 0 (velocity of 4kg block).
 From work energy principle,
(1/2) m × u1
2
– (1/2) m × 1
2
= – m × ug × s

2
1u
2
1 
= – 2 × 5  – 16 =
4
1u
2
1 
 64 × 10
–2
= u1
2
– 1  u1 = 6m/s
Since it is a perfectly elastic collision.
Let V1, V2  velocity of 2kg & 4kg block after collision.
m1V1 + m2V2 = m1v1 + m2v2
 2 × 0.6 + 4 × 0 = 2v1 + 4 v2  v1 + 2v2 = 0.6 …(1)
Again, V1 – V2 = – (u1 – u2) = – (0.6 – 0) = –0.6 …(2)
Subtracting (2) from (1)
3v2 = 1.2  v2 = 0.4 m/s.
 v1 = – 0.6 + 0.4 = – 0.2 m/s
 Putting work energy principle for 1
st
2kg block when come to rest.
(1/2) × 2 × 0
2
– (1/2) × 2 × (0.2)
2
= – 2 × 0.2 × 10 × s
 (1/2) × 2 × 0.2 × 0.2 = 2 × 0.2 × 10 × s  S1 = 1cm.
Putting work energy principle for 4kg block.
(1/2) × 4 × 0
2
– (1/2) × 4 × (0.4)
2
= – 4 × 0.2 × 10 × s
 2 × 0.4 × 0.4 = 4 × 0.2 × 10 × s  S2 = 4 cm.
Distance between 2kg & 4kg block = S1 + S2 = 1 + 4 = 5 cm.
60. The block ‘m’ will slide down the inclined plane of mass M with acceleration a1 g sin  (relative) to the
inclined plane.
The horizontal component of a1 will be, ax = g sin  cos , for which the block M will accelerate towards
left. Let, the acceleration be a2.
According to the concept of centre of mass, (in the horizontal direction external force is zero).
max = (M + m) a2
10 m/sm
A B
u = 0.1
u = 0.2
1 m/s
16cm
2kg 4kg
u = 0.2

Chapter 9
9.18
 a2 =
mM
max

=
mM
cossinmg


…(1)
So, the absolute (Resultant) acceleration of ‘m’ on the block ‘M’ along the direction of the incline will be,
a = g sin  - a2 cos 
= g sin  –
mM
cossinmg 2


= g sin 











mM
cosm
1
2
= g sin 










mM
cosmmM 2
So, a = g sin 










mM
sinmM 2
…(2)
Let, the time taken by the block ‘m’ to reach the bottom end be ‘t’.
Now, S = ut + (1/2) at
2

sin
h
= (1/2) at
2
 t =
sina
2
So, the velocity of the bigger block after time ‘t’ will be.
Vm = u + a2t =


sina
h2
mM
cossinmg
=


sina)mM(
cossinhgm2
2
2222
Now, subtracting the value of a from equation (2) we get,
VM =
2/1
22
2222
)sinmM(sing
)mM(
sin)mM(
cossinhgm2













or VM =
2/1
2
222
)sinmM)(mM(
coshgm2











61.
The mass ‘m’ is given a velocity ‘v’ over the larger mass M.
a) When the smaller block is travelling on the vertical part, let the velocity of the bigger block be v1
towards left.
From law of conservation of momentum, (in the horizontal direction)
mv = (M + m) v1
 v1 =
mM
mv

b) When the smaller block breaks off, let its resultant velocity is v2.
From law of conservation of energy,
(1/2) mv
2
= (1/2) Mv1
2
+ (1/2) mv2
2
+ mgh
 v2
2
= v
2
–
m
M
v1
2
– 2gh ..(1)
 v2
2 =









 2
2
2
)mM(
m
m
M
1v – 2gh
 v2 =
2/1
2
2
22
gh2v
)mM(
)mMmm(











g sin 
a2
h
M
m
m
a2

g sin 
a2 = g sin  cos 


m
h
M A
v
vyv2
B
C
v1
m
h1
v1

Chapter 9
9.19
e) Now, the vertical component of the velocity v2 of mass ‘m’ is given by,
vy
2
= v2
2
– v1
2
= 2
22
2
2
22
)mM(
vm
gh2v
)mM(
)mMmM(




[ v1 =
vM
mv

]
 vy
2
= gh2v
)mM(
mmMmM 2
2
222



 vy
2
= gh2
)mM(
Mv2


…(2)
To find the maximum height (from the ground), let us assume the body rises to a height ‘h’, over and
above ‘h’.
Now, (1/2)mvy
2
= mgh1  h1 =
g2
v 2
y
…(3)
So, Total height = h + h1 = h +
g2
v 2
y
= h +
g2)mM(
mv2

– h
[from equation (2) and (3)]
 H =
g2)mM(
mv2

d) Because, the smaller mass has also got a horizontal component of velocity ‘v1’ at the time it breaks
off from ‘M’ (which has a velocity v1), the block ‘m’ will again land on the block ‘M’ (bigger one).
Let us find out the time of flight of block ‘m’ after it breaks off.
During the upward motion (BC),
0 = vy – gt1
 t1 =
g
vy
=
2/1
2
gh2
)mM(
Mv
g
1










…(4) [from equation (2)]
So, the time for which the smaller block was in its flight is given by,
T = 2t1 =
2/1
2
)mM(
gh)mM(2Mv
g
2










So, the distance travelled by the bigger block during this time is,
S = v1T = 2/1
2/12
)mM(
]gh)mM(2Mv[
g
2
mM
mv




or S = 2/3
2/12
)mM(g
]gh)mM(2Mv[mv2


62. Given h < < < R.
Gmass = 6 I 10
24
kg.
Mb = 3 × 10
24
kg.
Let Ve  Velocity of earth
Vb  velocity of the block.
The two blocks are attracted by gravitational force of attraction. The gravitation potential energy stored
will be the K.E. of two blocks.








 hR
1
)2/h(R
1
Gpim
= (1/2) me × ve
2
+ (1/2) mb × vb
2
Again as the an internal force acts.
MeVe = mbVb  Ve =
e
bb
M
Vm
…(2)

Chapter 9
9.20
Putting in equation (1)
Gme × mb 







 hR
1
hR2
2
= (1/2) × Me × 2
e
2
b
2
b
M
vm
× ve
2
+ (1/2) Mb × Vb
2
= (1/2) × mb × Vb
2






1
M
M
e
b
 GM 







)hR)(hR2(
hR2h2R2
= (1/2) × Vb
2
×











1
106
103
24
24
 







22
hRh3R2
hGM
= (1/2) × Vb
2
×(3/2)
As h < < < R, if can be neglected
 2
R2
hGM
= (1/2) × Vb
2
×(3/2)  Vb =
3
gh2
63. Since it is not an head on collision, the two bodies move in different dimensions. Let V1, V2  velocities
of the bodies vector collision. Since, the collision is elastic. Applying law of conservation of momentum
on X-direction.
mu1 + mxo = mv1 cos  + mv2 cos 
 v1 cos a + v2 cos b = u1 …(1)
Putting law of conservation of momentum in y direction.
0 = mv1 sin  – mv2 sin   
 v1 sin  = v2 sin  …(2)
Again ½ m u1
2
+ 0 = ½ m v1
2
+ ½ m x v2
2
u1
2
= v1
2
+ v2
2
…(3)
Squaring equation(1)
u1
2
= v1
2
cos
2
 + v2
2
cos
2
 + 2 v1v2 cos  cos 
Equating (1) & (3)
v1
2
+ v2
2
= v1
2
cos
2
 + v2
2
cos
2
 + 2 v1v2 cos  cos 
 v1
2
sin
+ v2
2
sin
2
 = 
2 v1v2 cos  cos 
 2v1
2
sin
2
 = 2 × v1 ×


sin
sinv1
× cos  cos 
 sin  sin  = cos  cos   cos  cos  – sin  sin  = 0
 cos ( + ) = 0 = cos 90°  ( +  = 90°
64.
Let the mass of both the particle and the spherical body be ‘m’. The particle velocity ‘v’ has two
components, v cos  normal to the sphere and v sin  tangential to the sphere.
After the collision, they will exchange their velocities. So, the spherical body will have a velocity v cos 
and the particle will not have any component of velocity in this direction.
[The collision will due to the component v cos  in the normal direction. But, the tangential velocity, of
the particle v sin  will be unaffected]
So, velocity of the sphere = v cos  = 22
r
r
v
 [from (fig-2)]
And velocity of the particle = v sin  =
r
v
* * * * *
BlockEarth
m = 6 × 1024
m = 3 × 1024
h
R
m X
v1
u2 =
0
v2
u1
m 


v cos 
r
v 

v sin 
r


22
r 

10.1
CHAPTER – 10
1. 0 = 0 ;  = 100 rev/s ;  = 2 ;  = 200  rad/s
  = 0 = t
  = t
  = (200 )/4 = 50  rad /s
2
or 25 rev/s
2
  = 0t + 1/2 t
2
= 8 × 50  = 400  rad
  = 50  rad/s
2
or 25 rev/s
s
 = 400  rad.
2.  = 100 ; t = 5 sec
 = 1/2 t
2
 100 = 1/2  25
  = 8 × 5 = 40  rad/s = 20 rev/s
  = 8 rad/s
2
= 4 rev/s
2
 = 40 rad/s
2
= 20 rev/s
2
.
3. Area under the curve will decide the total angle rotated
 maximum angular velocity = 4 × 10 = 40 rad/s
Therefore, area under the curve = 1/2 × 10 × 40 + 40 × 10 + 1/2 × 40 × 10
= 800 rad
 Total angle rotated = 800 rad.
4.  = 1 rad/s
2
, 0 = 5 rad/s;  = 15 rad/s
 w = w0 + t
 t = ( – 0)/ = (15 – 5)/1 = 10 sec
Also,  = 0t + 1/2 t
2
= 5 ×10 + 1/2 × 1 × 100 = 100 rad.
5.  = 5 rev,  = 2 rev/s
2
, 0 = 0 ;  = ?

2
= (2  )
  = 522  = 52 rev/s.
or  = 10 rad,  = 4 rad/s
2
, 0 = 0,  = ?
 = 2 = 2 × 4 × 10
= 54 rad/s = 52 rev/s.
6. A disc of radius = 10 cm = 0.1 m
Angular velocity = 20 rad/s
 Linear velocity on the rim = r = 20 × 0.1 = 2 m/s
 Linear velocity at the middle of radius = r/2 = 20 × (0.1)/2 = 1 m/s.
7. t = 1 sec, r = 1 cm = 0.01 m
 = 4 rd/s
2
Therefore  = t = 4 rad/s
Therefore radial acceleration,
An = 
2
r = 0.16 m/s
2
= 16 cm/s
2
Therefore tangential acceleration, ar = r = 0.04 m/s
2
= 4 cm/s
2
.
8. The Block is moving the rim of the pulley
The pulley is moving at a  = 10 rad/s
Therefore the radius of the pulley = 20 cm
Therefore linear velocity on the rim = tangential velocity = r
= 20 × 20 = 200 cm/s = 2 m/s.
T
10
10 20

Chapter-10
10.2
9. Therefore, the  distance from the axis (AD) = 35102/3  cm.
Therefore moment of inertia about the axis BC will be
I = mr
2
= 200 K 2
)35( = 200 × 25 × 3
= 15000 gm – cm
2
= 1.5 × 10
–3
kg – m
2
.
b) The axis of rotation let pass through A and  to the plane of triangle
Therefore the torque will be produced by mass B and C
Therefore net moment of inertia = I = mr
2
+ mr
2
= 2 × 200 ×10
2
= 40000 gm-cm
2
= 4 ×10
–3
kg-m
2
.
10. Masses of 1 gm, 2 gm ……100 gm are kept at the marks 1 cm, 2 cm, ……1000 cm on he x axis
respectively. A perpendicular axis is passed at the 50
th
particle.
Therefore on the L.H.S. side of the axis there will be 49 particles and on
the R.H.S. side there are 50 particles.
Consider the two particles at the position 49 cm and 51 cm.
Moment inertial due to these two particle will be =
49 × 1
2
+ 51 + 1
2
= 100 gm-cm
2
Similarly if we consider 48
th
and 52
nd
term we will get 100 ×2
2
gm-cm
2
Therefore we will get 49 such set and one lone particle at 100 cm.
Therefore total moment of inertia =
100 {1
2
+ 2
2
+ 3
2
+ … + 49
2
} + 100(50)
2
.
= 100 × (50 × 51 × 101)/6 = 4292500 gm-cm
2
= 0.429 kg-m
2
= 0.43 kg-m
2
.
11. The two bodies of mass m and radius r are moving along the common tangent.
Therefore moment of inertia of the first body about XY tangent.
= mr
2
+ 2/5 mr
2
– Moment of inertia of the second body XY tangent = mr
2
+ 2/5 mr
2
= 7/5 mr
2
Therefore, net moment of inertia = 7/5 mr
2
+ 7/5 mr
2
= 14/5 mr
2
units.
12. Length of the rod = 1 m, mass of the rod = 0.5 kg
Let at a distance d from the center the rod is moving
Applying parallel axis theorem :
The moment of inertial about that point
 (mL
2
/ 12) + md
2
= 0.10
 (0.5 × 1
2
)/12 + 0.5 × d
2
= 0.10
 d
2
= 0.2 – 0.082 = 0.118
 d = 0.342 m from the centre.
13. Moment of inertia at the centre and perpendicular to the plane of the ring.
So, about a point on the rim of the ring and the axis  to the plane of the ring, the
moment of inertia
= mR
2
+ mR
2
= 2mR
2
(parallel axis theorem)
 mK
2
= 2mR
2
(K = radius of the gyration)
 K = R2R2 2
 .
14. The moment of inertia about the center and  to the plane of the disc of
radius r and mass m is = mr
2
.
According to the question the radius of gyration of the disc about a point =
radius of the disc.
Therefore mk
2
= ½ mr
2
+ md
2
(K = radius of gyration about acceleration point, d = distance of that point
from the centre)
 K
2
= r
2
/2 + d
2
 r
2
= r
2
/2 + d
2
( K = r)
 r
2
/2 = d
2
 d = 2/r .
CB
A
I1
D
1 10 20 30 40 50 60 70 80 90 100
48 49 51 52
1
x
2
y
r r
1m
ml2
/12
d
(ml2
/12)+md2
R
1/2 mr2
r
k
d
1/2 mr2
+md2

Chapter-10
10.3
15. Let a small cross sectional area is at a distance x from xx axis.
Therefore mass of that small section = m/a
2
× ax dx
Therefore moment of inertia about xx axis
= Ixx = 2/a
0
3
2/a
0
22
)]3/x)(a/m(2(x)adx()a/m(2 
= ma
2
/ 12
Therefore Ixx = Ixx + Iyy
= 2 × *ma
2
/12)= ma
2
/6
Since the two diagonals are  to each other
Therefore Izz = Ix’x’ + Iy’y’
 ma
2
/6 = 2 × Ix’x’ ( because Ix’x’ = Iy’y’)  Ix’x’ = ma
2
/12
16. The surface density of a circular disc of radius a depends upon the distance from the centre as
P(r) = A + Br
Therefore the mass of the ring of radius r will be
 = (A + Br) × 2r dr × r
2
Therefore moment of inertia about the centre will be
=  
a
0
4
a
0
3
a
0
drBr2drAr2drr2)BrA(
= 2A (r
4
/4) + 2 B(r
5
/5) a
0] = 2a
4
[(A/4) + (Ba/5)].
17. At the highest point total force acting on the particle id its weight acting downward.
Range of the particle = u
2
sin 2 / g
Therefore force is at a  distance,  (total range)/2 = (v
2
sin 2)/2g
(From the initial point)
Therefore  = F × r (  = angle of projection)
= mg × v
2
sin 2/2g (v = initial velocity)
= mv2
sin 2 / 2 = mv
2
sin  cos .
18. A simple of pendulum of length l is suspended from a rigid support. A bob of weight W is hanging on the
other point.
When the bob is at an angle  with the vertical, then total torque acting on the point of
suspension = i = F × r
 W r sin  = W l sin 
At the lowest point of suspension the torque will be zero as the force acting on the body
passes through the point of suspension.
19. A force of 6 N acting at an angle of 30° is just able to loosen the wrench at a distance 8 cm from it.
Therefore total torque acting at A about the point 0
= 6 sin 30° × (8/100)
Therefore total torque required at B about the point 0
= F × 16/100  F × 16/100 = 6 sin 30° × 8/100
 F = (8 × 3) / 16 = 1.5 N.
20. Torque about a point = Total force × perpendicular distance from the point to that force.
Let anticlockwise torque = + ve
And clockwise acting torque = –ve
Force acting at the point B is 15 N
Therefore torque at O due to this force
= 15 × 6 × 10
–2
× sin 37°
= 15 × 6 × 10
–2
× 3/5 = 0.54 N-m (anticlock wise)
Force acting at the point C is 10 N
Therefore, torque at O due to this force
= 10 × 4 × 10
–2
= 0.4 N-m (clockwise)
Force acting at the point A is 20 N
Therefore, Torque at O due to this force = 20 × 4 × 10
–2
× sin30°
= 20 × 4 × 10
–2
× 1/2 = 0.4 N-m (anticlockwise)
Therefore resultant torque acting at ‘O’ = 0.54 – 0.4 + 0.4 = 0.54 N-m.
y
O
y
x
D C
BA
x
x
x y
y
dxr

(v2
sin2) /2
I
B

A
A
10N
4cm
E B
C
18/5
30°
6cm
3cm
4cm
15N
2cm
5N
20N
30°

Chapter-10
10.4
21. The force mg acting on the body has two components mg sin  and mg cos 
and the body will exert a normal reaction. Let R =
Since R and mg cos  pass through the centre of the cube, there will be no torque
due to R and mg cos . The only torque will be produced by mg sin .
 i = F × r (r = a/2) (a = ages of the cube)
 i = mg sin  × a/2
= 1/2 mg a sin .
22. A rod of mass m and length L, lying horizontally, is free to rotate about a vertical axis passing through its
centre.
A force F is acting perpendicular to the rod at a distance L/4 from the centre.
Therefore torque about the centre due to this force
ii = F × r = FL/4.
This torque will produce a angular acceleration .
Therefore c = Ic × 
 ic = (mL
2
/ 12) ×  (Ic of a rod = mL
2
/ 12)
 F i/4 = (mL
2
/ 12) ×    = 3F/ml
Therefore  = 1/2 t
2
(initially at rest)
  = 1/2 × (3F / ml)t
2
= (3F/2ml)t
2
.
23. A square plate of mass 120 gm and edge 5 cm rotates about one of the edge.
Let take a small area of the square of width dx and length a which is at a distance x from the axis of
rotation.
Therefore mass of that small area
m/a
2
× a dx (m = mass of the square ; a = side of the plate)
I = a
0
3
a
0
22
)]3/x)(a/m(dxax)a/m( 
= ma
2
/3
Therefore torque produced = I ×  = (ma
2
/3) × 
= {(120 × 10
–3
× 5
2
× 10
–4
)/3} 0.2
= 0.2 × 10
–4
= 2 × 10
–5
N-m.
24. Moment of inertial of a square plate about its diagonal is ma
2
/12 (m = mass of the square plate)
a = edges of the square
Therefore torque produced = (ma
2
/12) × 
= {(120 × 10
–3
× 5
2
× 10
–4
)/12 × 0.2
= 0.5 × 10
–5
N-m.
25. A flywheel of moment of inertia 5 kg m is rotated at a speed of 60 rad/s. The flywheel comes to rest due
to the friction at the axle after 5 minutes.
Therefore, the angular deceleration produced due to frictional force =  = 0 + t
 0 = –t ( = 0+
  = –(60/5 × 60) = –1/5 rad/s
2
.
a) Therefore total workdone in stopping the wheel by frictional force
W = 1/2 i
2
= 1/2 × 5 × (60 × 60) = 9000 Joule = 9 KJ.
b) Therefore torque produced by the frictional force (R) is
IR = I ×  = 5 × (–1/5) = IN – m opposite to the rotation of wheel.
c) Angular velocity after 4 minutes
  = 0 + t = 60 – 240/5 = 12 rad/s
Therefore angular momentum about the centre = 1 ×  = 5 × 12 = 60 kg-m
2
/s.
mg sin

mg cos
mg sin
R
mg cos

A
B
B
t =sec
A
F
1/4
A
x
dx
xx
R1/2iW2

Chapter-10
10.5
26. The earth’s angular speed decreases by 0.0016 rad/day in 100 years.
Therefore the torque produced by the ocean water in decreasing earth’s angular velocity
 = I
= 2/5 mr
2
× ( – 0)/t
= 2/6 × 6 × 10
24
× 64
2
× 10
10
× [0.0016 /(26400
2
× 100 × 365)] (1 year = 365 days= 365 × 56400 sec)
= 5.678 × 10
20
N-m.
27. A wheel rotating at a speed of 600 rpm.
0 = 600 rpm = 10 revolutions per second.
T = 10 sec. (In 10 sec. it comes to rest)
 = 0
Therefore 0 = –t
  = –10/10 = –1 rev/s
2
  = 0 + t = 10 – 1 × 5 = 5 rev/s.
Therefore angular deacceleration = 1 rev/s
2
and angular velocity of after 5 sec is 5 rev/s.
28.  = 100 rev/min = 5/8 rev/s = 10/3 rad/s
 = 10 rev = 20  rad, r = 0.2 m
After 10 revolutions the wheel will come to rest by a tangential force
Therefore the angular deacceleration produced by the force =  = 
2
/2
Therefore the torque by which the wheel will come to an rest = Icm × 
 F × r = Icm ×   F × 0.2 = 1/2 mr
2
× [(10/3)
2
/ (2 × 20)]
 F = 1/2 × 10 × 0.2 × 100 
2
/ (9 × 2 × 20)
= 5 / 18 = 15.7/18 = 0.87 N.
29. A cylinder is moving with an angular velocity 50 rev/s brought in contact with another identical cylinder
in rest. The first and second cylinder has common acceleration and deacceleration as 1 rad/s
2
respectively.
Let after t sec their angular velocity will be same ‘’.
For the first cylinder  = 50 – t
 t = ( – 50)/–1
And for the 2
nd
cylinder  = 2t
 t = /I
So,  = ( – 50)/–1
 2 = 50   = 25 rev/s.
 t = 25/1 sec = 25 sec.
30. Initial angular velocity = 20 rad/s
Therefore  = 2 rad/s
2
 t1 = /1 = 20/2 = 10 sec
Therefore 10 sec it will come to rest.
Since the same torque is continues to act on the body it will produce same angular acceleration and
since the initial kinetic energy = the kinetic energy at a instant.
So initial angular velocity = angular velocity at that instant
Therefore time require to come to that angular velocity,
t2 = 2/2 = 20/2 = 10 sec
therefore time required = t1 + t2 = 20 sec.
31. Inet = Inet × 
 F1r1 – F2r2 = )rmrm( 2
22
2
11  ×  – 2 × 10 × 0.5
 5 × 10 × 0.5 = (5 × (1/2)
2
+ 2 × (1/2)
2
) × 
 15 = 7/4 
  = 60/7 = 8.57 rad/s
2
.
32. In this problem the rod has a mass 1 kg
a) net = Inet × 
 5 × 10 × 10.5 – 2 × 10 × 0.5
= (5 × (1/2)
2
+ 2 × (1/2)
2
+ 1/12) × 
50 rev/s
2 kg 5 kg
2 kg 5 kg

Chapter-10
10.6
 15 = (1.75 + 0.084) 
  = 1500/(175 + 8.4) = 1500/183.4 = 8.1 rad/s
2
(g = 10)
= 8.01 rad/s
2
(if g = 9.8)
b) T1 – m1g = m1a
 T1 = m1a + m1g = 2(a + g)
= 2(r + g) = 2(8 × 0.5 + 9.8)
= 27.6 N on the first body.
In the second body
 m2g – T2 = m2a  T2 = m2g – m2a
 T2 = 5(g – a) = 5(9.8 – 8 × 0.5) = 29 N.
33. According to the question
Mg – T1 = Ma …(1)
T2 = ma …(2)
(T1 – T2) = 1 a/r
2
…(3) [because a = r]…[T.r =I(a/r)]
If we add the equation 1 and 2 we will get
Mg + (T2 – T1) = Ma + ma …(4)
 Mg – Ia/r
2
= Ma + ma
 (M + m + I/r
2
)a = Mg
 a = Mg/(M + m + I/r
2
)
34. I = 0.20 kg-m
2
(Bigger pulley)
r = 10 cm = 0.1 m, smaller pulley is light
mass of the block, m = 2 kg
therefore mg – T = ma …(1)
 T = Ia/r
2
…(2)
 mg = (m + I/r
2
)a =>(2 × 9.8) / [2 + (0.2/0.01)]=a
= 19.6 / 22 = 0.89 m/s
2
Therefore, acceleration of the block = 0.89 m/s
2
.
35. m = 2 kg, i1 = 0.10 kg-m
2
, r1 = 5 cm = 0.05 m
i2 = 0.20 kg-m
2
, r2 = 10 cm = 0.1 m
Therefore mg – T1 = ma …(1)
(T1 – T2)r1 = I1 …(2)
T2r2 = I2 …(3)
Substituting the value of T2 in the equation (2), we get
 (t1 – I2 /r1)r2 = I1
 (T1 – I2 a /r1
2
) = I1a/r2
2
 T1 = [(I1/r1
2
) + I2/r2
2
)]a
Substituting the value of T1 in the equation (1), we get
 mg – [(I1/r1
2
) + I2/r2
2
)]a = ma
 a
m)]r/I()r/I[(
mg
2
22
2
11


 a =
2)01.0/2.0()0025.0/1.0(
8.92


= 0.316 m/s
2
 T2 = I2a/r2
2
=
01.0
316.020.0 
= 6.32 N.
Mg – T1 = Ma …(1)
(T2 – T1)R = Ia/R  (T2 – T1) = Ia/R
2
…(2)
(T2 – T3)R = Ia/R
2
…(3)
 T3 – mg = ma …(4)
By adding equation (2) and (3) we will get,
 (T1 – T3) = 2 Ia/R
2
…(5)
By adding equation (1) and (4) we will get
m1g
T1
2 kg 5 kg
T2
m2g
m
Mg
M
T1 T2
T1
T2
mg
2kg
T
T 10cm
mg
2kg
T1
T
I2
r2
T2
T2
T1
mg
M
T1
a
T2 T2
T1
mg
T3
T3

Chapter-10
10.7
– mg + Mg + (T3 – T1) = Ma + ma …(6)
Substituting the value for T3 – T1 we will get
 Mg – mg = Ma + ma + 2Ia/R
2
 a =
)R/I2mM(
G)mM(
2


37. A is light pulley and B is the descending pulley having I = 0.20 kg – m
2
and r = 0.2 m
Mass of the block = 1 kg
According to the equation
T1 = m1a …(1)
(T2 – T1)r = I …(2)
m2g – m2 a/2 = T1 + T2 …(3)
T2 – T1 = Ia/2R
2
= 5a/2 and T1 = a (because  = a/2R)
 T2 = 7/2 a
 m2g = m2a/2 + 7/2 a + a
 2I / r
2
g = 2I/r
2
a/2 + 9/2 a (1/2 mr
2
= I)
 98 = 5a + 4.5 a
 a = 98/9.5 = 10.3 ms
2
38. m1g sin  – T1 = m1a …(1)
(T1 – T2) = Ia/r
2
…(2)
T2 – m2g sin  = m2a …(3)
Adding the equations (1) and (3) we will get
m1g sin  + (T2 – T1) – m2g sin  = (m1 + m2)a
 (m1 – m2)g sin = (m1 + m2 + 1/r
2
)a
 a =
)r/1mm(
sing)mm(
2
21
21


= 0.248 = 0.25 ms
–2
.
39. m1 = 4 kg, m2 = 2 kg
Frictional co-efficient between 2 kg block and surface = 0.5
R = 10 cm = 0.1 m
I = 0.5 kg – m
2
m1g sin  – T1 = m1a …(1)
T2 – (m2g sin  +  m2g cos ) = m2a …(2)
(T1 – T2) = Ia/r
2
Adding equation (1) and (2) we will get
m1g sin  – (m2g sin  + m2g cos ) + (T2 – T1) = m1a + m2a
 4 × 9.8 × )2/1( – {(2 × 9.8 × )2/1( + 0.5 × 2 × 9.8 × )2/1( } = (4 + 2 + 0.5/0.01)a
 27.80 – (13.90 + 6.95) = 65 a  a = 0.125 ms
–2
.
m1 = 200 g, I = 1 m, m2 = 20 g
Therefore, (T1 × r1) – (T2 × r2) – (m1f × r3g) = 0
 T1 × 0.7 – T2 × 0.3 – 2 × 0.2 × g = 0
 7T1 – 3T2 = 3.92 …(1)
T1 + T2 = 0.2 × 9.8 + 0.02 × 9.8 = 2.156 …(2)
From the equation (1) and (2) we will get
10 T1 = 10.3
 T1 = 1.038 N = 1.04 N
Therefore T2 = 2.156 – 1.038 = 1.118 = 1.12 N.
41. R1 = R2, R2 = 16g + 60 g = 745 N
R1 × 10 cos 37° = 16g × 5 sin 37° + 60 g × 8 × sin 37°
 8R1 = 48g + 288 g
 R1 = 336g/8 = 412 N = f
Therefore  = R1 / R2 = 412/745 = 0.553.
m2
a
a/2
B
T1
A
m1
T2
aT1
mg2
cos

4 kg
T1
T2
T2
45° 45°
a
2 kg
T1
T2
45° 45°
4 kg
200kg
70cm
T1 T2
1m
200g
20g
f
60g
37°
16g
R1
R2

Chapter-10
10.8
42.  = 0.54, R2 = 16g + mg ; R1 = R2
 R1 × 10 cos 37° = 16g × 5 sin 37° + mg × 8 × sin 37°
 8R1 = 48g + 24/5 mg
 R2 =
54.08
mg5/24g48


 16g + mg =
54.040
m24240
m16
54.085
mg24g0.24





 m = 44 kg.
43. m = 60 kg, ladder length = 6.5 m, height of the wall = 6 m
Therefore torque due to the weight of the body
a)  = 600 × 6.5 / 2 sin  = i
  = 600 × 6.5 / 2 × ])5.6/6(1[ 2

  = 735 N-m.
b) R2 = mg = 60 × 9.8
R1 = R2  6.5 R1 cos  = 60g sin  × 6.5/2
 R1 = 60 g tan  = 60 g × (2.5/12)[because tan  = 2.5/6]
 R1 = (25/2) g = 122.5 N.
8g = F1 + F2 ; N1 = N2
Since, R1 = R2
Therefore F1 = F2
 2F1 = 8 g  F1 = 40
Let us take torque about the point B, we will get N1 × 4 = 8 g × 0.75.
 N1 = (80 × 3) / (4 × 4) = 15 N
Therefore 22
1
2
1
2
1 1540RNF(  = 42.72 = 43 N.
45. Rod has a length = L
It makes an angle  with the floor
The vertical wall has a height = h
R2 = mg – R1 cos  …(1)
R1 sin  = R2 …(2)
R1 cos  × (h/tan ) + R1 sin  × h = mg × 1/2 cos 
 R1 (cos
2
 / sin )h + R1 sin  h = mg × 1/2 cos 
 R1 =
}hsinh)sin/{(cos
cos2/Lmg
2


 R1 cos  =
}hsinh)sin/{(cos
sincos2/mgL
2
2


  = R1sin  / R2 =


22
cos2/1mgmg)}hsinh)sin/{(cos
sin.cos2/Lmg
 


sincosL)hsinh(cos2
sin2sin.cos2/L
222

 


sincosLh2
sincosL
2
2

46. A uniform rod of mass 300 grams and length 50 cm rotates with an uniform angular velocity = 2 rad/s
about an axis perpendicular to the rod through an end.
a) L = I
I at the end = mL
2
/3 = (0.3 × 0.5
2
}/3 = 0.025 kg-m
2
= 0.025 × 2 = 0.05 kg – m
2
/s
b) Speed of the centre of the rod
V = r = w × (50/2) = 50 cm/s = 0.5 m/s.
c) Its kinetic energy = 1/2 I
2
= (1/2) × 0.025 × 2
2
= 0.05 Joule.
f
60g
37°
16g
R1
R2

6.5m 
600
R1
R2
A
B
F1
R1
N1
R2
N2
8g
h

R1cosR1
R2

R1sin

mg

Chapter-10
10.9
47. I = 0.10 N-m; a = 10 cm = 0.1 m; m = 2 kg
Therefore (ma
2
/12) ×  = 0.10 N-m
  = 60 rad/s
Therefore  = 0 + t
  = 60 × 5 = 300 rad/s
Therefore angular momentum = I = (0.10 / 60) × 300 = 0.50 kg-m
2
/s
And 0 kinetic energy = 1/2 I
2
= 1/2 × (0.10 / 60) × 300
2
= 75 Joules.
48. Angular momentum of the earth about its axis is
= 2/5 mr
2
× (2 / 85400) (because, I = 2/5 mr
2
)
Angular momentum of the earth about sun’s axis
= mR
2
× (2 / 86400 × 365) (because, I = mR
2
)
Therefore, ratio of the angular momentum =
)36586400/(2mR
)86400/2(mr5/2
2
2


 (2r
2
× 365) / 5R
2
 (2.990 × 10
10
) / (1.125 × 10
17
) = 2.65 × 10
–7
.
49. Angular momentum due to the mass m1 at the centre of system is = m1 r
12
.
= m1 







 2
21
22
21
2
21
2
)mm(
rmm
mm
m
…(1)
Similarly the angular momentum due to the mass m2 at the centre of system is m2
r
112

= m2 







2
21
2
12
2
21
1
)mm(
mm
mm
rm
…(2)
Therefore net angular momentum = 2
21
22
12
2
21
22
21
)mm(
rmm
)mm(
rmm





 2
21
2
2121
)mm(
r)mm(mm


= 

22
21
21
rr
)mm(
mm
(proved)
50.  = I
 F × r = (mr
2
+ mr
2
)  5 × 0.25 = 2mr
2
× 
  = 20
25.0025.05.02
25.1


0 = 10 rad/s, t = 0.10 sec,  = 0 + t
  = 10 + 010 × 230 = 10 + 2 = 12 rad/s.
51. A wheel has
I = 0.500 Kg-m
2
, r = 0.2 m,  = 20 rad/s
Stationary particle = 0.2 kg
Therefore I11 = I22 (since external torque = 0)
 0.5 × 10 = (0.5 + 0.2 × 0.2
2
)2
 10/0.508 = 2 = 19.69 = 19.7 rad/s
52. I1 = 6 kg-m
2
, 1 = 2 rad/s , I2 = 5 kg-m
2
Since external torque = 0
Therefore I11 = I22
 2 = (6 × 2) / 5 = 2.4 rad/s
53. 1 = 120 rpm = 120 × (2 / 60) = 4 rad /s.
I1 = 6 kg – m
2
, I2 = 2 kgm
2
Since two balls are inside the system
Therefore, total external torque = 0
Therefore, I11 = I22
 6 × 4 = 22
 2 = 12  rad/s = 6 rev/s = 360 rev/minute.
a
l=0.10N-m
m1
m2
m2r
m1+m2
m1r
m1+m2
0.5kg0.5kg
r

Chapter-10
10.10
54. I1 = 2 × 10
–3
kg-m
2
; I2 = 3 × 10
–3
kg-m
2
; 1 = 2 rad/s
From the earth reference the umbrella has a angular velocity (1 – 2)
And the angular velocity of the man will be 2
Therefore I1(1 – 2) = I22
 2 × 10
–3
(2 – 2) = 3 × 10
–3
× 2
 52 = 4  2 = 0.8 rad/s. 
55. Wheel (1) has
I1 = 0.10 kg-m
2
, 1 = 160 rev/min
Wheel (2) has
I2 = ? ; 2 = 300 rev/min
Given that after they are coupled,  = 200 rev/min
Therefore if we take the two wheels to bean isolated system
Total external torque = 0
Therefore, I11 + I12 = (I1 + I1)
 0.10 × 160 + I2 × 300 = (0.10 + I2) × 200
 5I2 = 1 – 0.8  I2 = 0.04 kg-m
2
.
56. A kid of mass M stands at the edge of a platform of radius R which has a moment of inertia I. A ball of m
thrown to him and horizontal velocity of the ball v when he catches it.
Therefore if we take the total bodies as a system
Therefore mvR = {I + (M + m)R
2
}
(The moment of inertia of the kid and ball about the axis = (M + m)R
2
)
  = 2
R)mM(1
mvR

.
57. Initial angular momentum = Final angular momentum
(the total external torque = 0)
Initial angular momentum = mvR (m = mass of the ball, v = velocity of the ball, R = radius of platform)
Therefore angular momentum = I + MR
2

Therefore mVR = I + MR
2

  =
)MR1(
mVR
2

.
58. From a inertial frame of reference when we see the (man wheel) system, we can find that the wheel
moving at a speed of  and the man with ( + V/R) after the man has started walking.
( = angular velocity after walking,  = angular velocity of the wheel before walking.
Since I = 0
Extended torque = 0
Therefore (1 + MR
2
) = I + mR
2
( + V/R)
 (I + mR
2
) + I + mR
2
 + mVR
  =  –
)mR1(
mVR
2

.
59. A uniform rod of mass m length ℓ is struck at an end by a force F.  to the rod for a short time t
a) Speed of the centre of mass
mv = Ft  v =
m
Ft
b) The angular speed of the rod about the centre of mass
ℓ – r × p
 (mℓ
2
/ 12) ×  = (1/2) × mv
 mℓ
2
/ 12 ×  = (1/2) ℓ
2
  = 6Ft / mℓ
c) K.E. = (1/2) mv
2
+ (1/2) ℓ
2
= (1/2) × m(Ft / m)
2
(1/2) ℓ
2
= (1/2) × m × ( F
2
t
2
/m
2
) + (1/2) × (mℓ
2
/12) (36 ×( F
2
t
2
/m
2
ℓ
2
))
1
Earth reference
1–2 from earth
2
w
V/R of man w.r.t.
the platform

Chapter-10
10.11
= F
2
t
2
/ 2m + 3/2 (F
2
t
2
) / m = 2 F
2
t
2
/ m
d) Angular momentum about the centre of mass :-
L = mvr = m × Ft / m × (1/2) = F ℓ t / 2
60. Let the mass of the particle = m & the mass of the rod = M
Let the particle strikes the rod with a velocity V.
If we take the two body to be a system,
Therefore the net external torque & net external force = 0
Therefore Applying laws of conservation of linear momentum
MV = mV (V = velocity of the rod after striking)
 V / V = m / M
Again applying laws of conservation of angular momentum

2
mVR
= ℓ

2
mVR
=
t212
MR2

  t =
V12m
MR


Therefore distance travelled :-
V t = V 







12m
MR
=
12
R
m
M
M
m 
 =
12
R
61. a) If we take the two bodies as a system therefore total external force = 0
Applying L.C.L.M :-
mV = (M + m) v
 v =
mM
mv

b) Let the velocity of the particle w.r.t. the centre of mass = V
 v =
mM
Mv0m


 v =
mM
Mv

c) If the body moves towards the rod with a velocity of v, i.e. the rod is moving with a velocity – v
towards the particle.
Therefore the velocity of the rod w.r.t. the centre of mass = V
–
 V
–
=
mM
mv
mM
vmOM





d) The distance of the centre of mass from the particle
=
)mM(
2/lM
)mM(
Om2/lM





Therefore angular momentum of the particle before the collision
= l  = Mr
2
cm 
= m{m l/2) / (M + m)}
2
× V/ (l/2)
= (mM
2
vl) / 2(M + m)
Distance of the centre of mass from the centre of mass of the rod =
)mM(
)2/ml(
)mM(
)2/l(m0M
R1
cm




 
Therefore angular momentum of the rod about the centre of mass
= MVcm
1
cmR
= M × {(–mv) / (M + m)} {(ml/2) / (M + m)}
= 2
2
2
2
)mM(2
lvMm
)mM(2
lvMm




(If we consider the magnitude only)
e) Moment of inertia of the system = M.I. due to rod + M.I. due to particle
M
m
R
M
m, v
v

Chapter-10
10.12
= 2
2
2
22
)mM(
)s/Ml(m
)mM(
)2/ml(M
12
Ml




=
)mM(12
)m4M(Ml2


.
f) Velocity of the centre of mass
)mM(
mV
)mM(
mV0M
Vm





(Velocity of centre of mass of the system before the collision = Velocity of centre of mass of the
system after the collision)
(Because External force = 0)
Angular velocity of the system about the centre of mass,
Pcm = Icm 
  cmmmmM IrvmrVM


)mM(2
Ml
)mM(
Mv
m
)mM(2
ml
)mM(
mv
M







 = 


)mM(12
)m4M(Ml2
 





)mM(12
)m4M(Ml
)mM(2
vlmMvlMm 2
2
22
 





)mM(12
)mM(Ml
)mM(2
)mM/(Mm 2
2
 
 l)m4M(
mv6
62. Since external torque = 0
Therefore I11 = I2 2

4
ml
4
ml 22
 
2
ml2



4
ml
4
ml2 22
 
4
ml3 2

Therefore 2 =
2
11
I
I 
=
4
ml3
2
ml
2
2








=
3
2

63. Two balls A & B, each of mass m are joined rigidly to the ends of a light of rod of length L. The system
moves in a velocity v0 in a direction  to the rod. A particle P of mass m kept at rest on the surface
sticks to the ball A as the ball collides with it.
a) The light rod will exert a force on the ball B only along its length. So collision will not affect its
velocity.
B has a velocity = v0
If we consider the three bodies to be a system
Applying L.C.L.M.
Therefore mv0 = 2mv  v =
2
v0
Therefore A has velocity =
2
v0
b) if we consider the three bodies to be a system
Therefore, net external force = 0
Therefore Vcm =
m2m
2
v
m2vm 0
0








=
m3
mvmv 00 
=
3
v2 0
(along the initial velocity as before collision)
m
mm
L
L
vo
vo

Chapter-10
10.13
c) The velocity of (A + P) w.r.t. the centre of mass =
2
v
3
v2 00
 =
6
v0
&
The velocity of B w.r.t. the centre of mass
3
v2
v 0
0  =
3
v0
[Only magnitude has been taken]
Distance of the (A + P) from centre of mass = l/3 & for B it is 2 l/3.
Therefore Pcm = lcm × 
 












22
00
3
l2
m
3
1
m2
3
l2
3
v
m
3
1
6
v
m2
 
9
ml6
18
lmv6 0
  =
l2
v0

64. The system is kept rest in the horizontal position and a particle P falls from a height h and collides with
B and sticks to it.
Therefore, the velocity of the particle ‘’ before collision = gh2
If we consider the two bodies P and B to be a system. Net external torque and force = 0
Therefore, vm2gh2m 
 v = 2/)gh2(
Therefore angular momentum of the rod just after the collision
 2m (v × r) = 2m × 2/l2/)gh2(   2/)gh2(ml
 =
l3
gh8
l3
gh2
)4/ml24/ml(2
gh2ml
l
L
22



b) When the mass 2m will at the top most position and the mass m at the lowest point, they will
automatically rotate. In this position the total gain in potential energy = 2 mg × (l/2) – mg (l/2) =
mg(l/2)
Therefore  mg l/2 = l/2 l
2
 mg l/2 = (1/2 × 3ml
2
) / 4 × (8gh / 9gl
2
)
 h = 3l/2.
0.4g – T1 = 0.4 a …(1)
T2 – 0.2g = 0.2 a …(2)
(T1 – T2)r = Ia/r …(3)
From equation 1, 2 and 3
 a = 5/g
)4.0/6.12.04.0(
g)2.04.0(



Therefore (b) V = )5.0gl2(ah2 5

 )5/8.9()5/g(  = 1.4 m/s.
a) Total kinetic energy of the system
= 1/2 m1V
2
+ 1/2 m2V
2
+ 1/2 18
2
= (1/2 × 0.4 × 1.4
2
) + (1/2 × 0.2 × 1.4
2
) + (1/2 × (1.6/4) × 1.4
2
) = 0.98 Joule.
66. l = 0.2 kg-m
2
, r = 0.2 m, K = 50 N/m,
m = 1 kg, g = 10 ms
2
, h = 0.1 m
Therefore applying laws of conservation of energy
mgh = 1/2 mv
2
+ 1/2 kx
2
 1 = 1/2 × 1 × V
2
+ 1/2 × 0.2 × V
2
/0.04 + (1/2) × 50 × 0.01 (x = h)
 1 = 0.5 v
2
+ 2.5 v
2
+ 1/4
 3v
2
= 3/4
 v = 1/2 = 0.5 m/s
A
x
v
p,o
B
y
m
2m
200g
T1
T1
T2
T2
400g

Chapter-10
10.14
67. Let the mass of the rod = m
1/2 l
2
= mg l/2
 1/2 × M l
2
/3 × 
2
= mg 1/2
 
2
= 3g / l
  = l/g3 = 5.42 rad/s.
68. 1/2 I
2
– 0 = 0.1 × 10 × 1
  = 20
For collision
0.1 × 1
2
× 20 + 0 = [(0.24/3)×1
2
+ (0.1)
2
1
2
]
  = 20 /[10.(0.18)]
 0 – 1/2 
2
= –m1g l (1 – cos ) – m2g l/2 (1 – cos )
= 0.1 × 10 (1 – cos ) = 0.24 × 10 × 0.5 (1 – cos )
 1/2 × 0.18 × (20/3.24) = 2.2(1 – cos )
 (1 – cos ) = 1/(2.2 × 1.8)
 1 – cos  = 0.252
 cos  = 1 – 0.252 = 0.748
  = cos
–1
(0.748) = 41°.
69. Let l = length of the rod, and m = mass of the rod.
Applying energy principle
(1/2) l
2
– O = mg (1/2) (cos 37° – cos 60°)

3
ml
2
1 2
 
2
= mg × 






2
1
5
4
2
1
t
 
2
=
l10
g9
= 0.9 





l
g
Again 





3
2ml
 = mg 





2
1
sin 37° = mgl ×
5
3
  = 0.9 





l
g
= angular acceleration.
So, to find out the force on the particle at the tip of the rod
Fi = centrifugal force = (dm) 
2
l = 0.9 (dm) g
Ft = tangential force = (dm)  l = 0.9 (dm) g
So, total force F =  2
t
2
i FF  = 0.9 2 (dm) g
70. A cylinder rolls in a horizontal plane having centre velocity 25 m/s.
At its age the velocity is due to its rotation as well as due to its leniar motion & this two
velocities are same and acts in the same direction (v = r )
Therefore Net velocity at A = 25 m/s + 25 m/s = 50 m/s
71. A sphere having mass m rolls on a plane surface. Let its radius R. Its centre moves
with a velocity v
Therefore Kinetic energy = (1/2) l
2
+ (1/2) mv
2
= 2
2
2
2
mv
2
1
R
v
mR
5
2
2
1
 = 22
mv
2
1
mv
10
2
 =
10
52 
mv
2
=
10
7
mv
2

72. Let the radius of the disc = R
Therefore according to the question & figure
Mg – T = ma …(1)
& the torque about the centre
= T × R = I × 
 TR = (1/2) mR
2
×a/R
mg1m
mg
0.1kg
1m
1m
60°
37°
25 m/s
A
O

mg
mg
R

Chapter-10
10.15
 T = (1/2) ma
Putting this value in the equation (1) we get
 mg – (1/2) ma = ma
 mg = 3/2 ma  a = 2g/3
73. A small spherical ball is released from a point at a height on a rough track & the sphere does not slip.
Therefore potential energy it has gained w.r.t the surface will be converted to angular kinetic energy
about the centre & linear kinetic energy.
Therefore mgh = (1/2) l
2
+ (1/2) mv
2
 mgh =
5
2
2
1
 mR
2

2
+
2
1
mv
2
 gh =
5
1
v
2
+
2
1
v
2
 v
2
=
7
10
gh  v = gh
7
10

74. A disc is set rolling with a velocity V from right to left. Let it has attained a height h.
Therefore (1/2) mV
2
+ (1/2) l
2
= mgh
 (1/2) mV
2
+ (1/2) × (1/2) mR
2

2
=mgh
 (1/2) V
2
+ 1/4 V
2
= gh  (3/4) V
2
= gh
 h =
g
V
4
3 2
 
75. A sphere is rolling in inclined plane with inclination 
Therefore according to the principle
Mgl sin  = (1/2) l
2
+ (1/2) mv
2
 mgl sin  = 1/5 mv
2
+ (1/2) mv
2
Gl sin  = 7/10 
2
 v = singl
7
10

76. A hollow sphere is released from a top of an inclined plane of inclination .
To prevent sliding, the body will make only perfect rolling. In this condition,
mg sin  – f = ma …(1)
& torque about the centre
f × R =
3
2
mR
2
×
R
a
 f =
3
2
ma …(2)
Putting this value in equation (1) we get
 mg sin  –
3
2
ma = ma  a =
5
3
g sin 
 mg sin  – f =
5
3
mg sin  f =
5
2
mg sin 
  mg cos  =
5
2
mg sin   =
5
2
tan 
b)
5
1
tan  (mg cos ) R =
3
2
mR
2

  =
R
sing
10
3 

ac = g sin  –
5
g
sin  =
5
4
sin 
h
m
h
m
LL sin R
R

R
m
R 
mg cos
mg sin

Chapter-10
10.16
 t
2
=
ca
s2
=





 
5
sing4
l2
=
sing2
l5
Again,  = t
K.E. = (1/2) mv
2
+ (1/2) l
2
= (1/2) m(2as) +(1/2) l (
2
t
2
)
=
5
sing4
m
2
1 
 × 2 × l +




sing2
l5
R
sing
100
9
mR
3
2
2
1 22
2
=
40
sinmgl3
5
sinmgl4 


=
8
7
mgl sin 
77. Total normal force = mg +
rR
mv2

 mg (R – r) = (1/2) l
2
+ (1/2) mv
2
 mg (R – r) = 22
mv
2
1
mv
5
2
2
1


10
7
mv
2
= mg(R – r)  v
2
=
7
10
g(R – r)
Therefore total normal force = mg +
rR
)rR(g
7
10
mmg








= mg + mg 





7
10
=
7
17
mg
78. At the top most point
rR
mv2

= mg  v
2
= g(R – r)
Let the sphere is thrown with a velocity v
 (1/2) mv
2
+ (1/2) l
2
= mg 2 (R – r) + (1/2) mv
2
+ (1/2) l
2

10
7
v
2
= g 2(R – r) +
10
7
v
2
 v
2
=
7
20
g (R – r) + g (R – r) 
 v = )rR(g
7
27

79. a) Total kinetic energy y = (1/2) mv
2
+ (1/2) l
2
Therefore according to the question
mg H = (1/2) mv
2
+ (1/2) l
2
+ mg R (1 + cos )
 mg H – mg R (1 + cos ) = (1/2) mv
2
+ (1/2) l
2
 (1/2) mv
2
+ (1/2) l
2
= mg (H – R – R sin )
b) to find the acceleration components
 (1/2) mv
2
+ (1/2) l
2
= mg (H – R – R sin )
 7/10 mv
2
= mg (H – R – R sin )
R
v2
= 











sin1
R
H
g
7
10
 radical acceleration
 v
2
=
7
10
g (H – R) – R sin 
 2v
dt
dv
= –
7
10
g R cos 
dt
d
  R
dt
dv
= –
7
5
g R cos 
dt
d


dt
dv
= –
7
5
g cos   tangential acceleration
mg+mv2
/(R–r)
R
R
R
H R sin

Chapter-10
10.17
c) Normal force at  = 0

R
mv2
= 




 

1.0
1.06.0
10
7
10
1000
70
= 5N
Frictional force :-
f = mg - ma = m(g – a) = m (10 –
7
5
×10) = 0.07 




 
7
5070
=
100
1
×20 = 0.2N
80. Let the cue strikes at a height ‘h’ above the centre, for pure rolling, Vc = R
Applying law of conservation of angular momentum at a point A,
mvch – ℓ = 0
mvch =
3
2
mR
2
× 





R
vc

h =
3
R2
81. A uniform wheel of radius R is set into rotation about its axis (case-I) at an angular speed 
This rotating wheel is now placed on a rough horizontal. Because of its friction at contact, the wheel
accelerates forward and its rotation decelerates. As the rotation decelerates the frictional force will act
backward.
If we consider the net moment at A then it is zero.
Therefore the net angular momentum before pure rolling & after pure rolling remains
constant
Before rolling the wheel was only rotating around its axis.
Therefore Angular momentum = ℓ  = (1/2) MR
2
 …(1)
After pure rolling the velocity of the wheel let v
Therefore angular momentum = ℓcm  + m(V × R)
= (1/2) mR
2
(V/R) + mVR = 3/2 mVR …(2)
Because, Eq(1) and (2) are equal
Therefore, 3/2 mVR = ½ mR
2

 V =  R /3
82. The shell will move with a velocity nearly equal to v due to this motion a frictional force well act in the
background direction, for which after some time the shell attains a pure rolling. If we
consider moment about A, then it will be zero. Therefore, Net angular momentum
about A before pure rolling = net angular momentum after pure rolling.
Now, angular momentum before pure rolling about A = M (V × R) and angular
momentum after pure rolling :-
(2/3) MR
2
× (V0 / R) + M V0 R
(V0 = velocity after pure rolling)
 MVR = 2/3 MV0R + MV0R
 (5/3) V0 = V
 V0 = 3V/ 5
83. Taking moment about the centre of hollow sphere we will get
F × R =
3
2
MR
2

  =
MR2
F3
Again, 2 = (1/2) t
2
(From  = 0t + (1/2) t
2
)
 t
2
=
F3
MR8
 ac =
m
F
 X = (1/2) act
2
= (1/2) =
3
R4

R
h
x

vc

(1st
case)
 v
mg

(2nd
case)
R
A
(1st
case)
v

v

A
(2nd
case)

vo
R
mg
A
F
I

Chapter-10
10.18
84. If we take moment about the centre, then
F × R = ℓ × f × R
 F = 2/5 mR + mg …(1)
Again, F = mac –  mg …(2)
 ac =
m
mgF 
Putting the value ac in eq(1) we get
 mg
m
mgF
m
5
2





 
 
 2/5 (F +  mg) +  mg
 F = 105.0
7
2
105.0
5
2
F
5
2


5
F3
=
7
10
7
4
 = 2
 F =
3
25 
=
3
10
= 3.33 N
85. a) if we take moment at A then external torque will be zero
Therefore, the initial angular momentum = the angular momentum after rotation stops (i.e. only leniar
velocity exits)
MV × R – ℓ  = MVO × R
 MVR – 2/5 × MR
2
V / R = MVO R
 VO = 3V/5
b) Again, after some time pure rolling starts
therefore  M × vo × R = (2/5) MR
2
× (V/R) + MVR
 m × (3V/5) × R = (2/5) MVR + MVR
 V = 3V/7
86. When the solid sphere collides with the wall, it rebounds with velocity ‘v’ towards left but it continues to
rotate in the clockwise direction.
So, the angular momentum = mvR – (2/5) mR
2
× v/R
After rebounding, when pure rolling starts let the velocity be v
and the corresponding angular velocity is v / R
Therefore angular momentum = mvR + (2/5) mR
2
(v/R)
So, mvR – (2/5) mR
2
, v/R = mvR + (2/5) mR
2
(v/R)
mvR × (3/5) = mvR × (7/5)
v = 3v/7
So, the sphere will move with velocity 3v/7.
* * * *
a
mg
F
I
w=V/R
A
v
v
V/R
v

11.1
CHAPTER 11
1. Gravitational force of attraction,
F = 2
r
GMm
= 2
11
)1.0(
10101067.6  
= 6.67 × 10
–7
N
2. To calculate the gravitational force on ‘m’ at unline due to other mouse.
ODF = 22
)r/a(
m4mG 
= 2
2
a
Gm8
OIF = 22
)r/a(
m2mG 
= 2
2
a
Gm6
OBF = 22
)r/a(
m2mG 
= 2
2
a
Gm4
OAF = 22
)r/a(
mmG 
= 2
2
a
Gm2
Resultant OFF =
2
2
2
2
2
2
a
Gm
36
a
Gm
64

















= 10 2
2
a
Gm
Resultant OEF =
2
2
2
2
2
2
a
Gm
4
a
Gm
64 















= 2
2
a
Gm
52
The net resultant force will be,
F = 520
a
Gm
2
a
Gm
20
a
Gm
100 2
2
2
2
2
2
2
2



























=  540120
a
Gm
2
2
2









= )6.89120(
a
Gm
2
2
2









= 4.40
a
Gm
2
2
= 2
2
a
Gm
24
3. a) if ‘m’ is placed at mid point of a side
then OAF = 2
2
a
Gm4
in OA direction
OBF = 2
2
a
Gm4
in OB direction
Since equal & opposite cancel each other
ocF =
  23
2
a2/r
Gm
= 2
2
a3
Gm4
in OC direction
Net gravitational force on m = 2
2
a
Gm4
b) If placed at O (centroid)
the OAF =
)r/a(
Gm
3
2
= 2
2
a
Gm3
FD
BA E
C
2mm
4m 3m
m
B
A
C
m
m
m m
O
B
A
C
m
m
m m
O

Chapter 11
11.2
OBF = 2
2
a
Gm3
Resultant F

=
2
1
a
Gm3
2
a
Gm3
2
2
2
2
2
2
2
















= 2
2
a
Gm3
Since OCF = 2
2
a
Gm3
, equal & opposite to F, cancel
Net gravitational force = 0
4. CBF = jˆ60sin
a4
Gm
iˆ60cos
a4
Gm
2
2
2
2

CAF = jˆ60sin
a4
Gm
iˆ60cos
a4
Gm
2
2
2
2


F

= CBF + CAF
= jˆ60sin
a4
Gm2
2
2

=
2
r
a4
Gm2 3
2
2

= 2
2
3
a4
Gmr
5. Force on M at C due to gravitational attraction.
CBF = jˆ
R2
Gm
2
2
CDF = iˆ
R4
GM
2
2

CAF = jˆ45sin
R4
GM
jˆ45cos
R4
GM
2
2
2
2


So, resultant force on C,
 CF = CAF + CBF + CDF
= jˆ
2
1
2
R4
GM
iˆ
2
1
2
R4
GM
2
2
2
2













FC =  122
R4
GM
2
2

For moving along the circle, F

=
R
mv2
or  122
R4
GM
2
2
 =
R
MV2
or V =







 
4
122
R
GM
6.
 2
hR
GM

= 62
2211
10)10001740(
104.71067.6

 
= 6
11
1027402740
10358.49


= 13
11
1075.0
10358.49


= 65.8 × 10
–2
= 0.65 m/s
2
7. The linear momentum of 2 bodies is 0 initially. Since gravitational force is internal, final momentum is
also zero.
So (10 kg)v1 = (20 kg) v2
Or v1 = v2 …(1)
Since P.E. is conserved
Initial P.E. =
1
20101067.6 11
 
= –13.34×10
–9
J
When separation is 0.5 m,
M
B
C
A
D
B
C
A
R

Chapter 11
11.3
–13.34 × 10
–9
+ 0 =
)2/1(
1034.13 9

+ (1/2) × 10 v1
2
+ (1/2) × 20 v2
2
…(2)
 – 13.34 × 10
–9
= -26.68 ×10
–9
+ 5 v1
2
+ 10 v2
2
 – 13.34 × 10
–9
= -26.68 ×10
–9
+ 30 v2
2
 v2
2
=
30
1034.13 9

= 4.44 × 10
–10
 v2 = 2.1 × 10
–5
m/s.
So, v1 = 4.2 × 10
–5
m/s.
8. In the semicircle, we can consider, a small element of d then R d = (M/L) R d = dM.
F = 2
LR
mGMRd
dF3 = 2 dF since =
LR
GMm2
sin  d
F = 


2/
0
dsin
LR
GMm2
   2/
0cos
LR
GMm2 
 
 = –2 )1(
LR
GMm
 =
LR
GMm2
=
A/LL
GMm2

= 2
L
GMm2
9. A small section of rod is considered at ‘x’ distance mass of the element = (M/L). dx = dm
dE1 =
 22
xd
1)dm(G


= dE2
Resultant dE = 2 dE1 sin 
= 2 ×
   2222
xd
d
xd
)dm(G



=
   




 

2222
xdxdL
dxdGM2
Total gravitational field
E =
  
2/L
0
2/322
xdL
dxGmd2
Integrating the above equation it can be found that,
E =
22
d4Ld
GM2


10. The gravitational force on ‘m’ due to the shell of M2 is 0.
M is at a distance
2
RR 21 
Then the gravitational force due to M is given by
=
2/21
1
RR(
mGM

= 2
21
1
)RR(
mGM4

11. Man of earth M = (4/3) R
3

Man of the imaginary sphere, having
Radius = x, M = (4/3)x
3

or
M
M
= 3
3
R
x
 Gravitational force on F = 2
m
mMG 
or F = 23
3
xR
mGMx
= 3
R
GMmx

R

M
dd
m
L
a
dx
x
dE1
O
M

d
dE2
x

m
R1
M1
R2 m2
x
m

Chapter 11
11.4
12. Let d be the distance from centre of earth to man ‘m’ then
D = 








4
R
x
2
2
= (1/2) 22
Rx4 
M be the mass of the earth, M the mass of the sphere of radius d/2.
Then M = (4/3) R
3

M = (4/3)d
3

or
M
M
= 3
3
R
d
 Gravitational force is m,
F = 2
d
mmG 
= 23
3
dR
MmGd
= 3
R
GMmd
So, Normal force exerted by the wall = F cos.
=
d2
R
R
GMmd
3
 = 2
R2
GMm
(therefore I think normal force does not depend on x)
13. a) m is placed at a distance x from ‘O’.
If r < x , 2r, Let’s consider a thin shell of man
dm = 3
2
x
3
4
r)3/4(
m


= 3
3
r
mx
Thus dm = 3
3
r
mx
Then gravitational force F = 2
x
mmdG
= 2
33
x
r/Gmx
=  3
r
Gmx
b) 2r < x < 2R, then F is due to only the sphere.
 F =
 2
rx
mGm


c) if x > 2R, then Gravitational force is due to both sphere & shell, then due to shell,
F =
 2
Rx
mGM


due to the sphere =
 2
rx
mGm


So, Resultant force =
 2
rx
mGm


+
 2
Rx
mGM


14. At P1, Gravitational field due to sphere M =
 2
aa3
GM

= 2
a16
GM
At P2, Gravitational field is due to sphere & shell,
= 2
)aa4a(
GM

+ 2
)aa4(
GM

= 






25
1
36
1
a
GM
2
= 2
a
GM
900
61






15. We know in the thin spherical shell of uniform density has gravitational field at its internal point is zero.
At A and B point, field is equal and opposite and cancel each other so Net field is
zero.
Hence, EA = EB
16. Let 0.1 kg man is x m from 2kg mass and (2 – x) m from 4 kg mass.
 2
x
1.02
= – 2
)x2(
1.04


R/2
m
O
d
x

x
n
F
R/2

d
m
R
O
M
r
49
P1
a
P2
a
a
B
A
B
A

Chapter 11
11.5
or 2
x
2.0
= – 2
)x2(
4.0

or 2
x
1
= 2
)x2(
2

or (2 – x)
2
= 2 x
2
or 2 – x = 2 x or x(r2 + 1) = 2
or x =
414.2
2
= 0.83 m from 2kg mass.
17. Initially, the ride of  is a
To increase it to 2a,
work done =
a
Gm
a2
Gm 22
 =
a2
Gm3 2
18. Work done against gravitational force to take away the particle from sphere,
=
1.01.0
1.010G


= 1
11
101
11067.6




= 6.67 × 10
–10
J
19. E

= (5 N/kg) iˆ + (12 N/kg) jˆ
a) F

= E

m
= 2kg [(5 N/kg) iˆ + (12 N/kg) jˆ ] = (10 N) iˆ + (12 N) jˆ
F

= 576100  = 26 N
b) V

= E

r
At (12 m, 0), V

= – (60 J/kg) iˆ V

= 60 J
At (0, 5 m), V

= – (60 J/kg) jˆ V

= – 60 J
c)  V

=

)5,2,1(
)0,0(
mdrE

=    )5,12(
)0,0(rjˆ)N24(iˆ)N10( 
= – (120 J iˆ + 120 J iˆ ) = 240 J
d)  v = –   
 m5,0
0,m12)jˆN24iˆN10(r 
= –120 jˆ + 120 iˆ = 0
20. a) V = (20 N/kg) (x + y)
R
GM
= L
M
MLT 2
or
L
MTLM 1231 
=
M
TML 22 
Or M
0
L
2
T
–2
= M
0
L
2
T
–2
 L.H.S = R.H.S
b) )y,x(E

= – 20(N/kg) iˆ – 20(N/kg) jˆ
c) F

= E

m
= 0.5kg [– (20 N/kg) iˆ – (20 N/kg) jˆ = – 10N iˆ - 10 N jˆ
 |F|

= 100100  = 10 2 N
21. E

= 2 iˆ + 3 jˆ
The field is represented as
tan 1 = 3/2
Again the line 3y + 2x = 5 can be represented as
tan 2 = – 2/3
m1 m2 = –1
Since, the direction of field and the displacement are perpendicular, is done by the particle on the line.
m m
m
a
a
a
100g
10cm
10kg

2j
3j
5/2
5/3 2

Chapter 11
11.6
22. Let the height be h
(1/2) 2
R
GM
= 2
)hR(
GM

Or 2R
2
= (R + h)
2
Or 2 R = R + h
Or h = (r2 – 1)R
23. Let g be the acceleration due to gravity on mount everest.
g = 






R
h2
1g
=9.8 






6400000
17696
1 = 9.8 (1 – 0.00276) = 9.773 m/s
2
24. Let g be the acceleration due to gravity in mine.
Then g= g 






R
d
1
= 9.8 






 3
106400
640
1 = 9.8 × 0.9999 = 9.799 m/s
2
25. Let g be the acceleration due to gravity at equation & that of pole = g
g= g – 
2
R
= 9.81 – (7.3 × 10
–5
)
2
× 6400 × 10
3
= 9.81 – 0.034
= 9.776 m/s
2
mg = 1 kg × 9.776 m/s
2
= 9.776 N or 0.997 kg
The body will weigh 0.997 kg at equator.
26. At equator, g = g – 
2
R …(1)
Let at ‘h’ height above the south pole, the acceleration due to gravity is same.
Then, here g = g 






R
h2
1 …(2)
 g - 
2
R = g 






R
h2
1
or
g
R
1
2

 =
R
h2
1
or h =
g2
R22

=
   
81.92
106400103.7
2325

 
= 11125 N = 10Km (approximately) 
27. The apparent ‘g’ at equator becomes zero.
i.e. g = g – 
2
R = 0
or g = 
2
R
or  =
R
g
= 3
106400
8.9

= 6
105.1 
 = 1.2 × 10
–3
rad/s.
T =

2
= 3
102.1
14.32



= 1.5 × 10
–6
sec. = 1.41 hour
28. a) Speed of the ship due to rotation of earth v = R
b) T0 = mgr = mg – m
2
R
 T0 – mg = m
2
R
c) If the ship shifts at speed ‘v’
T = mg – m
2
R
A
To
A

Chapter 11
11.7
= T0 -
  R
R
Rv
2
2







 
= T0 –







 
R
Rv2Rv 222
m
 T = T0 + 2v m
29. According to Kepler’s laws of planetary motion,
T
2
 R
3
2
e
2
m
T
T
 3
es
3
ms
R
R

3
es
ms
R
R







2
1
88.1








es
ms
R
R
= (1.88)
2/3
= 1.52
30. T = 2
GM
r3
27.3 = 2 × 3.14
 
M1067.6
1084.3
11
35



or 2.73 × 2.73 =
 
M1067.6
1084.314.32
11
35



or M = 211
1532
)3.27(10335.3
10)84.3()14.3(2



= 6.02 × 10
24
kg
 mass of earth is found to be 6.02 × 10
24
kg.
31. T = 2
GM
r3
 27540 = 2 × 3.14
 
M1067.6
10104.9
11
333



or (27540)
2
= (6.28)
2  
M1067.6
104.9
11
26



or M = 211
1832
)27540(1067.6
10)4.9()28.6(



= 6.5 × 10
23
kg.
32. a) V =
hr
GM

=
hr
gr2

=
)24.6(10
)106400(8.9
6
23


= 6.9 × 10
3
m/s = 6.9 km/s
b) K.E. = (1/2) mv
2
= (1/2) 1000 × (47.6 × 10
6
) = 2.38 × 10
10
J
c) P.E. =
)hR(
GMm

= – 3
32411
10)20006400(
101061067.6

 
= –
8400
1040 13

= – 4.76 × 10
10
J
d) T =
V
)hr(2 
= 3
3
109.6
10840014.32


= 76.6 × 10
2
sec = 2.1 hour

Chapter 11
11.8
33. Angular speed f earth & the satellite will be same
eT
2
=
sT
2
or
360024
1

=
2
3
gR
)hR(
2
1


or 12 I 3600 = 3.14 2
3
gR
)hR( 
or 2
2
gR
)hR( 
= 2
2
)14.3(
)360012( 
or 62
93
10)6400(8.9
10)h6400(


= 2
2
)14.3(
)360012( 
or 9
93
106272
10)h6400(


= 432 × 10
4
or (6400 + h)
3
= 6272 × 432 × 10
4
or 6400 + h = (6272 × 432 × 10
4
)
1/3
or h = (6272 × 432 × 10
4
)
1/3
– 6400
= 42300 cm.
b) Time taken from north pole to equator = (1/2) t
= (1/2) × 6.28 62
3
10)6400(10
)640043200(


= 3.14 112
63
10)64(
10)497(


= 3.14 5
106464
497497497


= 6 hour.
34. For geo stationary satellite,
r = 4.2 × 10
4
km
h = 3.6 × 104
km
Given mg = 10 N
mgh = mg
  







 2
2
hR
R
= 10
 
  









233
23
103600106400
106400
=
17980
4096
= 0.23 N
35. T = 2 2
1
3
2
gR
R
Or T
2
= 4
2
1
3
2
gR
R
Or g = 2
1
3
2
2
2
R
R
T
4
 Acceleration due to gravity of the planet is = 2
1
3
2
2
2
R
R
T
4
36. The colattitude is given by .
OAB = 90° – ABO
Again OBC =  = OAB
 sin  =
42000
6400
=
53
8
 = sin
–1






53
8
= sin
–1
0.15. 
Colatitude
CB
A


O

Chapter 11
11.9
37. The particle attain maximum height = 6400 km.
On earth’s surface, its P.E. & K.E.
Ee = (1/2) mv
2
+ 




 
R
GMm
…(1)
In space, its P.E. & K.E.
Es = 







hR
GMm
+ 0
Es = 






R2
GMm
…(2) ( h = R)
Equating (1) & (2)
2
mv
2
1
R
GMm
 =
R2
GMm

Or (1/2) mv
2
= GMm 






R
1
R2
1
Or v
2
=
R
GM
= 3
2411
106400
1061067.6

 
= 6
13
104.6
1002.40


= 6.2 × 10
7
= 0.62 × 10
8
Or v = 8
1062.0  = 0.79 × 10
4
m/s = 7.9 km/s.
38. Initial velocity of the particle = 15km/s
Let its speed be ‘v’ at interstellar space.
(1/2) m[(15 × 10
3
)
2
– v
2
] =


R 2
dx
x
GMm
 (1/2) m[(15 × 10
3
)
2
– v
2
] = GMm








Rx
1
 (1/2) m[(225 × 10
6
) – v
2
] =
R
GMm
 225 × 10
6
– v
2
= 3
2411
106400
1061067.62

 
 v
2
= 225 × 10
6
–
32
02.40
× 10
8
 v
2
= 225 × 10
6
– 1.2 × 10
8
= 10
8
(1.05)
Or v = 1.01 × 10
4
m/s or
= 10 km/s
39. The man of the sphere = 6 × 10
24
kg.
Escape velocity = 3 × 10
8
m/s
Vc =
R
GM2
Or R = 2
cV
GM2
=
 28
2411
103
1061067.62

 
=
9
02.80
× 10
–3
= 8.89× 10
–3
m  9 mm.
    

12.1
CHAPTER 12
1. Given, r = 10cm.
At t = 0, x = 5 cm.
T = 6 sec.
So, w =
T
2
=
6
2
=
3

sec
–1
At, t = 0, x = 5 cm.
So, 5 = 10 sin (w × 0 + ) = 10 sin  [y = r sin wt]
Sin  = 1/2   =
6

 Equation of displacement x = (10cm) sin 




 
3
(ii) At t = 4 second
x = 10 sin 




 


6
4
3
= 10 sin 




 
6
8
= 10 sin 




 
2
3
= 10 sin 




 

2
= - 10 sin 




 
2
= -10
Acceleration a = – w
2
x = –







 
9
2
× (–10) = 10.9  0.11 cm/sec.
2. Given that, at a particular instant,
X = 2cm = 0.02m
V = 1 m/sec
A = 10 msec
–2
We know that a = 
2
x
  =
x
a
=
02.0
10
= 500 = 10 5
T =

2
=
510
2
=
236.210
14.32


= 0.28 seconds.
Again, amplitude r is given by v =  




  22
xr 
 v
2
= 
2
(r
2
– x
2
)
1 = 500 (r
2
– 0.0004)
 r = 0.0489  0.049 m
 r = 4.9 cm.
3. r = 10cm
Because, K.E. = P.E.
So (1/2) m 
2
(r
2
– y
2
) = (1/2) m 
2
y
2
r
2
– y
2
= y
2
 2y
2
= r
2
 y =
2
r
=
2
10
= 5 2 cm form the mean position.
4. vmax = 10 cm/sec.
 r = 10
 
2
= 2
r
100
…(1)
Amax = 
2
r = 50 cm/sec
 
2
=
y
50
=
r
50
…(2)

Chapter 12
12.2
 2
r
100
=
r
50
 r = 2 cm.
  = 2
r
100
= 5 sec
2
Again, to find out the positions where the speed is 8m/sec,
v
2
= 
2
(r
2
– y
2
)
 64 = 25 ( 4 – y
2
)
 4 – y
2
=
25
64
 y
2
= 1.44  y = 44.1  y = 1.2 cm from mean position.
5. x = (2.0cm)sin [(100s
–1
) t + (/6)]
m = 10g.
a) Amplitude = 2cm.
 = 100 sec
–1
 T =
100
2
=
50

sec = 0.063 sec.
We know that T = 2
k
m
 T
2
= 4
2
×
k
m
 k = m
T
4
2
2

= 10
5
dyne/cm = 100 N/m. [because  =
T
2
= 100 sec
–1
]
b) At t = 0
x = 2cm sin 




 
6
= 2 × (1/2) = 1 cm. from the mean position.
We know that x = A sin (t + )
v = A cos (t + )
= 2 × 100 cos (0 + /6) = 200 ×
2
3
= 100 3 sec
–1
= 1.73m/s
c) a = – 
2
x = 100
2
× 1 = 100 m/s
2

6. x = 5 sin (20t + /3)
a) Max. displacement from the mean position = Amplitude of the particle.
At the extreme position, the velocity becomes ‘0’.
 x = 5 = Amplitude.
 5 = 5 sin (20t + /3)
sin (20t + /3) = 1 = sin (/2)
 20t + /3 = /2
 t = /120 sec., So at /120 sec it first comes to rest.
b) a = 
2
x = 
2
[5 sin (20t + /3)]
For a = 0, 5 sin (20t + /3) = 0  sin (20t + /3) = sin ()
 20 t =  – /3 = 2/3
 t = /30 sec.
c) v = A  cos (t +/3) = 20 × 5 cos (20t + /3)
when, v is maximum i.e. cos (20t + /3) = –1 = cos 
 20t =  – /3 = 2/3
 t = /30 sec.
7. a) x = 2.0 cos (50t + tan
–1
0.75) = 2.0 cos (50t + 0.643)
v =
dt
dx
= – 100 sin (50t + 0.643)
 sin (50t + 0.643) = 0
As the particle comes to rest for the 1
st
time
 50t + 0.643 = 
 t = 1.6 × 10
–2
sec.

Chapter 12
12.3
b) Acceleration a =
dt
dv
= – 100 × 50  cos (50t + 0.643)
For maximum acceleration cos (50t + 0.643) = – 1 cos  (max) (so a is max)
 t = 1.6 × 10
–2
sec.
c) When the particle comes to rest for second time,
50t + 0.643 = 2
 t = 3.6 × 10
–2
s.
8. y1 =
2
r
, y2 = r (for the two given position)
Now, y1 = r sin t1

2
r
= r sin t1  sin t1 =
2
1
 t1 =
2


t
2
× t1 =
6

 t1

12
t
Again, y2 = r sin t2
 r = r sin t2  sin t2 = 1  t2 = /2  




 
t
2
t2 =
2

 t2 =
4
t
So, t2 – t1 =
12
t
4
t
 =
6
t

9. k = 0.1 N/m
T = 2
k
m
= 2 sec [Time period of pendulum of a clock = 2 sec]
So, 4
2+ 





k
m
= 4
 m = 2
k

=
10
1.0
= 0.01kg  10 gm.
10. Time period of simple pendulum = 2
g
1
Time period of spring is 2
k
m

Tp = Ts [Frequency is same]

g
1

k
m
 
k
m
g
1
 

k
mg

k
F
= x. (Because, restoring force = weight = F =mg)
 1 = x (proved)
11. x = r = 0.1 m
T = 0.314 sec
m = 0.5 kg.
Total force exerted on the block = weight of the block + spring force.
T = 2
k
m
 0.314 = 2
k
5.0
 k = 200 N/m
 Force exerted by the spring on the block is
F = kx = 201.1 × 0.1 = 20N
 Maximum force = F + weight = 20 + 5 = 25N
12. m = 2kg.
T = 4 sec.
T = 2
k
m
 4 = 2
K
2
 2 = 
K
2
x

0.5kg

Chapter 12
12.4
 4 = 
2






k
2
 k =
4
2 2

 k =
2
2

= 5 N/m
But, we know that F = mg = kx
 x =
k
mg
=
5
102
= 4
Potential Energy = (1/2) k x
2
= (1/2) × 5 × 16 = 5 × 8 = 40J
13. x = 25cm = 0.25m
E = 5J
f = 5
So, T = 1/5sec.
Now P.E. = (1/2) kx
2
(1/2) kx
2
= 5  (1/2) k (0.25)
2
= 5  k = 160 N/m.
Again, T = 2
k
m

5
1

160
m
 m = 0.16 kg.
14. a) From the free body diagram,
 R + m
2
x – mg = 0 …(1)
Resultant force m2
x = mg – R
 m
2
x = m 





 mM
k
 x =
mM
mkx

[  = )mM/(k  for spring mass system]
b) R = mg – m2
x = mg - m x
mM
k

= mg –
mM
mkx

For R to be smallest, m
2
x should be max. i.e. x is maximum.
The particle should be at the high point.
c) We have R = mg – m
2
x
The tow blocks may oscillates together in such a way that R is greater than 0. At limiting condition, R
= 0, mg = m
2
x
X = 2
m
mg

=
mk
)mM(mg 
So, the maximum amplitude is =
k
)mM(g 
15. a) At the equilibrium condition,
kx = (m1 + m2) g sin 
 x =
k
sing)mm( 21 
b) x1 =
k
2
(m1 + m2) g sin  (Given)
when the system is released, it will start to make SHM
where  =
21 mm
k

When the blocks lose contact, P = 0
So m2 g sin  = m2 x2 
2
= m2 x2 





 21 mm
k
 x2 = k
sing)mm( 21 
So the blocks will lose contact with each other when the springs attain its natural length.
A
B
M
K
x
m
mg
R
a= 2
x
m2
x
m2
x2
m1g
m1
x1
g
k
m2g
F
R
(m1 +m2)g
R
P
m2aa
m2g

Chapter 12
12.5
c) Let the common speed attained by both the blocks be v.
1/2 (m1 + m2) v
2
– 0 = 1/2 k(x1 + x2)
2
– (m1 + m2) g sin  (x + x1)
[ x + x1 = total compression]
 (1/2) (m1 + m2) v
2
= [(1/2) k (3/k) (m1 + m2) g sin  –(m1 + m2) g sin (x + x1)
 (1/2) (m1 + m2) v
2
= (1/2) (m1 + m2) g sin  × (3/k) (m1 + m2) g sin 
 v =
)mm(k
3
21 
g sin .
16. Given, k = 100 N/m, M = 1kg and F = 10 N
a) In the equilibrium position,
compression = F/k = 10/100 = 0.1 m = 10 cm
b) The blow imparts a speed of 2m/s to the block towards left.
P.E. + K.E. = 1/2 k
2
+ 1/2 Mv
2
= (1/2) × 100 × (0.1)
2
+ (1/2) × 1 × 4 = 0.5 + 2 = 2.5 J
c) Time period = 2
k
M
= 2
100
1
=
5

sec
d) Let the amplitude be ‘x’ which means the distance between the mean position and the extreme
position.
So, in the extreme position, compression of the spring is (x + ).
Since, in SHM, the total energy remains constant.
(1/2) k (x + )
2
= (1/2) k
2
+ (1/2) mv
2
+ Fx = 2.5 + 10x
[because (1/2) k
2
+ (1/2) mv
2
= 2.5]
So, 50(x + 0.1)
2
= 2.5 + 10x
 50 x
2
+ 0.5 + 10x = 2.5 + 10x
50x
2
=2  x
2
=
50
2
=
100
4
 x =
10
2
m = 20cm.
e) Potential Energy at the left extreme is given by,
P.E. = (1/2) k (x +)
2
= (1/2) × 100 (0.1 +0.2)
2
=50 × 0.09 = 4.5J
f) Potential Energy at the right extreme is given by,
P.E. = (1/2) k (x +)
2
– F(2x) [2x = distance between two extremes]
= 4.5 – 10(0.4) = 0.5J
The different values in (b) (e) and (f) do not violate law of conservation of energy as the work is done by
the external force 10N.
17. a) Equivalent spring constant k = k1 + k2 (parallel)
T = 2
k
M
= 2
21 kk
m

b) Let us, displace the block m towards left through displacement ‘x’ 
Resultant force F = F1 + F2 = (k1 + k2)x
Acceleration (F/m) =
m
x)kk( 21 
Time period T = 2
onAccelerati
ntdisplaceme
= 2
m
)kk(m
x
21 
= 2
21 kk
m


The equivalent spring constant k = k1 + k2
c) In series conn equivalent spring constant be k.
So,
k
1
=
1k
1
+
2k
1
=
21
12
kk
kk 
 k =
21
21
kk
kk

T = 2
k
M
= 2
21
21
kk
)kk(m 

M
F
k
M
(a)
parallel k2
k1
x-1
k2
mk1
k2
m
k1

Chapter 12
12.6
18. a) We have F = kx  x =
k
F
Acceleration =
m
F
onAccelerati
ntdisplaceme
= 2
m/F
k/F
= 2
k
m

Amplitude = max displacement = F/k
b) The energy stored in the spring when the block passes through the equilibrium position
(1/2) kx
2
= (1/2) k (F/k)
2
= (1/2) k (F
2
/k
2
) = (1/2) (F
2
/k)
c) At the mean position, P.E. is 0. K.E. is (1/2) kx
2
= (1/2) (F
2
/x)
19. Suppose the particle is pushed slightly against the spring ‘C’ through displacement ‘x’.
Total resultant force on the particle is kx due to spring C and
2
kx
due to spring A and B.
 Total Resultant force = kx +
22
2
kx
2
kx












= kx + kx = 2kx.
Acceleration =
m
kx2
onAccelerati
ntdisplaceme
= 2
m
kx2
x
= 2
k2
m
[Cause:- When the body pushed against ‘C’ the spring C, tries to pull the block towards
XL. At that moment the spring A and B tries to pull the block with force
2
kx
and
2
kx
respectively towards xy and xz respectively. So the total force on the block is due to the spring force
‘C’ as well as the component of two spring force A and B.]
20. In this case, if the particle ‘m’ is pushed against ’C’ a by distance ‘x’.
Total resultant force acting on man ‘m’ is given by,
F = kx +
2
kx
=
2
kx3
[Because net force A & B = 























120cos
2
kx
2
kx
2
2
kx
2
kx
22
=
2
kx
 a =
m
F
=
m2
kx3

x
a
=
m2
k3
= 
2
  =
m2
k3
Time period T =

2
= 2
k3
m2

21. K2 and K3 are in series.
Let equivalent spring constant be K4

4K
1
=
32 K
1
K
1
 =
32
32
KK
KK 
 K4 =
32
32
KK
KK

Now K4 and K1 are in parallel.
So equivalent spring constant k = k1 + k4 =
32
32
KK
KK

+ k1 =
32
312132
kk
kkkkkk


 T = 2
k
M
= 2
312132
32
kkkkkk
)kk(M


mK
k2
F
k1
Mk3
90°
45°
x m y
z
B
120°C
m
x
A
B x
2
kx
kx
45° 2
kx
120°
2
kx
kx
2
kx

Chapter 12
12.7
b) frequency =
T
1
=
)kk(M
kkkkkk
2
1
32
312132



c) Amplitude x =
k
F
=
313221
32
kkkkkk
)kk(F



22. k1, k2, k3 are in series,
k
1
=
321 k
1
k
1
k
1
  k =
313221
321
kkkkkk
kkk

k
m
= 2
321
313221
kkk
)kkkkkk(m 
= 2 






321 k
1
k
1
k
1
m
Now, Force = weight = mg.
 At k1 spring, x1 =
1k
mg
Similarly x2 =
2k
mg
and x3 =
3k
mg
PE1 = (1/2) k1 x1
2
=
2
1
1
k
Mg
k
2
1






= 1k
2
1
2
1
22
k
gm
=
1
22
k2
gm
Similarly PE2 =
2
22
k2
gm
and PE3 =
3
22
k2
gm
23. When only ‘m’ is hanging, let the extension in the spring be ‘ℓ’
So T1 = kℓ = mg.
When a force F is applied, let the further extension be ‘x’
T2 = k(x +ℓ)
Driving force = T2 –T1 = k(x + ℓ) – kℓ = kx
Acceleration =
m
K
T = 2
onAccelerati
ntdisplaceme
= 2
m
kx
x
= 2
k
m

24. Let us solve the problem by ‘energy method’.
Initial extension of the sprig in the mean position,
 =
k
mg
During oscillation, at any position ‘x’ below the equilibrium position, let the velocity of ‘m’ be v and
angular velocity of the pulley be ‘’. If r is the radius of the pulley, then v = r.
At any instant, Total Energy = constant (for SHM)
 (1/2) mv
2
+ (1/2)  
2
+ (1/2) k[(x +)
2
- 
2
] – mgx = Cosntant
 (1/2) mv
2
+ (1/2)  
2
+ (1/2) kx
2
– kx - mgx = Cosntant
 (1/2) mv
2
+ (1/2)  (v
2
/r
2
) + (1/2) kx
2
= Constant ( = mg/k)
Taking derivative of both sides eith respect to ‘t’,
mv
dt
dv
k
dt
dv
v
rdt
dv
2


 = 0
 




 
 2
r
ma = kx ( x =
dt
dx
and a =
dt
dx
)

x
a

2
r
m
k



 T = 2
k
r
m 2


Mk
T
Mk
I
k3
k2
k1
M

Chapter 12
12.8
25. The centre of mass of the system should not change during the motion. So, if the block ‘m’ on the left
moves towards right a distance ‘x’, the block on the right moves towards left a distance ‘x’. So, total
compression of the spring is 2x.
By energy method,
2
1
k (2x)
2
+
2
1
mv
2
+
2
1
mv
2
= C  mv
2
+ 2kx
2
= C.
Taking derivative of both sides with respect to ‘t’.
m × 2v
dt
dv
+ 2k × 2x
dt
dx
= 0
ma + 2kx = 0 [because v = dx/dt and a = dv/dt]

x
a
= –
m
k2
= 
2 
 =
m
k2
 Time period T = 2
k2
m 
26. Here we have to consider oscillation of centre of mass
Driving force F = mg sin 
Acceleration = a =
m
F
= g sin .
For small angle , sin  = .
 a = g  = g 





L
x
[where g and L are constant]
 a  x,
So the motion is simple Harmonic
onAccelerati
ntDisplaceme
= 2






L
gx
x
= 2
g
L

27. Amplitude = 0.1m
Total mass = 3 + 1 = 4kg (when both the blocks are moving together)
 T = 2
k
M
= 2
100
4
=
5
2
sec.
 Frequency =
2
5
Hz.
Again at the mean position, let 1kg block has velocity v.
KE. = (1/2) mv
2
= (1/2) mx
2
where x Amplitude = 0.1m.
 (1/2) ×(1 × v
2
) = (1/2) × 100 (0.1)
2
 v = 1m/sec …(1)
After the 3kg block is gently placed on the 1kg, then let, 1kg +3kg = 4kg block and the spring be one
system. For this mass spring system, there is so external force. (when oscillation takes place). The
momentum should be conserved. Let, 4kg block has velocity v.
 Initial momentum = Final momentum
 1 × v = 4 × v  v = 1/4 m/s (As v = 1m/s from equation (1))
Now the two blocks have velocity 1/4 m/s at its mean poison.
KEmass = (1/2) mv
2
= (1/2) 4 × (1/4)
2
= (1/2) × (1/4).
When the blocks are going to the extreme position, there will be only potential energy.
 PE = (1/2) k
2
= (1/2) × (1/4) where   new amplitude.
1/4 = 100 
2
  =
400
1
= 0.05m = 5cm.
So Amplitude = 5cm.
28. When the block A moves with velocity ‘V’ and collides with the block B, it transfers all energy to the
block B. (Because it is a elastic collision). The block A will move a distance ‘x’ against the spring, again
the block B will return to the original point and completes half of the oscillation.
k
x
mm
x
m
100N/m
3kg

Chapter 12
12.9
So, the time period of B is
2
k
m
2
=
k
m

The block B collides with the block A and comes to rest at that point.
The block A again moves a further distance ‘L’ to return to its original
position.
 Time taken by the block to move from M  N and N  M
is
V
L
V
L
 = 





V
L
2
 So time period of the periodic motion is
k
m
V
L
2 





29. Let the time taken to travel AB and BC be t1 and t2 respectively
Fro part AB, a1 = g sin 45°. s1 =
45sin
1.0
= 2m
Let, v = velocity at B
 v
2
– u
2
= 2a1 s1
 v
2
= 2 × g sin 45° ×
45sin
1.0
= 2
 v = 2 m/s
 t1 =
1a
uv 
=
2
g
02 
=
g
2
=
10
2
= 0.2 sec
Again for part BC, a2 = –g sin 60°, u = 2 , v = 0
t2 =










2
3
g
20
=
g3
22
=
10)732.1(
)414.1(2


= 0.165sec.
So, time period = 2 (t1 + t2) = 2(0.2 + 0.155) = 0.71sec
30. Let the amplitude of oscillation of ‘m’ and ‘M’ be x1 and x2 respectively.
a) From law of conservation of momentum,
mx1 = Mx2 …(1) [because only internal forces are present]
Again, (1/2) kx0
2
= (1/2) k (x1 + x2)
2
 x0 = x1 +x2 …(2)
[Block and mass oscillates in opposite direction. But x  stretched part]
From equation (1) and (2)
 x0 = x1 +
M
m
x1 = 




 
M
mM
x1
 x1
mM
Mx0

So, x2 = x0 – x1 = x0 







mM
M
1 =
mM
mx0

respectively.
b) At any position, let the velocities be v1 and v2 respectively.
Here, v1 = velocity of ‘m’ with respect to M.
By energy method
Total Energy = Constant
(1/2) Mv
2
+ (1/2) m(v1 –v2)
2
+ (1/2) k(x1 +x2)
2
= Constant …(i)
[v1 – v2 = Absolute velocity of mass ‘m’ as seen from the road.]
Again, from law of conservation of momentum,
m
v
m
mM
L
x
A BAB
R
A
B
C
45° 60°
10cm
x
m
k
M

Chapter 12
12.10
mx2 = mx1 x1 =
m
M
x2 ...(1)
mv2 = m(v1 –v2)  (v1 –v2) =
m
M
v2 …(2)
Putting the above values in equation (1), we get
2
1
Mv2
2
+
2
1
m 2
2
m
M
v2
2
+
2
1
kx2
2
2
m
M
1 





 = constant
 M 






m
M
1 v2 + k
2
m
M
1 





 x2
2
= Constant.
 mv2
2
+ k 






m
M
1 x2
2
= constant
Taking derivative of both sides,
M × 2v2
dt
dv2
+ k
 
m
mM 
– ex2
2
dt
dx2
= 0
 ma2 + k 




 
m
mM
x2 = 0 [because, v2 =
dt
dx2
]

2
2
x
a
= –
Mm
)mM(k 
= 
2
  =
Mm
)mM(k 
So, Time period, T = 2
)mM(k
Mm


31. Let ‘x’ be the displacement of the plank towards left. Now the centre of gravity is also displaced through ‘x’
In displaced position
R1 + R2 = mg.
Taking moment about G, we get
R1(ℓ/2 – x) = R2(ℓ/2 + x) = (mg – R1)(ℓ/2 + x) …(1)
So, R1 (ℓ/2 – x) = (mg – R1)(ℓ/2 + x)
 R1
2

– R1 x = mg
2

– R1 x + mgx – R1
2

 R1
2

+R1
2

= mg (x+
2

)
 R1 






22

= mg 




 
2
x2 
 R1 ℓ =
2
)x2(mg 
 R1 =


2
)x2(mg 
…(2)
Now F1 = R1 =


2
)x2(mg 

Similarly F2 =R2 =


2
)x2(mg 
Since, F1 > F2.  F1 –F2 =ma = x
mg2



x
a
=

g2
= 
2
  =

g2
 Time period = 2
rg2

v1
x2
x1
m
M
y2

Chapter 12
12.11
32. T = 2sec.
T = 2
g

 2 = 2
10


10

= 2
1

 ℓ = 1cm (
2
 10)
33. From the equation,
 =  sin [ sec
–1
t]
  =  sec
–1
(comparing with the equation of SHM)

T
2
=   T = 2 sec.
We know that T = 2
g

  2 = 2
g

  1 =
g

  ℓ = 1m.
 Length of the pendulum is 1m.
34. The pendulum of the clock has time period 2.04sec.
Now, No. or oscillation in 1 day =
2
360024
= 43200
But, in each oscillation it is slower by (2.04 – 2.00) = 0.04sec.
So, in one day it is slower by,
= 43200 × (0.04) = 12 sec = 28.8 min
So, the clock runs 28.8 minutes slower in one day.
35. For the pendulum,
2
1
T
T
=
1
2
g
g
Given that, T1 = 2sec, g1 = 9.8m/s
2
T2 =





 

2
24360024
360024
=
3599
3600
2
Now,
1
2
g
g
=
2
2
1
T
T






g2 = (9.8)
2
3600
3599






= 9.795m/s
2
36. L = 5m.
a) T = 2
g

= 2 5.0 = 2(0.7)
 In 2(0.7)sec, the body completes 1 oscillation,
In 1 second, the body will complete
)7.0(2
1

oscillation
f =
)7.0(2
1

=
14
10
=

70.0
times
b) When it is taken to the moon
T = 2
g

where g Acceleration in the moon.
= 2
67.1
5

f =
T
1
=
5
67.1
2
1

=
2
1
(0.577) =
32
1

times.

Chapter 12
12.12
37. The tension in the pendulum is maximum at the mean position and minimum on the extreme position.
Here (1/2) mv
2
– 0 = mg ℓ(1 – cos )
v
2
= 2gℓ(1 – cos)
Now, Tmax = mg + 2 mg (1 – cos ) [ T = mg +(mv
2
/ℓ)]
Again, Tmin = mg cos.
According to question, Tmax = 2Tmin
 mg + 2mg – 2mg cos= 2mg cos
 3mg = 4mg cos
 cos  = 3/4
= cos
–1
(3/4)
38. Given that, R = radius.
Let N = normal reaction.
Driving force F = mg sin.
Acceleration =a = g sin 
As, sin  is very small, sin  
Acceleration a = g
Let ‘x’ be the displacement from the mean position of the body,
  = x/R
 a = g = g(x/R)  (a/x) = (g/R)
So the body makes S.H.M.
T = 2
onAccelerati
ntDisplaceme
= 2
R/gx
x
= 2
g
R

39. Let the angular velocity of the system about the point os suspension at any time be ‘’
So, vc = (R – r)
Again vc = r1 [where, 1 = rotational velocity of the sphere]
1 =
r
vc
= 




 
r
rR
 …(1)
By Energy method, Total energy in SHM is constant.
So, mg(R – r)(1 – cos) + (1/2) mvc
2
+(1/2) I1
2
= constant
 mg(R – r) (1 – cos) +(1/2) m(R – r)
2

2
+(1/2) mr
2
2
r
rR





 

2
= constant
 g(R – r) 1 – cos) + (R – r)
2

2







5
1
2
1
= constant
Taking derivative, g(R – r) sin 
dt
d

10
7
R – r)
2
2
dt
d

 g sin  = 2 ×
10
7
(R – r)
 g sin  =
5
7
(R – r)
  =
)rR(7
sing5


=
)rR(7
g5





= 
2
=
)rR(7
g5


= constant
So the motion is S.H.M. Again  = 
)rR(7
g5

T = 2
g5
)rR(7 

40. Length of the pendulum = 40cm = 0.4m.
Let acceleration due to gravity be g at the depth of 1600km.
gd = g(1-d/R) = 9.8 






6400
1600
1 = 9.8 






4
1
1 = 9.8 ×
4
3
= 7.35m/s
2
mg
Tmin

x
mv
mx
2

mg
Tmin
ℓ
L

x
N
mg

R
mg
mg cos 
mg sin 
R–rR
w
A
B


mg
(R–r)cos 
R
B
(R – r)

Chapter 12
12.13
 Time period T = 2
g

= 2
35.7
4.0
= 2 054.0 = 2 × 0.23 = 2 × 3.14 × 0.23 = 1.465  1.47sec.
41. Let M be the total mass of the earth.
At any position x,

M
M
=
3
3
R
3
4
x
3
4














= 3
3
R
x
 M = 3
3
R
Mx
So force on the particle is given by,
FX = 2
x
mMG 
= x
R
GMm
3
…(1)
So, acceleration of the mass ‘M’ at that position is given by,
ax = x
R
GM
2

x
ax
= w
2
= 3
R
GM
=
R
g






 2
R
GM
g
So, T = 2
g
R
= Time period of oscillation.
a) Now, using velocity – displacement equation.
V =  )RA( 22
 [Where, A = amplitude]
Given when, y = R, v = gR ,  =
R
g
 gR =
R
g
)RA( 22
 [because  =
R
g
]
 R
2
= A
2
– R
2
 A = 2 R
[Now, the phase of the particle at the point P is greater than /2 but less than  and at Q is greater
than  but less than 3/2. Let the times taken by the particle to reach the positions P and Q be t1 & t2
respectively, then using displacement time equation]
y = r sin t
We have, R = 2 R sin t1  t1 = 3/4
& –R = 2 R sin t2  t2 = 5/4
So, (t2 – t1) = /2  t2 – t1 =


2
=
)g/R(2

Time taken by the particle to travel from P to Q is t2 – t1 =
)g/R(2

sec.
b) When the body is dropped from a height R, then applying conservation of energy, change in P.E. =
gain in K.E.

R2
GMm
R
GMm
 =
2
1
mv
2
 v = gR
Since, the velocity is same at P, as in part (a) the body will take same time to travel PQ.
c) When the body is projected vertically upward from P with a velocity gR , its velocity will be Zero at
the highest point.
The velocity of the body, when reaches P, again will be v = gR , hence, the body will take same
time
)g/R(2

to travel PQ.
Q
M
gR
P A
R
m
x

Chapter 12
12.14
42. M = 4/3 R
3
.
M
1
= 4/3 x1
3

M
1
= 





3
R
M
x1
3
a) F = Gravitational force exerted by the earth on the particle of mass ‘x’ is,
F = 2
1
1
x
mGM
= 2
1
3
1
3
x
x
R
GMm
= 13
x
R
GMm
=









4
R
x
R
GMm 2
2
3
b) Fy = F cos  =
1
3
1
x
x
R
GMmx
= 3
R
GMmx
Fx = F sin  =
1
3
1
x2
R
R
GMmx
= 2
R2
GMm
c) Fx = 2
R2
GMm
[since Normal force exerted by the wall N = Fx]
d) Resultant force = 3
R
GMmx
e) Acceleration =
mass
forceDriving
=
mR
GMmx
3
= 3
R
GMx
So, a  x (The body makes SHM)

x
a
= w
2
= 3
R
GM
 w = 3
R
GM
 T = 2
GM
R3

43. Here driving force F = m(g + a0) sin  …(1)
Acceleration a =
m
F
= (g + a0) sin  =

x)a(g 0
(Because when  is small sin   x/ℓ)
 a =

x)a(g 0
.
 acceleration is proportional to displacement.
So, the motion is SHM.
Now 
2
=

)a(g 0
 T = 2
0ag 

b) When the elevator is going downwards with acceleration a0
Driving force = F = m (g – a0) sin .
Acceleration = (g – a0) sin  =

x)ag( 0
= 
2
x
T =

2
= 2
0ag 

c) When moving with uniform velocity a0 = 0.
For, the simple pendulum, driving force =

mgx
 a =

gx

a
x
=
g

T = 2
onaccelerati
ntdisplaceme
= 2
g


M
R/2
m
x
A
x1
C
NFx
Fg

B
A
L
mg m(g+a0)sin 
mg
ma0

m(g+a0)sin 
mg
ma0
a0

B
A
L
mg
x

Chapter 12
12.15
44. Let the elevator be moving upward accelerating ‘a0’
Here driving force F = m(g + a0) sin 
Acceleration = (g + a0) sin 
= (g + a0) (sin   )
=
 

xag 0
= 
2
x
T = 2
0ag 

Given that, T = /3 sec, ℓ = 1ft and g = 32 ft/sec
2
3

= 2
0a32
1

9
1
= 4 





 a32
1
 32 + a =36  a = 36 – 32 = 4 ft/sec
2
45. When the car moving with uniform velocity
T = 2
g

 4 = 2
g

…(1)
When the car makes accelerated motion, let the acceleration be a0
T = 2 2
0
2
ag 

 3.99 = 2 2
0
2
ag 

Now
99.3
4
T
T


=
 
g
ag
4/12
0
2

Solving for ‘a0’ we can get a0 = g/10 ms
–2

46. From the freebody diagram,
T =








 2
2
2
r
mv
)mg(
= m 2
4
2
r
v
g  = ma, where a = acceleration =
2/1
2
4
2
r
v
g 








The time period of small accellations is given by,
T = 2
g

= 2 2/1
2
4
2
r
v
g










47. a) ℓ = 3cm = 0.03m.
T = 2
g

= 2
8.9
03.0
= 0.34 second.
b) When the lady sets on the Merry-go-round the ear rings also experience
centrepetal acceleration
a =
r
v2
=
2
42
= 8 m/s
2
Resultant Acceleration A = 22
ag  = 64100  = 12.8 m/s
2
A

= 2
8.12
03.0
= 0.30 second.
mg
a0

B
A
L
mg
x
ma0
ℓ
mg
mv2
/r
mg
T
mv2
/r
mg
g
v2
/r
A

Chapter 12
12.16
48. a) M.I. about the pt A = I = IC.G. + Mh
2
=
12
m 2

+ MH
2
=
12
m 2

+ m (0.3)
2
= 





 09.0
12
1
M = M 




 
12
08.11
= M 





12
08.2
 T = 2
mg
I
= 2
3.08.9m
m08.2

(ℓ = dis. between C.G. and pt. of suspension)
 1.52 sec.
b) Moment of in isertia about A
I = IC.G.+ mr
2
= mr
2
+ mr
2
= 2 mr
2
 Time period = 2
mg
I
= 2
mgr
mr2 2
= 2
g
r2
c) IZZ (corner) = m







 
3
aa 22
=
3
ma2 2
In the ABC, ℓ
2
+ ℓ
2
= a
2
 ℓ =
2
a
 T = 2
mg
I
= 2
mg3
ma2 2
= 2
2ga3
a2 2
= 2
g3
a8
d) h = r/2, ℓ = r/2 = Dist. Between C.G and suspension point.
M.I. about A, I = IC.G.+ Mh
2
=
22
2
r
n
2
mc






 = mr
2







4
1
2
1
=
4
3
mr
2
 T = 2
mg
I
 = 2
mg4
mr3 2
= 2






2
r
g4
r3 2
= 2
g2
r3

49. Let A  suspension of point.
B  Centre of Gravity.
ℓ = ℓ/2, h = ℓ/2
Moment of inertia about A is
I = IC.G. + mh
2
=
4
m
12
m 22

 =
3
m 2

 T = 2






2
mg
I

= 2
mgl3
m2 2

= 2
g3
2
Let, the time period ‘T’ is equal to the time period of simple pendulum of length ‘x’.
 T = 2
g
x
. So,
g3
2
=
g
x
 x =
3
2
 Length of the simple pendulum =
3
2

50. Suppose that the point is ‘x’ distance from C.G.
Let m = mass of the disc., Radius = r
Here ℓ = x
M.I. about A = IC.G. + mx
2
= mr
2
/2+mx
2
= m(r
2
/2 + x
2
)
T = 2
mg
I
= 2
mgx
x
2
r
m 2
2









= 2
 
mgx2
x2rm 22

= 2
gx2
x2r 22

…(1)
20cm
A
O
B
30cm
A
C.G
r
ℓ
ℓC.G

Chapter 12
12.17
For T is minimum
dx
dt2
=0

dx
d
T
2
=







 


gx2
x24
gx2
r4
dx
d 2222

g
4
x
1
g
r2 2
2
22









= 0

g
2
gx
r 2
2
22



 = 0

g
2
gx
r 2
2
22



 2x
2
= r
2
 x =
2
r
So putting the value of equation (1)
T = 2
gx2
2
r
2r
2
2









= 2
gx2
r2 2
= 2






2
r
g
r2
= 2
gr
r2 2
= 2
g
r2

51. According to Energy equation,
mgℓ (1 – cos ) + (1/2) I
2
= const.
mg(0.2) (1 – cos) + (1/2) I
2
= C. (I)
Again, I = 2/3 m(0.2)
2
+ m(0.2)
2
= m 





 04.0
3
008.0
= m 





3
1208.0
m. Where I  Moment of Inertia about the pt of suspension A
From equation
Differenting and putting the value of I and 1 is






 2
m
3
1208.0
2
1
)cos1)(2.0(mg
dt
d

dt
d
(C)
 mg (0.2) sin
dt
d
+ 





3
1208.0
2
1
m2
dt
d

 2 sin  =
3
1208.0
 [because, g = 10m/s
2
]



=
1208.0
6
= 
2
= 58.36
  = 7.3. So T =

2
= 0.89sec.
For simple pendulum T = 2
10
19.0
= 0.86sec.
% more =
89.0
86.089.0 
= 0.3.
 It is about 0.3% larger than the calculated value.
52. (For a compound pendulum)
a) T = 2
mg
I
= 2
mgr
I

The MI of the circular wire about the point of suspension is given by
 I = mr
2
+ mr
2
= 2 mr
2
is Moment of inertia about A.
2cm
1.8cm

A
B
A
r
mg
mg

Chapter 12
12.18
 2 = 2
mgrmr2 2
= 2
g
r2
 2
1
g
r2

  r = 2
2
g

= 0.5 = 50cm. (Ans)
b) (1/2) 
2
– 0 = mgr (1 – cos)
 (1/2) 2mr
2
– 
2
= mgr (1 – cos 2°)
 
2
= g/r (1 – cos 2°)
  = 0.11 rad/sec [putting the values of g and r]
 v =  × 2r = 11 cm/sec.
c) Acceleration at the end position will be centripetal.
= an = 
2
(2r) = (0.11)
2
× 100 = 1.2 cm/s
2
The direction of ‘an’ is towards the point of suspension.
d) At the extreme position the centrepetal acceleration will be zero. But, the particle will still have
acceleration due to the SHM.
Because, T = 2 sec.
Angular frequency  =
T
2
(= 3.14)
So, angular acceleration at the extreme position,
 = 
2
 = 
2
×
180
2
=
180
2 3

[1° =
180

radious]
So, tangential acceleration = (2r) =
180
2 3

× 100 = 34 cm/s
2
. 
53. M.I. of the centre of the disc. = mr
2
/2
T = 2
k
I
= 2
K2
mr2
[where K = Torsional constant]
T
2
= 4
2
K2
mr2
= 2
2
K
mr2
 2
2
mr
2
= KT
2
 2
22
T
mr2 

Torsional constant  2
22
T
mr2 
54. The M.I of the two ball system
I = 2m (L/2)
2
= m L
2
/2
At any position  during the oscillation, [fig-2]
Torque = k
So, work done during the displacement 0 to 0,
W = 


0
dk = k 0
2
/2
By work energy method,
(1/2) I
2
– 0 = Work done = k 0
2
/2
 
2
=
I2
k
2
0
= 2
2
0
mL
k
Now, from the freebody diagram of the rod,
T2 = 222
)mg()Lm( 
= 22
2
2
2
0
gmL
mL
k
m 










= 22
2
4
0
2
gm
L
k



L
Fig-1
mm



Chapter 12
12.19
55. The particle is subjected to two SHMs of same time period in the same direction/
Given, r1 = 3cm, r2 = 4cm and  = phase difference.
Resultant amplitude = R =  cosrr2rr 21
2
2
2
1
a) When  = 0°,
R =  0cos43243( 22
= 7 cm
b) When  = 60°
R =  60cos43243( 22
= 6.1 cm
c) When 
R =  90cos43243( 22
= 5 cm
56. Three SHMs of equal amplitudes ‘A’ and equal time periods in the same dirction combine.
The vectors representing the three SHMs are shown it the figure.
Using vector method,
Resultant amplitude = Vector sum of the three vectors
= A + A cos 60° + A cso 60° = A + A/2 + A/2 = 2A
So the amplitude of the resultant motion is 2A.
57. x1 = 2 sin 100 t
x2 = w sin (120t + /3)
So, resultant displacement is given by,
x = x1 + x2 = 2 [sin (100t) + sin (120t + /3)]
a) At t = 0.0125s,
x = 2 [sin (100× 0.0125) + sin (120 ×0.0125 + /3)]
= 2 [sin 5+ sin (3/2 + /3)]
= 2 [(–0.707) + (–0.5)] = – 2.41cm.
b) At t = 0.025s.
x = 2 [sin (100× 0.025) + sin (120 ×0.025 + /3)]
= 2 [sin 5+ sin (3 + /3)]
=2[1+(–0.8666)] = 0.27 cm.
58. The particle is subjected to two simple harmonic motions represented by,
x = x0 sin wt
s = s0 sin wt
and, angle between two motions =  = 45°
Resultant motion will be given by,
R = )45cosxs2sx( 22

= )}2/1(wtxsinsx2wtsinswtsinx{ 2
00
22
0
22
0 
= [x0
2
+s0
2
= 2 x0s0]
1/2
sin wt
 Resultant amplitude = [x0
2
+s0
2
= 2 x0s0]
1/2
♠ ♠ ♠ ♠ ♠
Y1
Y2
Y3
A
A
A
60°
60°

13.1
CHAPTER 13
1. p = h  g
It is necessary to specify that the tap is closed. Otherwise pressure will gradually decrease, as h
decrease, because, of the tap is open, the pressure at the tap is atmospheric.
2. a) Pressure at the bottom of the tube should be same when considered for both limbs.
From the figure are shown,
pg + Hg × h2 × g = pa + Hg × h1 × g
 pg = pa + Hg × g(h1 – h2)
b) Pressure of mercury at the bottom of u tube
p = pa + Hg h1 × g
3. From the figure shown
pa + hg = pa + mg/A
 hg = mg/A
 h =
Ap
m

4. a) Force exerted at the bottom.
= Force due to cylindrical water colum + atm. Force
= A × h × w × g + pa × A
= A(h w g + pa)
b) To find out the resultant force exerted by the sides of the glass, from the freebody, diagram of water
inside the glass
pa × A + mg = A × h × w × g + Fs + pa × A
 mg = A × h × w × g + Fs
This force is provided by the sides of the glass.
5. If the glass will be covered by a jar and the air is pumped out, the atmospheric pressure has no effect.
So,
a) Force exerted on the bottom.
= (h w g) × A
b) mg = h × w × g × A × Fs.
c) It glass of different shape is used provided the volume, height and area remain same, no change in
answer will occur.
6. Standard atmospheric pressure is always pressure exerted by 76 cm Hg column
= (76 × 13.6 × g) Dyne/cm
2
.
If water is used in the barometer.
Let h  height of water column.
 h × w × g
7. a) F = P × A = (h w × g) A
b) The force does not depend on the orientation of the rock as long as the surface area remains same.
8. a) F = A h  g.
b) The force exerted by water on the strip of width x as shown,
dF = p × A
= (xg) × A
c) Inside the liquid force act in every direction due to adhesion.
di = F × r
d) The total force by the water on that side is given by
F = 
1
0
20000 xx  F = 20,000 1
0
2
]2/x[
e) The torque by the water on that side will be,
Pa
Pa
Gas
Pa
h pa
45 kg
A =900 cm2

Chapter-13
13.2
i = 
1
0
20000 xx (1 – x)  20,000 1
0
32
]3/x2/x[  
9. Here, m0 = mAu + mcu = 36 g …(1)
Let V be the volume of the ornament in cm
3
So, V × w × g = 2 × g
 (Vau + Vcu) × w × g = 2 × g
 g
mm
w
auau











= 2 × g
 1
9.8
m
3.19
m AuAu






 = 2
 8.9 mAu + 19.3 mcu = 2 × 19.3 × 8.9 = 343.54 …(2)
From equation (1) and (2), 8.9 mAu + 19.3 mcu = 343.54

g225.2m
369.8)mm(9.8
cu
cuAu


So, the amount of copper in the ornament is 2.2 g.
10. gV
M
wc
Au
Au










= 2 × g (where Vc = volume of cavity)
11. mg = U + R (where U = Upward thrust)
 mg – U = R
 R = mg – v w g (because, U = vwg)
= mg –

m
× w × g
12. a) Let Vi  volume of boat inside water = volume of water displace in m
3
.
Since, weight of the boat is balanced by the buoyant force.
 mg = Vi × w × g
b) Let, v1
 volume of boat filled with water before water starts coming in from the sides.
mg + v1
w × g = V × w × g.
13. Let x  minimum edge of the ice block in cm.
So, mg + Wice = U. (where U = Upward thrust)
 0.5 × g + x
3
× ice × g = x
3
× w × g
14. Vice = Vk + Vw
Vice × ice × g = Vk × k × g + Vw × w × g
 (Vk + Vw) × ice = Vk × k + Vw × w
 1
V
V
k
w
 .
15. Viig = V w g
16. (mw + mpb)g = (Vw + Vpb)  × g
 (mw + mpb) = 










 pb
pb
w
w
mm
17. Mg = w  (mw + mpb)g = Vw ×  × g
18. Given, x = 12 cm
Length of the edge of the block Hg = 13.6 gm/cc
Given that, initially 1/5 of block is inside mercuty.
Let b  density of block in gm/cc.
 (x)
3
× b × g = (x)
2
× (x/5) × Hg × g
 123
× b = 122
× 12/5 × 13.6
 b =
5
6.13
gm/cc

Chapter-13
13.3
After water poured, let x = height of water column.
Vb = VHg + Vw = 12
3
Where VHg and Vw are volume of block inside mercury and water respectively
(Vb × b × g) = (VHg × Hg × g) + (Vw × w × g)
 (VHg + Vw)b = VHg × Hg + Vw × w.
 (VHg + Vw) ×
5
6.13
= VHg × 13.6 + Vw × 1
 (12)
3
×
5
6.13
= (12 – x) × (12)
2
× 13.6 + (x) × (12)
2
× 1
 x = 10.4 cm
19. Here, Mg = Upward thrust
 Vg = (V/2) (w) × g (where w = density of water)
 w
3
2
3
1
3
2 r
3
4
2
1
r
3
4
r
3
4



















 1r
2
1
)rr(
3
2
3
1
3
2  = 865 kg/m
3
.
20. W1 + W2 = U.
 mg + V × s × g = V × w × g (where s = density of sphere in gm/cc)
 1 – s = 0.19
 s = 1 – (0.19) = 0.8 gm/cc
So, specific gravity of the material is 0.8.
21. Wi = mg – Vi air × g = g
m
m air
i











Ww = mg – Vw air g = g
m
m air
w










 
22. Driving force U = Vwg
 a = r2
(X) × w g  T =
onAccelerati
ntdisplaceme
2
23. a) F + U = mg (where F = kx)
 kx + Vwg = mg
b) F = kX + Vw × g
 ma = kX + r
2
× (X) × w × g = (k + r
2
× w × g)X
 
2
× (X) = )X(
m
)grk( w
2


 T =
grK
m
2
w
2


24. a) mg = kX + Vwg
b) a = kx/m
w
2
x = kx/m
T = k/m2
25. Let x  edge of ice block
When it just leaves contact with the bottom of the glass.
h  height of water melted from ice
W = U
 x
3
× ice × g = x
2
× h × w × g
Again, volume of water formed, from melting of ice is given by,
4
3
– x
3
=  × r
2
× h – x
2
h ( because amount of water = (r
2
– x
2
)h)
 4
3
– x
3
=  × 3
2
× h – x
2
h
Putting h = 0.9 x  x = 2.26 cm.

Chapter-13
13.4
26. If pa  atm. Pressure
A  area of cross section
h  increase in hright
paA + A × L ×  × a0 = pa
A
+ hg × A
 hg = a0L  a0L/g
27. Volume of water, discharged from Alkananda + vol are of water discharged from Bhagirathi = Volume of
water flow in Ganga.
28. a) aA × VA = QA
b) aA × VA = aB × VB
c) 1/2 vA
2
+ pA = 1/2 vB
2
+ pB
 (pA – pB) = 1/2  (vB
2
– vA
2
)
29. From Bernoulli’s equation, 1/2 vA
2
+ ghA + pA
= 1/2 vB
2
+ ghB + pB.
 PA – PB = (1/2)  (vB
2
– vA
2
) + g (hB – hA)
30. 1/2 vB
2
+ ghB + pB = 1/2 vA
2
+ ghA + pA
31. 1/2 vA
2
+ ghA + pA =1/2 vB
2
+ ghB + pB
 PB – PA = 1/2 (vA
2
– vB
2
) + g (hA – hB)
32. BBAA avav 

 1/2 vA
2
+ ghA + pA = 1/2 vB
2
+ ghB + pB
 1/2 vA
2
+ pA = 1/2 vB
2
+ pB
 PA – PB = 1/2 (vB
2
– vB
2
)
Rate of flow = va × aA
33. VA aA = vB aB 
A
BA
a
a
B
v

5vA = 2vB  vB = (5/2)vA
1/2 vA
2
+ ghA + pA = 1/2 vB
2
+ ghB + pB
 PA – PB = 1/2  (vB
2
– vB
2
) (because PA – PB = hmg)
34. PA + (1/2)vA
2
= PB + (1/2) vB
2
 pA – pB = (1/2)vB
2
{vA = 0}
 gh = (1/2) vB
2
{pA = patm + gh}
 vB = gh2
a) v = gh2
b) v = gh)2/h(g2 
c) v = gh2
v = av × dt
AV = av
 A × 2gha
dt
dh
  dh =
A
dtgh2a 
d) dh =
A
dtgh2a 
 T = ]HH[
g
2
a
A
21 
35. v = )hH(g2 
t = g/h2
x = v × t = g/h2)hH(g2  = )hHh(4 2

So,  0)hHh(
dh
d 2






 0 = H – 2h  h = H/2.
   

14.1
CHAPTER 14
1. F = mg
Stress =
F
A
Strain =
L
L

Y =
FL L F
A L L YA

 

2.  = stress = mg/A
e = strain = /Y
Compression L = eL
3. y =
F L FL
L
A L AY
  

4. Lsteel = Lcu and Asteel = Acu
a)
gcu
cu g
AFStress of cu
Stress of st A F
 = cu
st
F
1
F

b) Strain = st st cu cu
st st cu cu
F L A YLst
lcu A Y F I

 

( Lcu = Ist ; Acu = Ast)
5.
st st
L F
L AY
 
 
 
cu cu
L F
L AY
 
 
 
cu cu
cu st
st st
AY Ystrain steel wire F
( A A )
Strain om copper wire AY F Y
   
6. Stress in lower rod = 1 1
1 1
T m g g
A A
 
  w = 14 kg
Stress in upper rod = 2 2 1
u u
T m g m g wg
A A
 
  w = .18 kg
For same stress, the max load that can be put is 14 kg. If the load is increased the lower wire will break
first.
1 1
1 1
T m g g
A A
 
 = 8  10
8
 w = 14 kg
2 2 1
u u
T m g m g g
A A
  
 = 8  10
8
 0 = 2 kg
The maximum load that can be put is 2 kg. Upper wire will break first if load is increased.
7.
F L
Y
A L


8.
F L
Y
A L



YA L
F
L


9. m2g – T = m2a …(1)
and T – F = m1a …(2)
 a = 2
1 2
m g F
m m



Chapter-14
14.2
From equation (1) and (2), we get 2
1 2
m g
2(m m )
Again, T = F + m1a
 2 2
1
1 2
m g m g
T m
2 2(m m )
 


2
2 1 2
1 2
m g 2m m g
2(m m )


Now Y =
FL L F
A L L AY

 


2
2 1 2 2 2 1
1 2 1 2
(m 2m m )g m g(m 2m )L
L 2(m m )AY 2AY(m m )
 
 
 
10. At equilibrium  T = mg
When it moves to an angle , and released, the tension the T at lowest point is
 T = mg +
2
mv
r
The change in tension is due to centrifugal force T =
2
mv
r
…(1)
 Again, by work energy principle,
 21
mv
2
– 0 = mgr(1 – cos)
 v
2
= 2gr (1 – cos) …(2)
So,
m[2gr(1 cos )]
T 2mg(1 cos )
r
 
    
 F = T
 F =
YA L
L

= 2mg – 2mg cos   2mg cos  = 2mg –
YA L
L

= cos  = 1 –
YA L
L(2mg)


11. From figure cos  =
2 2
x
x l
=
1/ 22
2
x x
1
l l

 
 
 
= x / l … (1)
Increase in length L = (AC + CB) – AB
Here, AC = (l
2
+ x
2
)
1/2
So, L = 2(l
2
+ x
2
)
1/2
– 100 …(2)
Y =
F l
A l
…(3)
From equation (1), (2) and (3) and the freebody diagram,
2l cos = mg.
12. Y =
FL
A L

L F
L Ay


 =
D/D
L /L



D L
D L
 

Again,
A 2 r
A r
 


2 r
A
r

  
m1
m2
m2g
a
T
T
a
F

BA
T T
l l
T x
mg
C
L L

Chapter-14
14.3
13. B =
Pv
v
 P =
v
B
v
 
 
 
14. 0
0 d
m m
V V
  
so, d 0
0 d
V
V



…(1)
vol.strain = 0 d
0
V V
V

B = 0
0 d 0
gh
(V V )/ V


 1 – d
0
V
V
= 0gh
B

 0
0
ghvD
1
v B
 
  
 
…(2)
Putting value of (2) in equation (1), we get
d
0 0
1
1 gh/B


  
 d 0
0
1
(1 gh/B)
  
 
15.
F
A
 

Lateral displacement = l.
16. F = T l
17. a)
Hg2T
P
r
 b)
g4T
P
r
 c)
g2T
P
r

18. a) F = P0A
b) Pressure = P0 + (2T/r)
F = PA = (P0 + (2T/r)A
c) P = 2T/r
F = PA =
2T
A
r
19. a) A
A
2Tcos
h
r g


 
b) B
B
2Tcos
h
r g



c) C
C
2Tcos
h
r g



20.
Hg Hg
Hg
Hg
2T cos
h
r g



2T cos
h
r g
 





where, the symbols have their usual meanings.
Hg
Hg Hg Hg
h T cos
h T cos
  

 
  
 
21.
2Tcos
h
r g



22. P =
2T
r
P = F/r
23. A = r
2
24. 3 34 4
R r 8
3 3
   
 r = R/2 = 2
Increase in surface energy = TA – TA

Chapter-14
14.4
25. h =
2Tcos
r g


, h =
2Tcos
r g


 cos  =
h r g
2T
 
So,  = cos
–1
(1/2) = 60°.
26. a) h =
2Tcos
r g


b) T  2r cos  = r
2
h    g
  cos  =
hr g
2T


27. T(2l) = [1  (10
–3
)  h]g
28. Surface area = 4r
2
29. The length of small element = r d 
dF = T  r d 
considering symmetric elements,
dFy = 2T rd . sin [dFx = 0]
so, F =
/ 2
0
2Tr sin d

  = / 2
02Tr[cos ]
 = T  2 r
Tension  2T1 = T  2r  T1 = Tr
30. a) Viscous force = 6rv
b) Hydrostatic force = B = 34
r g
3
 
  
 
c) 6 rv + 34
r g
3
 
  
 
= mg
v =
2
2 r ( )g
9
  


3
2
m
g
(4/3) r2
r
9 n
 
  
 
31. To find the terminal velocity of rain drops, the forces acting on the drop are,
i) The weight (4/3) r
3
g downward.
ii) Force of buoyancy (4/3) r
3
g upward.
iii) Force of viscosity 6  r v upward.
Because,  of air is very small, the force of buoyancy may be neglected.
Thus,
6  r v = 24
r g
3
 
  
 
or v =
2
2r g
9



32. v =
R
D


 R =
v D

   

15.1
CHAPTER 15
1. v = 40 cm/sec
As velocity of a wave is constant location of maximum after 5 sec
= 40  5 = 200 cm along negative x-axis.
2. Given y =
2
[(x / a) (t / T)]
Ae 
a) [A] = [M
0
L
1
T
0
], [T] = [M
0
L
0
T
1
]
[a] = [M
0
L
1
T
0
]
b) Wave speed, v = /T = a/T [Wave length  = a]
c) If y = f(t – x/v)  wave is traveling in positive direction
and if y = f( t + x/v)  wave is traveling in negative direction
So, y =
2
[(x / a) (t / T)]
Ae 
=
2
x
(1/ T) t
a / T
Ae
 
   
=
2
x
(1/ T) t
v
Ae
 
   
i.e. y = f{t + (x / v)}
d) Wave speed, v = a/T
 Max. of pulse at t = T is (a/T)  T = a (negative x-axis)
Max. of pulse at t = 2T = (a/T)  2T = 2a (along negative x-axis)
So, the wave travels in negative x-direction.
3. At t = 1 sec, s1 = vt = 10  1 = 10 cm
t = 2 sec, s2 = vt = 10  2 = 20 cm
t = 3 sec, s3 = vt = 10  3 = 30 cm
4. The pulse is given by, y = [(a
3
) / {(x – vt)
2
+ a
2
}]
a = 5 mm = 0.5 cm, v = 20 cm/s
At t = 0s, y = a
3
/ (x
2
+ a
2
)
The graph between y and x can be plotted by taking different values of x.
(left as exercise for the student)
similarly, at t = 1 s, y = a
3
/ {(x – v)
2
+ a
2
}
and at t = 2 s, y = a
3
/ {(x – 2v)
2
+ a
2
}
5. At x = 0, f(t) = a sin (t/T)
Wave speed = v
  = wavelength = vT (T = Time period)
So, general equation of wave
Y = A sin [(t/T) – (x/vT)] [because y = f((t/T) – (x/))
6. At t = 0, g(x) = A sin (x/a)
a) [M
0
L
1
T
0
] = [L]
a = [M
0
L
1
T
0
] = [L]
b) Wave speed = v
 Time period, T = a/v (a = wave length = )
 General equation of wave
y = A sin {(x/a) – t/(a/v)}
= A sin {(x – vt) / a}
7. At t = t0, g(x, t0) = A sin (x/a) …(1)
For a wave traveling in the positive x-direction, the general equation is given by
y =
x t
f
a T
 
 
 
Putting t = –t0 and comparing with equation (1), we get
 g(x, 0) = A sin {(x/a) + (t0/T)}
 g(x, t) = A sin {(x/a) + (t0/T) – (t/T)}
x
y

Chapter 15
15.2
As T = a/v (a = wave length, v = speed of the wave)
 y = 0tx t
A sin
a (a / v) (a / v)
 
  
 
= 0x v(t t)
Asin
a
  
 
 
 y = 0x v(t t )
Asin
a
  
  
8. The equation of the wave is given by
y = (0.1 mm) sin [(31.4 m
–1
)x +(314 s
–1
)t] y = r sin {(2x / )} + t)
a) Negative x-direction
b) k = 31.4 m
–1
 2/ = 31.4   = 2/31.4 = 0.2 mt = 20 cm
Again,  = 314 s
–1
 2f = 314  f = 314 / 2 = 314 / (2  (3/14)} = 50 sec
–1
 wave speed, v = f = 20  50 = 1000 cm/s
c) Max. displacement = 0.10 mm
 Max. velocity = a = 0.1  10
–1
 314 = 3.14 cm/sec.
9. Wave speed, v = 20 m/s
A = 0.20 cm
  = 2 cm
a) Equation of wave along the x-axis
y = A sin (kx – wt)
 k = 2/ = 2/2 =  cm
–1
T = /v = 2/2000 = 1/1000 sec = 10
–3
sec
  = 2/T = 2  10
–3
sec
–1
So, the wave equation is,
 y = (0.2 cm)sin[( cm
–1
)x – (2  10
3
sec
–1
)t]
b) At x = 2 cm, and t = 0,
 y = (0.2 cm) sin (/2) = 0
 v = r cos x = 0.2  2000   cos 2 = 400 
= 400  (3.14) = 1256 cm/s
= 400  cm/s = 4 m/s
10. Y = (1 mm) sin 
x t
2cm 0.01sec
 
  
a) T = 2  0.01 = 0.02 sec = 20 ms
 = 2  2 = 4 cm
b) v = dy/dt = d/dt [sin 2 (x/4 – t/0.02)] = –cos2 {x/4) – (t/0.02)}  1/(0.02)
 v = –50 cos 2 {(x/4) – (t/0.02)}
at x = 1 and t = 0.01 sec, v = –50 cos 2* [(1/4) – (1/2)] = 0
c) i) at x = 3 cm, t = 0.01 sec
v = –50 cos 2 (3/4 – ½) = 0
ii) at x = 5 cm, t = 0.01 sec, v = 0 (putting the values)
iii) at x = 7 cm, t = 0.01 sec, v = 0
at x = 1 cm and t = 0.011 sec
v = –50 cos 2 {(1/4) – (0.011/0.02)} = –50 cos (3/5) = –9.7 cm/sec
(similarly the other two can be calculated)
11. Time period, T = 4  5 ms = 20  10
–3
= 2  10
–2
s
 = 2  2 cm = 4 cm
frequency, f = 1/T = 1/(2  10
–2
) = 50 s
–1
= 50 Hz
Wave speed = f = 4  50 m/s = 2000 m/s = 2 m/s

Chapter 15
15.3
12. Given that, v = 200 m/s
a) Amplitude, A = 1 mm
b) Wave length,  = 4 cm
c) wave number, n = 2/ = (2  3.14)/4 = 1.57 cm
–1
(wave number = k)
d) frequency, f = 1/T = (26/)/20 = 20/4 = 5 Hz
(where time period T = /v)
13. Wave speed = v = 10 m/sec
Time period = T = 20 ms = 20  10
–3
= 2  10
–2
sec
a) wave length,  = vT = 10  2  10
–2
= 0.2 m = 20 cm
b) wave length,  = 20 cm
 phase diff
n
= (2/) x = (2 / 20)  10 =  rad
 y1 = a sin (t – kx)  1.5 = a sin (t – kx)
So, the displacement of the particle at a distance x = 10 cm.
[ =
2 x 2 10
20
 
  

] is given by
y2 = a sin (t – kx + )  –a sin(t – kx) = –1.5 mm
 displacement = –1.5 mm
14. mass = 5 g, length l = 64 cm
 mass per unit length = m = 5/64 g/cm
 Tension, T = 8N = 8  10
5
dyne
V = 5
(T /m) (8 10 64)/5 3200    cm/s = 32 m/s
15.
a) Velocity of the wave, v = 5
(T /m) (16 10 )/0.4 2000   cm/sec
 Time taken to reach to the other end = 20/2000 = 0.01 sec
Time taken to see the pulse again in the original position = 0.01  2 = 0.02 sec
b) At t = 0.01 s, there will be a ‘though’ at the right end as it is reflected.
16. The crest reflects as a crest here, as the wire is traveling from denser to rarer medium.
 phase change = 0
a) To again original shape distance travelled by the wave S = 20 + 20 = 40 cm.
Wave speed, v = 20 m/s  time = s/v = 40/20 = 2 sec
b) The wave regains its shape, after traveling a periodic distance = 230 = 60 cm
 Time period = 60/20 = 3 sec.
c) Frequency, n = (1/3 sec
–1
)
n = (1/2l) (T /m) m = mass per unit length = 0.5 g/cm
 1/3 = 1/(2  30) (T /0.5)
 T = 400  0.5 = 200 dyne = 2  10
–3
Newton.
17. Let v1 = velocity in the 1
st
string
 v1 = 1(T /m )
Because m1 = mass per unit length = (1a1l1 / l1) = 1a1 where a1 = Area of cross section
 v1 = 1 1(T / a ) …(1)
Let v2 = velocity in the second string
 v2 = 2
(T /m )
 v2 = 2 2(T / a ) …(2)
Given that, v1 = 2v2
 1 1(T / a ) = 2 2 2(T / a )  (T/a11) = 4(T/a22)
 1/2 = 1/4  1 : 2 = 1 : 4 (because a1 = a2)
20 cm
30 cm

Chapter 15
15.4
18. m = mass per unit length = 1.2  10
–4
kg/mt
Y = (0.02m) sin [(1.0 m
–1
)x + (30 s
–1
)t]
Here, k = 1 m
–1
= 2/
 = 30 s
–1
= 2f
 velocity of the wave in the stretched string
v = f = /k = 30/I = 30 m/s
 v = T /m  4
30 (T /1.2) 10 N)

 T = 10.8  10
–2
N  T = 1.08  10
–1
Newton.
19. Amplitude, A = 1 cm, Tension T = 90 N
Frequency, f = 200/2 = 100 Hz
Mass per unit length, m = 0.1 kg/mt
a)  V = T /m = 30 m/s
 = V/f = 30/100 = 0.3 m = 30 cm
b) The wave equation y = (1 cm) cos 2 (t/0.01 s) – (x/30 cm)
[because at x = 0, displacement is maximum]
c) y = 1 cos 2(x/30 – t/0.01)
 v = dy/dt = (1/0.01)2sin 2 {(x/30) – (t/0.01)}
a = dv/dt = – {4
2
/ (0.01)
2
} cos 2 {(x/30) – (t/0.01)}
When, x = 50 cm, t = 10 ms = 10  10
–3
s
x = (2 / 0.01) sin 2 {(5/3) – (0.01/0.01)}
= (p/0.01) sin (2  2 / 3) = (1/0.01) sin (4/3) = –200  sin (/3) = –200 x ( 3 / 2)
= 544 cm/s = 5.4 m/s
Similarly
a = {4
2
/ (0.01)
2
} cos 2 {(5/3) – 1}
= 4
2
 10
4
 ½  2  10
5
cm/s
2
 2 km/s
2
20. l = 40 cm, mass = 10 g
 mass per unit length, m = 10 / 40 = 1/4 (g/cm)
spring constant K = 160 N/m
deflection = x = 1 cm = 0.01 m
 T = kx = 160  0.01 = 1.6 N = 16  10
4
dyne
Again v = (T /m) = 4
(16 10 /(1/ 4) = 8  10
2
cm/s = 800 cm/s
 Time taken by the pulse to reach the spring
t = 40/800 = 1/20 = 0/05 sec.
21. m1 = m2 = 3.2 kg
mass per unit length of AB = 10 g/mt = 0.01 kg.mt
mass per unit length of CD = 8 g/mt = 0.008 kg/mt
for the string CD, T = 3.2  g
 v = (T /m) = 3
(3.2 10)/0.008 (32 10 )/8   = 2 10 10 = 20  3.14 = 63 m/s
for the string AB, T = 2  3.2 g = 6.4  g = 64 N
 v = (T /m) = (64/0.01) 6400 = 80 m/s
22. Total length of string 2 + 0.25 = 2.25 mt
Mass per unit length m =
3
4.5 10
2.25


= 2  10
–3
kg/m
T = 2g = 20 N
Wave speed, v = (T /m) = 3 4
20 /(2 10 ) 10
  = 10
2
m/s = 100 m/s
Time taken to reach the pully, t = (s/v) = 2/100 = 0.02 sec.
23. m = 19.2  10
–3
kg/m
from the freebody diagram,
T – 4g – 4a = 0
 T = 4(a + g) = 48 N
wave speed, v = (T /m) = 50 m/s
D
C
B
A
m2
m1
2mt
2g
T
25 cm
2kg
4a
a = 2 m/s2
4 kg
4g

Chapter 15
15.5
24. Let M = mass of the heavy ball
(m = mass per unit length)
Wave speed, v1 = (T /m) = (Mg/m) (because T = Mg)
 60 = (Mg/m)  Mg/ m = 60
2
…(1)
From the freebody diagram (2),
v2 = (T'/m)
 v2 =
2 2 1/ 4
1/ 2
[(Ma) (Mg) ]
m

(because T’ = 2 2
(Ma) (Mg) )
 62 =
2 2 1/ 4
1/ 2
[(Ma) (Mg) ]
m


2 2
(Ma) (Mg)
m

= 62
2
…(2)
Eq(1) + Eq(2)  (Mg/m)  [m / 2 2
(Ma) (Mg) ] = 3600 / 3844
 g / 2 2
(a g ) = 0.936  g
2
/ (a
2
+ g
2
) = 0.876
 (a
2
+ 100) 0.876 = 100
 a
2
 0.876 = 100 – 87.6 = 12.4
 a
2
= 12.4 / 0.876 = 14.15  a = 3.76 m/s
2
 Acce
n
of the car = 3.7 m/s
2
25. m = mass per unit length of the string
R = Radius of the loop
 = angular velocity, V = linear velocity of the string
Consider one half of the string as shown in figure.
The half loop experiences cetrifugal force at every point, away from
centre, which is balanced by tension 2T.
Consider an element of angular part d at angle . Consider another
element symmetric to this centrifugal force experienced by the element
= (mRd)
2
R.
(…Length of element = Rd, mass = mRd)
Resolving into rectangular components net force on the two symmetric elements,
DF = 2mR
2
d
2
sin  [horizontal components cancels each other]
So, total F =
/ 2
2 2
0
2mR sin d

   = 2mR
2

2
[– cos]  2mR
2

2
Again, 2T = 2mR
2

2
 T = mR
2

2
Velocity of transverse vibration V = T /m = R = V
So, the speed of the disturbance will be V.
26. a) m  mass per unit of length of string
consider an element at distance ‘x’ from lower end.
Here wt acting down ward = (mx)g = Tension in the string of upper part
Velocity of transverse vibration = v = T /m = (mgx /m) (gx)
b) For small displacement dx, dt = dx / (gx)
Total time T =
L
0
dx / gx (4L / g)
c) Suppose after time ‘t’ from start the pulse meet the particle at distance y from lower end.
t =
y
0
dx / gx (4y / g)
 Distance travelled by the particle in this time is (L – y)
Mg
(Rest)
T
a
Mg
(Motion)
T
Ma

T
(mRd)w2
R

T
c
d
y
L-y
A
B
TA
TB
x
4xl

Chapter 15
15.6
 S – ut + 1/2 gt
2
 L – y (1/2)g  2
{ (4y / g) } {u = 0}
 L – y = 2y  3y = L
 y = L/3. So, the particle meet at distance L/3 from lower end.
27. mA = 1.2  10
–2
kg/m, TA = 4.8 N
 VA = T /m = 20 m/s
mB = 1.2  10
–2
kg/m, TB = 7.5 N
 VB = T /m = 25 m/s
t = 0 in string A
t1 = 0 + 20 ms = 20  10
–3
= 0.02 sec
In 0.02 sec A has travelled 20  0.02 = 0.4 mt
Relative speed between A and B = 25 – 20 = 5 m/s
Time taken for B for overtake A = s/v = 0.4/5 = 0.08 sec
28. r = 0.5 mm = 0.5  10
–3
mt
f = 100 Hz, T = 100 N
v = 100 m/s
v = T /m  v
2
= (T/m)  m = (T/v
2
) = 0.01 kg/m
Pave = 2
2
mvr
2
f
2
= 2(3.14)
2
(0.01)  100  (0.5  10
–3
)
2
 (100)
2
 49  10
–3
watt = 49 mW.
29. A = 1 mm = 10
–3
m, m = 6 g/m = 6  10
–3
kg/m
T = 60 N, f = 200 Hz
 V = T /m = 100 m/s
a) Paverage = 2
2
mv A
2
f
2
= 0.47 W
b) Length of the string is 2 m. So, t = 2/100 = 0.02 sec.
Energy = 2
2
mvf
2
A
2
t = 9.46 mJ.
30. f = 440 Hz, m = 0.01 kg/m, T = 49 N, r = 0.5  10
–3
m
a) v = T /m = 70 m/s
b) v = f   = v/f = 16 cm
c) Paverage = 2
2
mvr
2
f
2
= 0.67 W.
31. Phase difference  = /2
f and  are same. So,  is same.
y1 = r sin wt, y2 = rsin(wt + /2)
From the principle of superposition
y = y1 + y2  = r sin wt + r sin (wt + /2)
= r[sin wt + sin(wt + /2)]
= r[2sin{(wt + wt + /2)/2} cos {(wt – wt – /2)/2}]
 y = 2r sin (wt + /4) cos (–/4)
Resultant amplitude = 2 r = 4 2 mm (because r = 4 mm)
32. The distance travelled by the pulses are shown below.
t = 4 ms = 4  10
–3
s s = vt = 50  10  4  10
–3
= 2 mm
t = 8 ms = 8  10
–3
s s = vt = 50  10  8  10
–3
= 4 mm
t = 6 ms = 6  10
–3
s s = 3 mm
t = 12 ms = 12  10
–3
s s = 50  10  12  10
–3
= 6 mm
The shape of the string at different times are shown in the figure.
33. f = 100 Hz,  = 2 cm = 2  10
–2
m
 wave speed, v = f = 2 m/s
a) in 0.015 sec 1
st
wave has travelled
x = 0.015  2 = 0.03 m = path diff
n
 corresponding phase difference,  = 2x/ = {2 / (2  10
–2
)}  0.03 = 3.
b) Path different x = 4 cm = 0.04 m
2 6
10 14

Chapter 15
15.7
  = (2/)x = {(2/2  10
–2
)  0.04} = 4.
c) The waves have same frequency, same wavelength and same amplitude.
Let, y1 = r sin wt, y2 = r sin (wt + )
 y = y1 + y2 = r[sin wt + (wt + )]
= 2r sin (wt + /2) cos (/2)
 resultant amplitude = 2r cos /2
So, when  = 3, r = 2  10
–3
m
Rres = 2  (2  10
–3
) cos (3/2) = 0
Again, when  = 4, Rres = 2  (2  10
–3
) cos (4/2) = 4 mm.
34. l = 1 m, V = 60 m/s
 fundamental frequency, f0 = V/2l = 30 sec
–1
= 30 Hz.
35. l = 2m, f0 = 100 Hz, T = 160 N
f0 = 1/ 2l (T /m)
 m = 1 g/m. So, the linear mass density is 1 g/m.
36. m = (4/80) g/ cm = 0.005 kg/m
T = 50 N, l = 80 cm = 0.8 m
v = (T /m) = 100 m/s
fundamental frequency f0 = 1/ 2l (T /m) = 62.5 Hz
First harmonic = 62.5 Hz
f4 = frequency of fourth harmonic = 4f0 = F3 = 250 Hz
V = f4 4  4 = (v/f4) = 40 cm.
37. l = 90 cm = 0.9 m
m = (6/90) g/cm = (6/900) kg/mt
f = 261.63 Hz
f = 1/ 2l (T /m)  T = 1478.52 N = 1480 N.
38. First harmonic be f0, second harmonic be f1
 f1 = 2f0
 f0 = f1/2
f1 = 256 Hz
 1
st
harmonic or fundamental frequency
f0 = f1/2 = 256 / 2 = 128 Hz
/2 = 1.5 m   = 3m (when fundamental wave is produced)
 Wave speed = V = f0Ql = 384 m/s.
39. l = 1.5 m, mass – 12 g
 m = 12/1.5 g/m = 8  10
–3
kg/m
T = 9  g = 90 N
 = 1.5 m, f1 = 2/2l T /m
[for, second harmonic two loops are produced]
f1 = 2f0  70 Hz.
40. A string of mass 40 g is attached to the tuning fork
m = (40  10
–3
) kg/m
The fork vibrates with f = 128 Hz
 = 0.5 m
v = f = 128  0.5 = 64 m/s
v = T /m  T = v
2
m = 163.84 N  164 N.
41. This wire makes a resonant frequency of 240 Hz and 320 Hz.
The fundamental frequency of the wire must be divisible by both 240 Hz and 320 Hz.
a) So, the maximum value of fundamental frequency is 80 Hz.
b) Wave speed, v = 40 m/s
 80 = (1/2l)  40  0.25 m.
1.5 cm
9 kg 9 kg
l

Chapter 15
15.8
42. Let there be ‘n’ loops in the 1
st
case
 length of the wire, l = (n1)/2 [1 = 2  2 = 4 cm]
So there are (n + 1) loops with the 2nd
case
 length of the wire, l = {(n+1)2/2 [ = 2  1.6 = 3.2 cm]
 n1/2 = 2(n 1)
2
 
 n  4 = (n + 1) (3.2)  n = 4
 length of the string, l = (n1)/2 = 8 cm.
43. Frequency of the tuning fork, f = 660 Hz
Wave speed, v = 220 m/s   = v/f = 1/3 m
No.of loops = 3
a) So, f = (3/2l)v  l = 50 cm
b) The equation of resultant stationary wave is given by
y = 2A cos (2x/Ql) sin (2vt/)
 y = (0.5 cm) cos (0.06  cm
–1
) sin (1320 s
–1
t)
44. l1 = 30 cm = 0.3 m
f1 = 196 Hz, f2 = 220 Hz
We know f  (1/l) (as V is constant for a medium)
 1 2
2 1
f l
f l
  l2 = 26.7 cm
Again f3 = 247 Hz
 3 1
1 3
f l
f l
 
3
0.3
l
 l3 = 0.224 m = 22.4 cm and l3 = 20 cm
45. Fundamental frequency f1 = 200 Hz
Let l4 Hz be nth harmonic
 F2/F1 = 14000/200
 NF1/F1 = 70  N = 70
 The highest harmonic audible is 70
th
harmonic.
46. The resonant frequencies of a string are
f1 = 90 Hz, f2 = 150 Hz, f3 = 120 Hz
a) The highest possible fundamental frequency of the string is f = 30 Hz
[because f1, f2 and f3 are integral multiple of 30 Hz]
b) The frequencies are f1 = 3f, f2 = 5f, f3 = 7f
So, f1, f2 and f3 are 3
rd
harmonic, 5
th
harmonic and 7
th
harmonic respectively.
c) The frequencies in the string are f, 2f, 3f, 4f, 5f, ……….
So, 3f = 2
nd
overtone and 3
rd
harmonic
5f = 4
th
overtone and 5
th
harmonic
7f = 6th
overtone and 7
th
harmonic
d) length of the string is l = 80 cm
 f1 = (3/2l)v (v = velocity of the wave)
 90 = {3/(280)}  K
 K = (90  2  80) / 3 = 4800 cm/s = 48 m/s.
47. Frequency f = 1
1
1 1 1
T1 T 1
f
lD l D
 
 
2
2
2 2 2
T1
f
l n
 

Given that, T1/T2 = 2, r1 / r2 = 3 = D1/D2
1
2
1
2



So, 1 2 2 1 2
2 1 1 2 1
f l D T
f l D T



(l1 = l2 = length of string)
 f1 : f2 = 2 : 3
l
2 cm
l
1.6 cm
l

Chapter 15
15.9
48. Length of the rod = L = 40 cm = 0.4 m
Mass of the rod m = 1.2 kg
Let the 4.8 kg mass be placed at a distance
‘x’ from the left end.
Given that, fl = 2fr
 l rT T1 2
2l m 2l m

 l
r
T
T
= 2  l
r
T
T
= 4 …(1)
Tl + Tr = 60 N
 4Tr +Tr = 60 N
 Tr = 12 N and Tl = 48 N
Now taking moment about point A,
Tr  (0.4) = 48x + 12 (0.2)  x = 5 cm
So, the mass should be placed at a distance 5 cm from the left end.
49. s = 7.8 g/cm
3
, A = 2.6 g/cm
3
ms = s As = 7.8  10
–2
g/cm (m = mass per unit length)
mA = A AA = 2.6  10
–2
 3 g/cm = 7.8  10
–3
kg/m
A node is always placed in the joint. Since aluminium and steel rod has
same mass per unit length, velocity of wave in both of them is same.
 v = T /m  500/7 m/x
For minimum frequency there would be maximum wavelength for maximum wavelength minimum no of
loops are to be produced.
 maximum distance of a loop = 20 cm
 wavelength =  = 2  20 = 40 cm = 0.4 m
 f = v/ = 180 Hz.
50. Fundamental frequency
V = 1/2l T /m  T /m = v2l [ T /m = velocity of wave]
a) wavelength,  = velocity / frequency = v2l / v = 2l
and wave number = K = 2/ = 2/2l = /l
b) Therefore, equation of the stationary wave is
 y = A cos (2x/) sin (2Vt / L)
= A cos (2x / 2l) sin (2Vt / 2L)
v = V/2L [because v = (v/2l)]
51. V = 200 m/s, 2A = 0.5 m
a) The string is vibrating in its 1st
overtone
  = 1 = 2m
 f = v/ = 100 Hz
b) The stationary wave equation is given by
y = 2A
2 x 2 Vt
cos sin
 
 
= (0.5 cm) cos [(m
–1
)x] sin [(200 s
–1
)t]
52. The stationary wave equation is given by
y = (0.4 cm) sin [(0.314 cm – 1)x] cos [(6.00 s
–1
)t]
a)  = 600   2f = 600   f = 300 Hz
wavelength,  = 2/0.314 = (2  3.14) / 0.314 = 20 cm
b) Therefore nodes are located at, 0, 10 cm, 20 cm, 30 cm
c) Length of the string = 3/2 = 3  20/2 = 30 cm
d) y = 0.4 sin (0.314 x) cos (600 t)  0.4 sin {(/10)x} cos (600 t)
 since,  and v are the wavelength and velocity of the waves that interfere to give this vibration  = 20
cm
40 cm
A
Tl Tr
B C
Tl
Tr
B
A C
48N12N
80 cm 60 cm
20 cmSteel
Aluminium
L
l = 2 m
0
l
10 20 30

Chapter 15
15.10
v= /k = 6000 cm/sec = 60 m/s
53. The equation of the standing wave is given by
y = (0.4 cm) sin [(0.314 cm
–1
)x] cos [(6.00 s
–1
)t]
 k = 0.314 = /10
 2/ = /10   = 20 cm
for smallest length of the string, as wavelength remains constant, the string
should vibrate in fundamental frequency
 l = /2 = 20 cm / 2 = 10 cm
54. L = 40 cm = 0.4 m, mass = 3.2 kg = 3.2  10
–3
kg
 mass per unit length, m = (3.2)/(0.4) = 8  10
–3
kg/m
change in length, L = 40.05 – 40 = 0.05  10
–2
m
strain = L/L = 0.125  10
–2
m
f = 220 Hz
f = 3
1 T 1 T
2l' m 2 (0.4005) 8 10

 
 T = 248.19 N
Strain = 248.19/1 mm
2
= 248.19  10
6
Y = stress / strain = 1.985  10
11
N/m
2
55. Let,   density of the block
Weight  Vg where V = volume of block
The same turning fork resonates with the string in the two cases
f10 = w wT Vg ( )Vg10 11
2l m 2l m
    

As the f of tuning fork is same,
w
10 11
( )Vg10 Vg 11
f f
2l m 2l m
  
  
 w10 1 100
11 m 121
    
  

(because, w = 1 gm/cc)
 100 = 121  – 121  5.8  10
3
kg/m
3
56. l = length of rope = 2 m
M = mass = 80 gm = 0.8 kg
mass per unit length = m = 0.08/2 = 0.04 kg/m
Tension T = 256 N
Velocity, V = T /m = 80 m/s
For fundamental frequency,
l = /4   = 4l = 8 m
 f = 80/8 = 10 Hz
a) Therefore, the frequency of 1
st
two overtones are
1
st
overtone = 3f = 30 Hz
2
nd
overtone = 5f = 50 Hz
b) 1 = 4l = 8 m
1 = V/ f1 = 2.67 m
2 = V/f2 = 1.6 mt
so, the wavelengths are 8 m, 2.67 m and 1.6 m respectively.
57. Initially because the end A is free, an antinode will be formed.
So, l = Ql1 / 4
Again, if the movable support is pushed to right by 10 m, so that the joint is placed on the pulley, a node
will be formed there.
So, l = 2 / 2
Since, the tension remains same in both the cases, velocity remains same.
As the wavelength is reduced by half, the frequency will become twice as that of 120 Hz i.e. 240 Hz.
  
L
L
string
rope
l = /4
Initial position
Final position

16.1
CHAPTER – 16
1. Vair= 230 m/s. Vs = 5200 m/s. Here S = 7 m
So, t = t1 – t2 = 






5200
1
330
1
= 2.75 × 10–3
sec = 2.75 ms.
2. Here given S = 80 m × 2 = 160 m.
v = 320 m/s
So the maximum time interval will be
t = 5/v = 160/320 = 0.5 seconds.
3. He has to clap 10 times in 3 seconds.
So time interval between two clap = (3/10 second).
So the time taken go the wall = (3/2 × 10) = 3/20 seconds.
= 333 m/s.
4. a) for maximum wavelength n = 20 Hz.
as 







1
b) for minimum wavelength, n = 20 kHz
  = 360/ (20 × 10
3
) = 18 × 10
–3
m = 18 mm
 x = (v/n) = 360/20 = 18 m.
5. a) for minimum wavelength n = 20 KHz
 v = n   = 





 3
1020
1450
= 7.25 cm.
b) for maximum wavelength n should be minium
 v = n   = v/n  1450 / 20 = 72.5 m.
6. According to the question,
a)  = 20 cm × 10 = 200 cm = 2 m
 v = 340 m/s
so, n = v/ = 340/2 = 170 Hz.
N = v/  2
102
340


= 17.000 Hz = 17 KH2 (because  = 2 cm = 2 × 10–2
m)
7. a) Given Vair = 340 m/s , n = 4.5 ×10
6
Hz
 air = (340 / 4.5) × 10–6
= 7.36 × 10–5
m.
b) Vtissue = 1500 m/s  t = (1500 / 4.5) × 10
–6
= 3.3 × 10
–4
m.
8. Here given ry = 6.0 × 10
–5
m
a) Given 2/ = 1.8   = (2/1.8)
 So,





2
s/m10)8.1(0.6r 5
y
= 1.7 × 10–5
m
b) Let, velocity amplitude = Vy
V = dy/dt = 3600 cos (600 t – 1.8) × 10
–5
m/s
Here Vy = 3600 × 10
–5
m/s
Again,  = 2/1.8 and T = 2/600  wave speed = v = /T = 600/1.8 = 1000 / 3 m/s.
So the ratio of (Vy/v) =
1000
1033600 5

.
9. a) Here given n = 100, v = 350 m/s
  =
100
350
n
v
 = 3.5 m.
In 2.5 ms, the distance travelled by the particle is given by
x = 350 × 2.5 × 10
–3

Chapter 16
16.2
So, phase difference  = x
2



 )2/(105.2350
)100/350(
2 3

 
.
b) In the second case, Given  = 10 cm = 10
–1
m
 So,  = 35/2
)100/350(
102
x
x
2 1



 
.
10. a) Given x = 10 cm,  = 5.0 cm
  = 

2
= 

410
5
2
.
So phase difference is zero.
b) Zero, as the particle is in same phase because of having same path.
11. Given that p = 1.0 × 105
N/m2
, T = 273 K, M = 32 g = 32 × 10–3
kg
V = 22.4 litre = 22.4 × 10
–3
m
3
C/Cv = r = 3.5 R / 2.5 R = 1.4
 V =
4.22/32
100.14.1
f
rp 5

 = 310 m/s (because  = m/v)
12. V1 = 330 m/s, V2 = ?
T1 = 273 + 17 = 290 K, T2 = 272 + 32 = 305 K
We know v  T
1
21
2
2
1
2
1
T
TV
V
T
T
V
V 

=
290
305
340 = 349 m/s.
13. T1 = 273 V2 = 2V1
V1 = v T2 = ?
We know that V  T  2
1
2
2
1
2
V
V
T
T
  T2 = 273 × 2
2
= 4 × 273 K
So temperature will be (4 × 273) – 273 = 819°c.
14. The variation of temperature is given by
T = T1 +
d
)TT( 22 
x …(1)
We know that V  T 
273
T
V
VT
  VT =
273
T
v
 dt =
T
273
V
du
V
dx
T

 t =  
d
0
2/1
121 ]x)d/)TT(T[
dx
V
273
= d
0
12
1
12
]x
d
TT
T[
TT
d2
V
273 


 = 12
12
TT
TT
273
V
d2
















= T =
12 TT
273
V
d2

Putting the given value we get
=
310280
273
330
332



= 96 ms.
x
cm
20 cm
10cm
A
B
d
x
T1 T2

Chapter 16
16.3
15. We know that v = /K
Where K = bulk modulus of elasticity
 K = v2
 = (1330)2
× 800 N/m2
We know K = 





 V/V
A/F
 V =
80013301330
102
K
essuresPr 5



So, V = 0.15 cm
3
16. We know that,
Bulk modulus B =
0
0
S2
P
)V/V(
p





Where P0 = pressure amplitude  P0 = 1.0 × 10
5
S0 = displacement amplitude  S0 = 5.5 × 10–6
m
 B =
m10)5.5(2
m103514
6
2




= 1.4 × 10
5
N/m
2
.
17. a) Here given Vair = 340 m/s., Power = E/t = 20 W
f = 2,000 Hz,  = 1.2 kg/m
3
So, intensity I = E/t.A
= 44
64
20
r4
20
22




mw/m2
(because r = 6m)
b) We know that I =
air
2
0
V2
P

 air0 V21P 
 = 3
10443402.12 
 = 6.0 N/m2
.
c) We know that I = VvS2 22
0
2
 where S0 = displacement amplitude
 S0 =
air
22
V
I

Putting the value we get Sg = 1.2 × 10–6
m.
18. Here I1 = 1.0 × 10
–8
W1/m
2
; I2 = ?
r1 = 5.0 m, r2 = 25 m.
We know that I  2
r
1
 I1r1
2
= I2r2
2
 I2 = 2
2
2
11
r
rI
=
625
25100.1 8
 
= 4.0 × 10–10
W/m2
.
19. We know that  = 10 log10 





0I
I
A =
o
A
I
I
log10 , B =
o
B
I
I
log10
 IA / I0 = )10/( A
10 
 IB/Io = )10/( B
10 

2
2
A
2
B
B
A
5
50
r
r
I
I






  2)(
1010 BA

 2
10
BA


 20BA 
 B = 40 – 20 = 20 d.

Chapter 16
16.4
20. We know that,  = 10 log10 J/I0
According to the questions
A = 10 log10 (2I/I0)
 B – A = 10 log (2I/I) = 10 × 0.3010 = 3 dB.
21. If sound level = 120 dB, then I = intensity = 1 W/m
2
Given that, audio output = 2W
Let the closest distance be x.
So, intensity = (2 / 4x
2
) = 1  x
2
= (2/2)  x = 0.4 m = 40 cm.
22. 1 = 50 dB, 2 = 60 dB
 I1 = 10
–7
W/m
2
, I2 = 10
–6
W/m
2
(because  = 10 log10 (I/I0), where I0 = 10–12
W/m2
)
Again, I2/I1 = (p2/p1)
2
=(10
–6
/10
–7
) = 10 (where p = pressure amplitude).
 (p2 / p1) = 10 .
23. Let the intensity of each student be I.
According to the question
A =
0
10
I
I50
log10 ; B = 





0
10
I
I100
log10
 B – A =
0
10
I
I50
log10 – 





0
10
I
I100
log10
= 32log10
I50
I100
log10 10 





So, A = 50 + 3 = 53 dB.
24. Distance between tow maximum to a minimum is given by, /4 = 2.50 cm
  = 10 cm = 10
–1
m
We know, V = nx
 n = 1
10
340V



= 3400 Hz = 3.4 kHz.
25. a) According to the data
/4 = 16.5 mm   = 66 mm = 66 × 10–6=3
m
 n = 3
1066
330V




= 5 kHz.
b) Iminimum = K(A1 – A2)
2
= I  A1 – A2 = 11
Imaximum = K(A1 + A2)
2
= 9  A1 + A2 = 31
So,
4
3
AA
AA
21
21



 A1/A2 = 2/1
So, the ratio amplitudes is 2.
26. The path difference of the two sound waves is given by
L = 6.4 – 6.0 = 0.4 m
The wavelength of either wave =  =



320V
(m/s)
For destructive interference L =
2
)1n2( 
where n is an integers.
or 0.4 m =


 320
2
1n2
  = n = Hz
2
1n2
800
4.0
320 
 = (2n + 1) 400 Hz
Thus the frequency within the specified range which cause destructive interference are 1200 Hz,
2000 Hz, 2800 Hz, 3600 Hz and 4400 Hz.

Chapter 16
16.5
27. According to the given data
V = 336 m/s,
/4 = distance between maximum and minimum intensity
= (20 cm)   = 80 cm
 n = frequency = 2
1080
336V




= 420 Hz.
28. Here given  = d/2
Initial path difference is given by = dd2
2
d
2 2
2






If it is now shifted a distance x then path difference will be
= 











4
d
d2
4
d
d)xd2(
2
d
2 2
2

64
d169
)xd2(
2
d 2
2
2






 2
d
64
153
  xd2 1.54 d  x = 1.54 d – 1.414 d = 0.13 d.
29. As shown in the figure the path differences 2.4 = x = 2.3)4.2()2.3( 22

Again, the wavelength of the either sound waves =

320
We know, destructive interference will be occur
If x =
2
)1n2( 




320
2
)1n2(
)2.3()4.2()2.3( 22
Solving we get
 V = )1n2(200
2
400)1n2(


where n = 1, 2, 3, …… 49. (audible region)
30. According to the data
 = 20 cm, S1S2 = 20 cm, BD = 20 cm
Let the detector is shifted to left for a distance x for hearing the
minimum sound.
So path difference AI = BC – AB
= 2222
)x10()20()x10()20( 
So the minimum distances hearing for minimum
=
2
20
22
)1n2(




= 10 cm
 2222
)x10()20()x10()20(  = 10 solving we get x = 12.0 cm.
31.
Given, F = 600 Hz, and v = 330 m/s   = v/f = 330/600 = 0.55 mm
S
D
=x/4
20cm
2 d
D
S 22
d2)2/d( 
x
d
22
)4.2()2.3( 
A
AA
B
A
x
C
20cm
20cm
X
S
 P
R
1m
Q
y
S1
S2
O
1m
D
S1
S2
O 
P
Q
X

Chapter 16
16.6
Let OP = D, PQ = y   = y/R …(1)
Now path difference is given by, x = S2Q – S1Q = yd/D
Where d = 2m
[The proof of x = yd/D is discussed in interference of light waves]
a) For minimum intensity, x = (2n + 1)(/2)
 yd/D = /2 [for minimum y, x = /2]
 y/D =  = /2 = 0.55 / 4 = 0.1375 rad = 0.1375 × (57.1)° = 7.9°
b) For minimum intensity, x = 2n(/2)
yd/D =   y/D =  = /D = 0.55/2 = 0.275 rad
  = 16°
c) For more maxima,
yd/D = 2, 3, 4, …
 y/D =  = 32°, 64°, 128°
But since, the maximum value of  can be 90°, he will hear two more maximum i.e. at 32° and 64°.

32.
Because the 3 sources have equal intensity, amplitude are equal
So, A1 = A2 = A3
As shown in the figure, amplitude of the resultant = 0 (vector method)
So, the resultant, intensity at B is zero.
33. The two sources of sound S1 and S2 vibrate at same phase and frequency.
Resultant intensity at P = I0
a) Let the amplitude of the waves at S1 and S2 be ‘r’.
When  = 45°, path difference = S1P – S2P = 0 (because S1P = S2P)
So, when source is switched off, intensity of sound at P is I0/4.
b) When  = 60°, path difference is also 0.
Similarly it can be proved that, the intensity at P is I0 / 4 when one is switched off.
34. If V = 340 m/s, I = 20 cm = 20 × 10
–2
m
Fundamental frequency = 2
10202
340
21
V


 = 850 Hz
We know first over tone = 2
10202
3402
21
V2



 (for open pipe) = 1750 Hz
Second over tone = 3 (V/21) = 3 × 850 = 2500 Hz.
35. According to the questions V = 340 m/s, n = 500 Hz
We know that V/4I (for closed pipe)
 I =
5004
340

m = 17 cm.
36. Here given distance between two nodes is = 4. 0 cm,
  = 2 × 4.0 = 8 cm
We know that v = n
  = 2
108
328


= 4.1 Hz.
37. V = 340 m/s
Distances between two nodes or antinodes
 /4 = 25 cm
  = 100 cm = 1 m
 n = v/ = 340 Hz.
38. Here given that 1 = 50 cm, v = 340 m/s
As it is an open organ pipe, the fundamental frequency f1 = (v/21)
= 2
10502
340


= 340 Hz.
PS1 S2
S3
P
A2
120°
A3
120°
S2S1
P
 

Chapter 16
16.7
So, the harmonies are
f3 = 3 × 340 = 1020 Hz
f5 = 5 × 340 = 1700, f6 = 6 × 340 = 2040 Hz
so, the possible frequencies are between 1000 Hz and 2000 Hz are 1020, 1360, 1700.
39. Here given I2 = 0.67 m, l1 = 0.2 m, f = 400 Hz
We know that
 = 2(l2 – l1)   = 2(62 – 20) = 84 cm = 0.84 m.
So, v = n = 0.84 × 400 = 336 m/s
We know from above that,
l1 + d = /4  d = /4 – l1 = 21 – 20 = 1 cm.
40. According to the questions
f1 first overtone of a closed organ pipe P1 = 3v/4l =
304
V3


f2 fundamental frequency of a open organ pipe P2 =
2l2
V
Here given
2l2
V
304
V3


 l2 = 20 cm
 length of the pipe P2 will be 20 cm.
41. Length of the wire = 1.0 m
For fundamental frequency /2 = l
  = 2l = 2 × 1 = 2 m
Here given n = 3.8 km/s = 3800 m/s
We know  v = n  n = 3800 / 2 = 1.9 kH.
So standing frequency between 20 Hz and 20 kHz which will be heard are
= n × 1.9 kHz where n = 0, 1, 2, 3, … 10.
42. Let the length will be l.
Here given that V = 340 m/s and n = 20 Hz
Here /2 = l   = 2l
We know V = n  l = 5.8
4
34
202
340
n
V


 cm (for maximum wavelength, the frequency is minimum).
43. a) Here given l = 5 cm = 5 × 10
–2
m, v = 340 m/s
 n = 2
1052
340
l2
V


 = 3.4 KHz
b) If the fundamental frequency = 3.4 KHz
 then the highest harmonic in the audible range (20 Hz – 20 KHz)
=
3400
20000
= 5.8 = 5 (integral multiple of 3.4 KHz).
44. The resonance column apparatus is equivalent to a closed organ pipe.
Here l = 80 cm = 10 × 10
–2
m ; v = 320 m/s
 n0 = v/4l = 2
10504
320


= 100 Hz
So the frequency of the other harmonics are odd multiple of n0 = (2n + 1) 100 Hz
According to the question, the harmonic should be between 20 Hz and 2 KHz.
45. Let the length of the resonating column will be = 1
Here V = 320 m/s
Then the two successive resonance frequencies are
l4
nv
and
l4
v)1n( 
Here given
l4
v)1n( 
= 2592 ;  =
l4
nv
= 1944

l4
nv
l4
v)1n(


= 2592 – 1944 = 548 cm = 25 cm.

Chapter 16
16.8
46. Let, the piston resonates at length l1 and l2
Here, l = 32 cm; v = ?, n = 512 Hz
Now  512 = v/
 v = 512 × 0.64 = 328 m/s.

47. Let the length of the longer tube be L2 and smaller will be L1.
According to the data 440 =
2L4
3303


…(1) (first over tone)
and 440 =
1L4
330

…(2) (fundamental)
solving equation we get L2 = 56.3 cm and L1 = 18.8 cm.
48. Let n0 = frequency of the turning fork, T = tension of the string
L = 40 cm = 0.4 m, m = 4g = 4 × 10–3
kg
So, m = Mass/Unit length = 10
–2
kg/m
n0 =
m
T
l2
1
.
So, 2nd
harmonic 2n0 = m/T)l2/2(
As it is unison with fundamental frequency of vibration in the air column
 2n0 =
14
340

= 85 Hz
 85 =
14
T
4.02
2

 T = 85
2
× (0.4)
2
× 10
–2
= 11.6 Newton.
49. Given, m = 10 g = 10 × 10–3
kg, l = 30 cm = 0.3 m
Let the tension in the string will be = T
 = mass / unit length = 33 × 10
–3
kg
The fundamental frequency  n0 =

T
l2
1
…(1)
The fundamental frequency of closed pipe
 n0 = (v/4l) 2
10504
340

= 170 Hz …(2)
According equations (1) × (2) we get
170 = 32
1033
T
10302
1




 T = 347 Newton.
50. We know that f  T
According to the question f + f  T + T

T
Tt
f
ff 


 1 + ...
T
T
2
1
1
T
T
1
f
f
2/1







 


(neglecting other terms)

T
T
)2/1(
f
f 


.
51. We know that the frequency = f, T = temperatures
f  T
So
2
1
2
1
T
T
f
f
 
295
293
f
293
2

 f2 =
293
295293 
= 294
(I2 -I1)
I2
I1

Chapter 16
16.9
52. Vrod = ?, Vair = 340 m/s, Lr = 25 × 10–2
, d2 = 5 × 10–2
metres
a
r
a
r
D
L2
V
V
  Vr = 2
2
105
21025340




= 3400 m/s.
53. a) Here given, Lr = 1.0/2 = 0.5 m, da = 6.5 cm = 6.5 × 10
–2
m
As Kundt’s tube apparatus is a closed organ pipe, its fundamental frequency
 n =
r
r
L4
V
 Vr = 2600 × 4 × 0.5 = 5200 m/s.
b)
a
r
a
r
d
L2
V
V
  va =
5.02
105.65200 2

 
= 338 m/s.
54. As the tunning fork produces 2 beats with the adjustable frequency the frequency of the tunning fork will
be  n = (476 + 480) / 2 = 478.
55. A tuning fork produces 4 beats with a known tuning fork whose frequency = 256 Hz
So the frequency of unknown tuning fork = either 256 – 4 = 252 or 256 + 4 = 260 Hz
Now as the first one is load its mass/unit length increases. So, its frequency decreases.
As it produces 6 beats now original frequency must be 252 Hz.
260 Hz is not possible as on decreasing the frequency the beats decrease which is not allowed here.
56. Group – I Group – II
Given V = 350 v = 350
1 = 32 cm 2 = 32.2 cm
= 32 × 10–2
m = 32.2 × 10–2
m
So 1 = frequency = 1093 Hz 2 = 350 / 32.2 × 10
–2
= 1086 Hz
So beat frequency = 1093 – 1086 = 7 Hz.
57. Given length of the closed organ pipe, l = 40 cm = 40 × 10–2
m
Vair = 320
So, its frequency  =
l4
V
= 2
10404
320


= 200 Hertz.
As the tuning fork produces 5 beats with the closed pipe, its frequency must be 195 Hz or 205 Hz.
Given that, as it is loaded its frequency decreases.
So, the frequency of tuning fork = 205 Hz.
58. Here given nB = 600 =
14
TB
l2
1
As the tension increases frequency increases
It is given that 6 beats are produces when tension in A is increases.
So, nA  606 =
M
TA
l2
1

TA
TB
)M/TA()l2/1(
)M/TB()l2/1(
606
600
n
n
B
A


600
606
T
T
B
A
 = 1.01 
B
A
T
T
= 1.02.
59. Given that, l = 25 cm = 25 × 10–2
m
By shortening the wire the frequency increases, [f = )M/TB()l2/1( ]
As the vibrating wire produces 4 beats with 256 Hz, its frequency must be 252 Hz or 260 Hz.
Its frequency must be 252 Hz, because beat frequency decreases by shortening the wire.
So, 252 =
M
T
10252
1
2

…(1)
Let length of the wire will be l, after it is slightly shortened,

Chapter 16
16.10
 256 =
M
T
l2
1
1
…(2)
Dividing (1) by (2) we get
260
10252252
l
10252
l
256
252 2
12
1





 = 0.2431 m
So, it should be shorten by (25 – 24.61) = 0.39 cm.
60. Let u = velocity of sound; Vm = velocity of the medium;
vo = velocity of the observer; va = velocity of the sources.
f = F
vVv
vvu
sm
om









using sign conventions in Doppler’s effect,
Vm = 0, u = 340 m/s, vs = 0 and ov

= –10 m (36 km/h = 10 m/s)
= KHz2
00340
)10(0340








= 350/340 × 2 KHz = 2.06 KHz.
61. f
1
= f
vvu
vvu
sm
om










[18 km/h = 5 m/s]
using sign conventions,
app. Frequency = 2400
50340
00340








= 2436 Hz.
62.
a) Given vs = 72 km/hour = 20 m/s,  = 1250
apparent frequency = 1250
200340
00340



= 1328 H2
b) For second case apparent frequency will be = 1250
)20(0340
00340



= 1181 Hz.
63. Here given, apparent frequency = 1620 Hz
So original frequency of the train is given by
1620 = f
15332
00332








 f = 




 
332
3171620
Hz
So, apparent frequency of the train observed by the observer in
f
1
= f
15332
00332








× 




 
332
3171620
= 1620
347
317
 = 1480 Hz.
64. Let, the bat be flying between the walls W1 and W2.
So it will listen two frequency reflecting from walls W2 and W1.
So, apparent frequency, as received by wall W = fw2 = f
6330
00330



= 330/324
Therefore, apparent frequency received by the bat from wall W2 is given by
2BF of wall W1 = 



















324
330
330
336
f
00330
)6(0330
2w f
Similarly the apparent frequency received by the bat from wall W1 is
1Bf (324/336)f
So the beat frequency heard by the bat will be = 4.47 × 104
= 4.3430 × 104
= 3270 Hz.
65. Let the frequency of the bullet will be f
Given, u = 330 m/s, vs = 220 m/s
(36km/h = 10m/s)
100 m/s
18km/h = 5m/s
I II
bat
w2w1

Chapter 16
16.11
a) Apparent frequency before crossing = f = f
220330
330







= 3f
b) Apparent frequency after crossing = f = f
220530
330







= 0.6 f
So,
f3
f6.0
f
f








= 0.2
Therefore, fractional change = 1 – 0.2 = 0.8.
 The person will receive, the sound in the directions BA and CA making an angle  with the track.
Here,  = tan
–1
(0.5/2.4) = 22°
So the velocity of the sources will be ‘v cos ’ when heard by the observer.
So the apparent frequency received by the man from train B.
f = 529500
22cosv340
00340








Hz
And the apparent frequency heard but the man from train C,
f = 500
22cosv340
00340








= 476 Hz.
67. Let the velocity of the sources is = vs
a) The beat heard by the standing man = 4
So, frequency = 440 + 4 = 444 Hz or 436 Hz
 440 = 400
v340
00340
s








On solving we get Vs = 3.06 m/s = 11 km/hour.
b) The sitting man will listen less no.of beats than 4.
68. Here given velocity of the sources vs = 0
Velocity of the observer v0 = 3 m/s
So, the apparent frequency heard by the man = 




 
332
3332
× 256 = 258.3 Hz.
from the approaching tuning form = f
f = [(332–3)/332] × 256 = 253.7 Hz.
So, beat produced by them = 258.3 – 253.7 = 4.6 Hz.
69. According to the data, Vs = 5.5 m/s for each turning fork.
So, the apparent frequency heard from the tuning fork on the left,
f = 512
5.5330
330







= 527.36 Hz = 527.5 Hz
similarly, apparent frequency from the tunning fork on the right,
f = 512
5.5330
330







= 510 Hz
So, beats produced 527.5 – 510 = 17.5 Hz.
70. According to the given data
Radius of the circle = 100/ × 10
–2
m = (1/) metres;  = 5 rev/sec.
So the linear speed v = r = 5/ = 1.59
So, velocity of the source Vs = 1.59 m/s
As shown in the figure at the position A the observer will listen maximum
and at the position B it will listen minimum frequency.
So, apparent frequency at A =
59.1332
332

× 500 = 515 Hz
Apparent frequency at B =
59.1332
332

× 500 = 485 Hz.
1.2km
v cos 

1.2km
0.5km
1 2
1 2
(A)
S2
(B) S1

Chapter 16
16.12
71. According to the given data Vs = 90 km/hour = 25 m/sec.
v0 = 25 m/sec
So, apparent frequency heard by the observer in train B or
observer in = 







25350
25350
× 500 = 577 Hz.
72. Here given fs = 16 × 10
3
Hz
Apparent frequency f = 20 × 103
Hz (greater than that value)
Let the velocity of the observer = vo
Given vs = 0
So 20 × 10
3
= 







0330
v330 o
× 16 × 10
3
 (330 + vo) =
16
33020
 vo = s/m
4
330
4
3301633020


= 297 km/h
b) This speed is not practically attainable ordinary cars.
73. According to the questions velocity of car A = VA = 108 km/h = 30 m/s
VB = 72 km/h = 20 m/s, f = 800 Hz
So, the apparent frequency heard by the car B is given by,
f = 







30330
20330
× 800  826.9 = 827 Hz.
74. a) According to the questions, v = 1500 m/s, f = 2000 Hz, vs = 10 m/s, vo = 15 m/s
So, the apparent frequency heard by the submarine B,
= 







101500
151500
× 2000 = 2034 Hz
b) Apparent frequency received by submarine A,
= 







151500
101500
× 2034 = 2068 Hz.
75. Given that, r = 0.17 m, F = 800 Hz, u = 340 m/s
Frequency band = f1 – f2 = 6 Hz
Where f1 and f2 correspond to the maximum and minimum apparent frequencies (both will occur at the
mean position because the velocity is maximum).
Now, f1 = 





 sv340
340
f and f2 = 





 sv340
340
f
 f1 – f2 = 8
 340 f 8
v340
1
v340
1
ss










800340
8
v340
v2
2
s
2
s



 340
2
– vs
2
= 68000 vs
Solving for vs we get, vs = 1.695 m/s
For SHM, vs = r   = (1.695/0.17) = 10
So, T = 2 /  = /5 = 0.63 sec.
76. u = 334 m/s, vb = 24 m/s, vo = 0
so, vs = Vb cos  = )2/1(24  = 4 m/s.
so, the apparent frequency f = 1650
4334
334
f
cosvu
0u
b















= 1670 Hz.
BA
BA 30m/s
B
Vs
10m/s
A
15m/s
Vs
N
E
s/m24
W
S
v cos 
45°
VS
0.17 m
VS
O BA
D

Chapter 16
16.13
77. u = 330 m/s, v0 = 26 m/s
a) Apparent frequency at, y = – 336
m = f
sinuv
v







= 660
23sin26330
330







[because,  = tan–1
(140/336) = 23°] = 680 Hz.
b) At the point y = 0 the source and listener are on a x-axis so no apparent change
in frequency is seen. So, f = 660 Hz.
c) As shown in the figure  = tan
–1
(140/336) = 23°
Here given, = 330 m/s ; v = V sin 23° = 10.6 m/s
So, F = 660
23sinvu
u


= 640 Hz.
78. Vtrain or Vs = 108 km/h = 30 m/s; u = 340 m/s
a) The frequency by the passenger sitting near the open window is 500 Hz, he is inside the train and
does not hair any relative motion.
b) After the train has passed the apparent frequency heard by a person standing near the track will be,
so f = 500
30340
0340








= 459 Hz
c) The person inside the source will listen the original frequency of the train.
Here, given Vm = 10 m/s
For the person standing near the track
Apparent frequency = 500
)V(Vu
0Vu
sm
m



= 458 Hz.
79. To find out the apparent frequency received by the wall,
a) Vs = 12 km/h = 10/3 = m/s
Vo = 0, u = 330 m/s
So, the apparent frequency is given by = f = 1600
3/10330
330







= 1616 Hz
b) The reflected sound from the wall whistles now act as a sources whose frequency is 1616 Hz.
So, u = 330 m/s, Vs = 0, Vo = 10/3 m/s
So, the frequency by the man from the wall,
 f = 1616
330
3/10330





 
= 1632 m/s.
80. Here given, u = 330 m/s, f = 1600 Hz
So, apparent frequency received by the car
f = 1600
330
20330
f
Vu
Vu
s
o





 








Hz … [Vo = 20 m/s, Vs = 0]
The reflected sound from the car acts as the source for the person.
Here, Vs = –20 m/s, Vo = 0
So f = 160
330
310
350
330
f
20330
0330








= 1417 Hz.
 This is the frequency heard by the person from the car.
81. a) f = 400 Hz,, u = 335 m/s
  (v/f) = (335/400) = 0.8 m = 80 cm
b) The frequency received and reflected by the wall,
f = 400
320
335
f
Vu
Vu
s
o










…[Vs = 54 m/s and Vo = 0]
336
140m
V0 26m/s

L
S 336
140

20m/s
f f

Chapter 16
16.14
 x = (v/f) =
400335
335320


= 0.8 m = 80 cm
c) The frequency received by the person sitting inside the car from reflected wave,
f = 400
320
335
f
15335
0335








= 467 [Vs = 0 and Vo = –15 m/s]
d) Because, the difference between the original frequency and the apparent frequency from the wall is
very high (437 – 440 = 37 Hz), he will not hear any beats.mm)
82. f = 400 Hz, u = 324 m/s, f = 400
324
v324
f
)0(u
)v(u





…(1)
for the reflected wave,
f = 410 = f
vu
0u



 410 = 400
324
v324
v324
324




 810 v = 324 × 10
 v =
810
10324 
= 4 m/s.
83. f = 2 kHz, v = 330 m/s, u = 22 m/s
At t = 0, the source crosses P
a) Time taken to reach at Q is
t =
330
330
v
S
 = 1 sec
b) The frequency heard by the listner is
f = 





 cosuv
v
f
since,  = 90°
f = 2 × (v/u) = 2 KHz.
c) After 1 sec, the source is at 22 m from P towards right.
84. t = 4000 Hz, u = 22 m/s
Let ‘t’ be the time taken by the source to reach at ‘O’. Since observer hears the sound at the instant it
crosses the ‘O’, ‘t’ is also time taken to the sound to reach at P.
 OQ = ut and QP = vt
Cos  = u/v
Velocity of the sound along QP is (u cos ).
f =































22
2
2
uv
v
f
v
u
v
v
f
cosuv
0v
f
Putting the values in the above equation, f = 4000 × 22
2
22330
330

= 4017.8 = 4018 Hz.
85. a) Given that, f = 1200 Hz, u = 170 m/s, L = 200 m, v = 340 m/s
From Doppler’s equation (as in problem no.84)
f =








 22
2
uv
v
f = 1200 × 22
2
170340
340

= 1600 Hz.
b) v = velocity of sound, u = velocity of source
let, t be the time taken by the sound to reach at D
DO = vt = L, and SO = ut
t = L/V
PS
330m
Q
u=22m/s
660m/s
S
P
O
S

uS O
(Detector)
D
L=vt
ut

Chapter 16
16.15
SD = 222
2
2
222
vu
v
L
L
v
L
uDOOS 
Putting the values in the above equation, we get
SD = 22
340170
340
220
 = 223.6 m.
86. Given that, r = 1.6 m, f = 500 Hz, u = 330 m/s
a) At A, velocity of the particle is given by
vA = 4106.1rg  m/s
and at C, vc = 9.8106.15rg5  m/s
So, maximum frequency at C,
fc = 85.513500
9.8330
330
f
vu
u
s




Hz.
Similarly, maximum frequency at A is given by 494)500(
4330
330
f
)v(u
u
f
s
A 



 Hz.
b) Velocity at B = 92.6106.13rg3  m/s
So, frequency at B is given by,
fB = 500
92.6330
330
f
vu
u
s




= 490 Hz
and frequency at D is given by,
fD = 500
92.6330
330
f
vu
u
s




87. Let the distance between the source and the observer is ‘x’ (initially)
So, time taken for the first pulse to reach the observer is t1 = x/v
and the second pulse starts after T (where, T = 1/v)
and it should travel a distance 





 2
aT
2
1
x .
So, t2 =
v
aT2/1x
T
2


t2 – t1 =
v
aT
2
1
T
v
x
v
aT2/1x
T
22



Putting = T = 1/v, we get
t2 – t1 = 2
vv2
auv2 
so, frequency heard =
auv2
vv2 2

(because, f =
12 tt
1

)
    
D
C
B
A
r
vD
vC
vA
vB
vD
vB
x – ½ at2
t=0
x
S S
t=T

17.1
CHAPTER 17
1. Given that, 400 m <  < 700 nm.
1 1 1
700nm 400nm
 


8 8
7 7 7 7
1 1 1 3 10 c 3 10
7 10 4 10 7 10 4 10   
 
    
    
(Where, c = speed of light = 3  108
m/s)
 4.3  10
14
< c/ < 7.5  10
14
 4.3  10
14
Hz < f < 7.5  10
14
Hz.
2. Given that, for sodium light,  = 589 nm = 589  10–9
m
a) fa =
8
9
3 10
589 10


= 5.09  10
14 1 c
sec f  
  

b) a w w
w9
w a
1
1.33 589 10
  
    
  
= 443 nm
c) fw = fa = 5.09  10
14
sec
–1
[Frequency does not change]
d)
8
a a aw
w
w a w
vv 3 10
v
v 1.33
  
   
 
= 2.25  10
8
m/sec.
3. We know that, 2 1
1 2
v
v



So,
8
8
400
400
1472 3 10
v 2.04 10 m/ sec.
1 v

   
[because, for air,  = 1 and v = 3  10
8
m/s]
Again,
8
8
760
760
1452 3 10
v 2.07 10 m/ sec.
1 v

    
4.
8
t 8
1 3 10
1.25
(2.4) 10
 
  

velocity of light in vaccum
since, =
velocity of light in the given medium
 
 
 
5. Given that, d = 1 cm = 10
–2
m,  = 5  10
–7
m and D = 1 m
a) Separation between two consecutive maxima is equal to fringe width.
So,  =
7
2
D 5 10 1
d 10


  
 m = 5  10
–5
m = 0.05 mm.
b) When,  = 1 mm = 10–3
m
10
–3
m =
7
5 10 1
D

 
 D = 5  10
–4
m = 0.50 mm.
6. Given that,  = 1 mm = 10–3
m, D = 2.t m and d = 1 mm = 10–3
m
So, 10
–3
m = 3
25
10
 
  = 4  10
–7
m = 400 nm.
7. Given that, d = 1 mm = 10
–3
m, D = 1 m.
So, fringe with =
D
d

= 0.5 mm.
a) So, distance of centre of first minimum from centre of central maximum = 0.5/2 mm = 0.25 mm
b) No. of fringes = 10 / 0.5 = 20.
8. Given that, d = 0.8 mm = 0.8  10
–3
m,  = 589 nm = 589  10
–9
m and D = 2 m.
So,  =
D
d

=
9
3
589 10 2
0.8 10


 

= 1.47  10
–3
m = 147 mm.

Chapter 17
17.2
9. Given that,  = 500 nm = 500  10–9
m and d = 2  10–3
m
As shown in the figure, angular separation  =
D
D dD d
  
 
So,  =
9
3
500 10
D d 2 10


  
 

= 250  10–6
= 25  10
–5
radian = 0.014 degree.
10. We know that, the first maximum (next to central maximum) occurs at y =
D
d

Given that, 1 = 480 nm, 2 = 600 nm, D = 150 cm = 1.5 m and d = 0.25 mm = 0.25  10–3
m
So, y1 =
9
1
3
D 1.5 480 10
d 0.25 10


  


= 2.88 mm
y2 =
9
3
1.5 600 10
0.25 10


 

= 3.6 mm.
So, the separation between these two bright fringes is given by,
 separation = y2 – y1 = 3.60 – 2.88 = 0.72 mm.
11. Let m
th
bright fringe of violet light overlaps with n
th
bright fringe of red light.

m 400nm D n 700nm D m 7
d d n 4
   
  
 7th
bright fringe of violet light overlaps with 4th
bright fringe of red light (minimum). Also, it can be
seen that 14th
violet fringe will overlap 8th
red fringe.
Because, m/n = 7/4 = 14/8.
12. Let, t = thickness of the plate
Given, optical path difference = ( – 1)t = /2
 t =
2( 1)

 

13. a) Change in the optical path = t – t = ( – 1)t
b) To have a dark fringe at the centre the pattern should shift by one half of a fringe.
 ( – 1)t = t
2 2( 1)
 
 
 
.
14. Given that,  = 1.45, t = 0.02 mm = 0.02  10–3
m and  = 620 nm = 620  10–9
m
We know, when the transparent paper is pasted in one of the slits, the optical path changes by ( – 1)t.
Again, for shift of one fringe, the optical path should be changed by .
So, no. of fringes crossing through the centre is given by,
n =
3
9
( 1)t 0.45 0.02 10
620 10


   

 
= 14.5
15. In the given Young’s double slit experiment,
 = 1.6, t = 1.964 micron = 1.964  10
–6
m
We know, number of fringes shifted =
( 1)t 

So, the corresponding shift = No.of fringes shifted  fringe width
=
( 1)t D ( 1)tD
d d
    
 

… (1)
Again, when the distance between the screen and the slits is doubled,
Fringe width =
(2D)
d

…(2)
From (1) and (2),
( 1)tD
d
 
=
(2D)
d

  =
( 1)t 

=
6
(1.6 1) (1.964) 10
2

  
= 589.2  10–9
= 589.2 nm.
B
S1
S2

D

Chapter 17
17.3
16. Given that, t1 = t2 = 0.5 mm = 0.5  10–3
m, m = 1.58 and p = 1.55,
 = 590 nm = 590  10
–9
m, d = 0.12 cm = 12  10
–4
m, D = 1 m
a) Fringe width =
9
4
D 1 590 10
d 12 10


  


= 4.91  10
–4
m.
b) When both the strips are fitted, the optical path changes by
x = (m – 1)t1 – (p – 1)t2 = (m – p)t
= (1.58 – 1.55)  (0.5)(10
–3
) = 0.015  10
–13
m.
So, No. of fringes shifted =
3
3
0.015 10
590 10




= 25.43.
 There are 25 fringes and 0.43 th of a fringe.
 There are 13 bright fringes and 12 dark fringes and 0.43 th of a dark fringe.
So, position of first maximum on both sides will be given by
 x = 0.43  4.91  10–4
= 0.021 cm
x = (1 – 0.43)  4.91  10
–4
= 0.028 cm (since, fringe width = 4.91  10
–4
m)
17. The change in path difference due to the two slabs is (1 – 2)t (as in problem no. 16).
For having a minimum at P0, the path difference should change by /2.
So,  /2 = (1 –2)t  t =
1 22( )

  
.
18. Given that, t = 0.02 mm = 0.02  10
–3
m, 1 = 1.45,  = 600 nm = 600  10
–9
m
a) Let, I1 = Intensity of source without paper = I
b) Then I2 = Intensity of source with paper = (4/9)I
 1 1
2 2
I r9 3
I 4 r 2
   [because I  r
2
]
where, r1 and r2 are corresponding amplitudes.
So,
2
max 1 2
2
min 1 2
I (r r )
I (r r )



= 25 : 1
b) No. of fringes that will cross the origin is given by,
n =
( 1)t 

=
3
9
(1.45 1) 0.02 10
600 10


  

= 15.
19. Given that, d = 0.28 mm = 0.28  10–3
m, D = 48 cm = 0.48 m, a = 700 nm in vacuum
Let, w = wavelength of red light in water
Since, the fringe width of the pattern is given by,
=
9
w
3
D 525 10 0.48
d 0.28 10


  


= 9  10–4
m = 0.90 mm.
20. It can be seen from the figure that the wavefronts reaching O from S1 and S2 will
have a path difference of S2X.
In the  S1S2X,
sin = 2
1 2
S X
S S
So, path difference = S2 X = S1S2 sin = d sin = d  /2d = /2
As the path difference is an odd multiple of /2, there will be a dark fringe at point P0.
21. a) Since, there is a phase difference of  between direct light and
reflecting light, the intensity just above the mirror will be zero.
b) Here, 2d = equivalent slit separation
D = Distance between slit and screen.
We know for bright fringe, x =
y 2d
D

= n
But as there is a phase reversal of /2.

y 2d
D

+
2

= n 
y 2d
D

= n –
2

 y =
D
4d


mica
Screen
polysterene
S2
S1

P0
x
S2
S1
Dark
fringe(1 – 0.43)
0.43
S2
S1
2d
D
Screen

Chapter 17
17.4
22. Given that, D = 1 m,  = 700 nm = 700  10–9
m
Since, a = 2 mm, d = 2a = 2mm = 2  10–3
m (L loyd’s mirror experiment)
Fringe width =
9
3
D 700 10 m 1m
d 2 10 m


  


= 0.35 mm.
23. Given that, the mirror reflects 64% of energy (intensity) of the light.
So, 1 1
2 2
I r16 4
0.64
I 25 r 5
   
So,
2
max 1 2
2
min 1 2
I (r r )
I (r r )



= 81 : 1.
24. It can be seen from the figure that, the apparent distance of the screen from the slits is,
D = 2D1 + D2
So, Fringe width = 1 2(2D D )D
d d
 

25. Given that,  = (400 nm to 700 nm), d = 0.5 mm = 0.5  10–3
m,
D = 50 cm = 0.5 m and on the screen yn = 1 mm = 1  10–3
m
a) We know that for zero intensity (dark fringe)
yn = nD2n 1
2 d
 
 
 
where n = 0, 1, 2, …….
 n =
3 3
6 3nd2 2 10 0.5 10 2 2
10 m 10 nm
(2n 1) D 2n 1 0.5 (2n 1) (2n 1)
 
  
     
   
If n = 1, 1 = (2/3)  1000 = 667 nm
If n = 1, 2 = (2/5)  1000 = 400 nm
So, the light waves of wavelengths 400 nm and 667 nm will be absent from the out coming light.
b) For strong intensity (bright fringes) at the hole
 yn = n n
n
n D y d
d nD

  
When, n = 1, 1 = ny d
D
=
3 3
610 0.5 10
10 m 1000nm
0.5
 
 
  .
1000 nm is not present in the range 400 nm – 700 nm
Again, where n = 2, 2 = ny d
2D
= 500 nm
So, the only wavelength which will have strong intensity is 500 nm.
26. From the diagram, it can be seen that at point O.
Path difference = (AB + BO) – (AC + CO)
= 2(AB – AC) [Since, AB = BO and AC = CO] = 2 2
2( d D D) 
For dark fringe, path difference should be odd multiple of /2.
So, 2 2
2( d D D)  = (2n + 1)(/2)
 2 2
d D = D + (2n + 1) /4
 D
2
+ d
2
= D
2
+ (2n+1)
2

2
/16 + (2n + 1) D/2
Neglecting, (2n+1)
2

2
/16, as it is very small
We get, d =
D
(2n 1)
2


For minimum ‘d’, putting n = 0  dmin =
D
2

.
C O
P
B
d
A
x
DD
D
yn
d=0.5mm
50cm
1 mm

Chapter 17
17.5
27. For minimum intensity
 S1P – S2P = x = (2n +1) /2
From the figure, we get
 2 2
Z (2 ) Z (2n 1)
2

    

2
2 2 2 2
Z 4 Z (2n 1) Z(2n 1)
4

       
 Z =
2 2 2 2 2 2
4 (2n 1) ( / 4) 16 (2n 1)
(2n 1) 4(2n 1)
       

   
…(1)
Putting, n = 0  Z = 15/4 n = –1  Z = –15/4
n = 1  Z = 7/12 n = 2  Z = –9/20
 Z = 7/12 is the smallest distance for which there will be minimum intensity.
28. Since S1, S2 are in same phase, at O there will be maximum intensity.
Given that, there will be a maximum intensity at P.
 path difference = x = n
From the figure,
(S1P)
2
– (S2P)
2
= 2 2 2 2 2 2
( D X ) ( (D 2 ) X )    
= 4D – 4
2
= 4 D (
2
is so small and can be neglected)
 S1P – S2P =
2 2
4 D
2 x D


= n

2 2
2D
x D

 n
2
(X
2
+ D
2
) = 4D
2
= X = 2D
4 n
n

when n = 1, x = 3 D (1
st
order)
n = 2, x = 0 (2nd
order)
 When X = 3 D, at P there will be maximum intensity.
(S1P)
2
= (PX)
2
+ (S1X)
2
…(1)
(S2P)2
= (PX)2
+ (S2X)2
…(2)
From (1) and (2),
(S1P)
2
– (S2P)
2
= (S1X)
2
– (S2X)
2
= (1.5  + R cos )2
– (R cos  – 15 )2
= 6 R cos 
 (S1P – S2P) =
6 Rcos
2R
 
= 3 cos .
For constructive interference,
(S1P – S2P)
2
= x = 3 cos  = n
 cos  = n/3   = cos–1
(n/3), where n = 0, 1, 2, ….
  = 0°, 48.2°, 70.5°, 90° and similar points in other quadrants.
30. a) As shown in the figure, BP0 – AP0 = /3
 2 2
(D d ) D /3   
 D
2
+ d
2
= D
2
+ (
2
/ 9) + (2D)/3
 d = (2 D)/3 (neglecting the term 
2
/9 as it is very small)
b) To find the intensity at P0, we have to consider the interference of light
waves coming from all the three slits.
Here, CP0 – AP0 = 2 2
D 4d D 
S2
Z
S1
P
2
Screen
P0
x
d
d
D
C
B
A
Screen
x2
OS2S1
P
D

xS1 1.5 O
P
S2
R

Chapter 17
17.6
=  
1/ 2
2 8 D 8
D D D 1 D
3 3D
 
    
=  8 4
D 1 ...... D
3D 2 3
 
   

[using binomial expansion]
So, the corresponding phase difference between waves from C and A is,
c =
2 x 2 4 8 2 2
2
3 3 3 3
      
      
   
…(1)
Again, B =
2 x 2
3 3
 


…(2)
So, it can be said that light from B and C are in same phase as they have some phase difference
with respect to A.
So, R = 2 2
(2r) r 2 2r r cos(2 /3)     (using vector method)
= 2 2 2
4r r 2r 3r  
 0
2 2
PI K( 3r) 3Kr 3I  
As, the resulting amplitude is 3 times, the intensity will be three times the intensity due to individual slits. 
31. Given that, d = 2 mm = 2  10
–3
m,  = 600 nm = 6  10
–7
m, Imax = 0.20 W/m
2
, D = 2m
For the point, y = 0.5 cm
We know, path difference = x =
2 3
yd 0.5 10 2 10
D 2
 
  
 = 5  10–6
m
So, the corresponding phase difference is,
 =
6
7
2 x 2 5 10
6 10


  

 

50 2
16
3 3
 
     =
2
3

So, the amplitude of the resulting wave at the point y = 0.5 cm is,
A = 2 2 2 2 2 2
r r 2r cos(2 /3) r r r      = r
Since,
2
2
max
I A
I (2r)
 [since, maximum amplitude = 2r]

2 2
2 2
I A r
0.2 4r 4r
 

0.2
I 0.05
4
  W/m
2
.
32. i) When intensity is half the maximum
max
I 1
I 2


2 2
2
4a cos ( / 2) 1
24a


 2
cos ( / 2) 1/ 2 cos( / 2) 1/ 2    
 /2 = /4   = /2
 Path difference, x = /4
 y = xD/d = D/4d
ii) When intensity is 1/4th
of the maximum
max
I 1
I 4


2 2
2
4a cos ( / 2) 1
44a


 2
cos ( / 2) 1/ 4 cos( / 2) 1/ 2    
 /2 = /3   = 2/3
 Path difference, x = /3
 y = xD/d = D/3d

Chapter 17
17.7
33. Given that, D = 1 m, d = 1 mm = 10–3
m,  = 500 nm = 5  10–7
m
For intensity to be half the maximum intensity.
y =
D
4d

(As in problem no. 32)
 y =
7
3
5 10 1
4 10


 

 y = 1.25  10–4
m.
34. The line width of a bright fringe is sometimes defined as the separation between the points on the two
sides of the central line where the intensity falls to half the maximum.
We know that, for intensity to be half the maximum
y = ±
D
4d

 Line width =
D
4d

+
D
4d

=
D
2d

.
35. i) When, z = D/2d, at S4, minimum intensity occurs (dark fringe)
 Amplitude = 0,
At S3, path difference = 0
 Maximum intensity occurs.
 Amplitude = 2r.
So, on 2 screen,
2
max
2
min
I (2r 0)
I (2r 0)



= 1
ii) When, z = D/2d, At S4, minimum intensity occurs. (dark fringe)
 Amplitude = 0.
At S3, path difference = 0
 Maximum intensity occurs.
 Amplitude = 2r.
So, on 2 screen,

2
max
2
min
I (2r 2r)
I (2r 0)

  

iii) When, z = D/4d, At S4, intensity = Imax / 2
 Amplitude = 2r .
 At S3, intensity is maximum.
 Amplitude = 2r

2
max
2
min
I (2r 2r)
I (2r 2r)



= 34.
36. a) When, z = D/d
So, OS3 = OS4 = D/2d  Dark fringe at S3 and S4.
 At S3, intensity at S3 = 0  I1 = 0
At S4, intensity at S4 = 0  I2 = 0
At P, path difference = 0  Phase difference = 0.
 I = I1 + I2 + 1 2I I cos 0° = 0 + 0 + 0 = 0  Intensity at P = 0.
b) Given that, when z = D/2d, intensity at P = I
Here, OS3 = OS4 = y = D/4d
  =
2 x 2 yd 2 D d
D 4d D 2
    
     
  
. [Since, x = path difference = yd/D]
Let, intensity at S3 and S4 = I
 At P, phase difference = 0
So, I + I + 2I cos 0° = I.
 4I = I  I = 1/4.
D
S1
S2
x
d
1
D
S3
S4
 2
O
P
D
S1
S2
z
d
D
S3
S4

Chapter 17
17.8
When, z =
3D
2d

,  y =
3D
4d

  =
2 x 2 yd 2 3D d 3
D 4d D 2
    
     
  
Let, I be the intensity at S3 and S4 when,  = 3/2
Now comparing,
2 2 2 2
2 2 2 2
I a a 2a cos(3 / 2) 2a
1
I a a 2a cos / 2 2a
   
  
  
 I = I = I/4.
 Intensity at P = I/4 + I/4 + 2  (I/4) cos 0° = I/2 + I/2 = I.
c) When z = 2D/d
 y = OS3 = OS4 = D/d
  =
2 x 2 yd 2 D d
2
D d D
   
      
  
.
Let, I = intensity at S3 and S4 when,  = 2.
2 2 2 2
2 2 2 2
I a a 2a cos2 4a
2
I a a 2a cos / 2 2a
   
  
   

 I = 2I = 2(I/4) = I/2
At P, Iresultant = I/2 + I/2 + 2(I/2) cos 0° = I + I = 2I.
So, the resultant intensity at P will be 2I.
37. Given d = 0.0011  10–3
m
For minimum reflection of light, 2d = n
  =
9
7
n 2n 580 10 2n 5.8
(2n)
2d 4d 444 11 10


   
  
 
= 0.132 (2n)
Given that,  has a value in between 1.2 and 1.5.
 When, n = 5,  = 0.132  10 = 1.32.
38. Given that,  = 560  10–9
m,  = 1.4.
For strong reflection, 2d = (2n + 1)/2  d =
(2n 1)
4d
 
For minimum thickness, putting n = 0.
 d =
4d

 d =
9
560 10
14


= 10–7
m = 100 nm. 
39. For strong transmission, 2 d = n   =
2 d
n

Given that,  = 1.33, d = 1  10
–4
cm = 1  10
–6
m.
  =
6 9
2 1.33 1 10 2660 10
m
n n
 
   

when, n = 4, 1 = 665 nm
n = 5, 2 = 532 nm
n = 6, 3 = 443 nm
40. For the thin oil film,
d = 1  10–4
cm = 10–6
m, oil = 1.25 and x = 1.50
 =
6 6
2 d 2 10 1.25 2 5 10 m
(n 1/ 2) 2n 1 2n 1
 
    

  
  =
5000 nm
2n 1
For the wavelengths in the region (400 nm – 750 nm)
When, n = 3,  =
5000 5000
2 3 1 7

 
= 714.3 nm

Chapter 17
17.9
When, n = 4,  =
5000 5000
2 4 1 9

 
= 555.6 nm
When, n = 5,  =
5000 5000
2 5 1 11

 
= 454.5 nm
41. For first minimum diffraction, b sin  = 
Here,  = 30°, b = 5 cm
  = 5  sin 30° = 5/2 = 2.5 cm.
42.  = 560 nm = 560  10
–9
m, b = 0.20 mm = 2  10
–4
m, D = 2 m
Since, R =
D
1.22
b

=
9
4
560 10 2
1.22
2 10


 


= 6.832  10
–3
M = 0.683 cm.
So, Diameter = 2R = 1.37 cm.
43.  = 620 nm = 620  10
–9
m,
D = 20 cm = 20  10–2
m, b = 8 cm = 8  10–2
m
 R =
4 2
2
620 10 20 10
1.22
8 10
 

  


= 1891  10
–9
= 1.9  10
–6
m
So, diameter = 2R = 3.8  10
–6
m
  

18.1
CHAPTER – 18
SIGN CONVENTION :
1) The direction of incident ray (from object to the mirror or lens) is taken as positive direction.
2) All measurements are taken from pole (mirror) or optical centre (lens) as the case may be.
1. u = –30 cm, R = – 40 cm
From the mirror equation,
1 1 2
v u R
 

1 2 1 2 1
v R u 40 30
   
 
=
1
60

or, v = –60 cm
So, the image will be formed at a distance of 60 cm in front of the mirror.
2. Given that,
H1 = 20 cm, v = –5 m = –500 cm, h2 = 50 cm
Since, 2
1
hv
u h


or
500 50
u 20
  (because the image in inverted)
or u =
500 2
5

 = –200 cm = – 2 m
1 1 1
v u f
  or
1 1 1
5 2 f
 
 
or f =
10
7

= –1.44 m
So, the focal length is 1.44 m.
3. For the concave mirror, f = –20 cm, M = –v/u = 2
 v = –2u
1st
case 2nd
case
1 1 1
v u f
 
1 1 1
2u u f

  

1 1 1
2u u f
   
3 1
2u f

 u = f/2 = 10 cm  u = 3f/2 = 30 cm
 The positions are 10 cm or 30 cm from the concave mirror.
4. m = –v/u = 0.6 and f = 7.5 cm = 15/2 cm
From mirror equation,

1 1 1
v u f
  
1 1 1
0.6u u f
 
 u = 5 cm
5. Height of the object AB = 1.6 cm
Diameter of the ball bearing = d = 0.4 cm
 R = 0.2 cm
Given, u = 20 cm
We know,
1 1 2
u v R
 
30cm
P
C S
– Sign convertion+ve

40cm
P
A 
B 
F

h1
A
B
500cm
h2
Case II(Real)
P
A 
B 
A
B
A 
B 
A
B
Case I (Virtual)
C
P

20cm0.2cm

Chapter 18
18.2
Putting the values according to sign conventions
1 1 2
20 v 0.2
 


1 1 201
10
v 20 20
    v = 0.1 cm = 1 mm inside the ball bearing.
Magnification = m =
A B v 0.1 1
AB u 20 200
 
    

 AB =
AB 16
200 200
 = +0.008 cm = +0.8 mm.
6. Given AB = 3 cm, u = –7.5 cm, f = 6 cm.
Using
1 1 1
v u f
  
1 1 1
v f u
 
Putting values according to sign conventions,
1 1 1 3
v 6 7.5 10
  

 v = 10/3 cm
 magnification = m =
v 10
u 7.5 3
 


A B 10 100 4
A B 1.33
AB 7.5 3 72 3
 
     

cm.
 Image will form at a distance of 10/3 cm. From the pole and image is 1.33 cm (virtual and erect).
7. R = 20 cm, f = R/2 = –10 cm
For part AB, PB = 30 + 10 = 40 cm
So, u = –40 cm 
1 1 1 1 1 3
v f u 10 40 40
 
        
 v =
40
3
 = –13.3 cm.
So, PB = 13.3 cm
m =
A B v 13.3 1
AB u 40 3
     
           
 AB = –10/3 = –3.33 cm
For part CD, PC = 30, So, u = –30 cm
1 1 1 1 1 1
v f u 10 30 15
 
        
 
 v = –15 cm = PC
So, m =
C D v 15 1
CD u 30 2
   
       
 C D  = 5 cm
BC = PC – PB = 15 – 13.3 = 17 cm
So, total length AB + BC + CD = 3.3 + 1.7 + 5 = 10 cm.
8. u = –25 cm
m =
A B v v 14 v
1.4
AB u 25 10 25
   
        
 v =
25 14
10

= 35 cm.
Now,
1 1 1
v u f
 

1 1 1 5 7 2
f 35 25 175 175
 
      
 f = –87.5 cm.
So, focal length of the concave mirror is 87.5 cm.
6cm
F
+ve

A 
B 
C
7.5cm
10cm A
B
C P
A 
B 
D
30cm
D
C

Chapter 18
18.3
9. u = –3.8  105
km
diameter of moon = 3450 km ; f = –7.6 m

1 1 1
v u f
   5
1 1 1
v 7.63.8 10
   
      
   
Since, distance of moon from earth is very large as compared to focal
length it can be taken as .
 Image will be formed at focus, which is inverted.

1 1
v 7.6
v 7.6
 
     
 
m.
m =
image
object
dv
u d
  
image
8 3
d( 7.6)
( 3.8 10 ) 3450 10
 

  
dimage =
3
8
3450 7.6 10
3.8 10
 

= 0.069 m = 6.9 cm.
10. u = –30 cm, f = –20 cm
We know,
1 1 1
v u f
 

1 1 1
v 30 20
   
      
   
 v = –60 cm.
Image of the circle is formed at a distance 60 cm in front of the mirror.
 m =
image
object
Rv
u R
  
imageR60
30 2

 

 Rimage = 4 cm
Radius of image of the circle is 4 cm.
11. Let the object be placed at a height x above the surface of the water.
The apparent position of the object with respect to mirror should be at the centre
of curvature so that the image is formed at the same position.
Since,
Real depth 1
Apparent depth


(with respect to mirror)
Now,
x 1 R h
x
R h

  
  
.
12. Both the mirrors have equal focal length f.
They will produce one image under two conditions.
Case I : When the source is at distance ‘2f’ from each mirror i.e. the source is at
centre of curvature of the mirrors, the image will be produced at the same point
S. So, d = 2f + 2f = 4f.
Case II : When the source S is at distance ‘f’ from each mirror, the rays from the
source after reflecting from one mirror will become parallel and so these parallel
rays after the reflection from the other mirror the object itself. So, only sine
image is formed.
Here, d = f + f = 2f.
13. As shown in figure, for 1st
reflection in M1, u = –30 cm, f = –20 cm

1 1 1
v 30 20
  

 v = –60 cm.
So, for 2nd
reflection in M2
u = 60 – (30 + x) = 30 – x
v = –x ; f = 20 cm
 21 1 1
x 10x 600 0
30 x x 20
     

7.6cm
F
+ve

F
Robject
20cm
30cm
RImage
C
R
O 
h 
x
(R–h)
S2f  2f 
S
f  f 
– for M1
30cm
S
+ve

+ve

– for M2
x

Chapter 18
18.4
 x =
10 50 40
2 2

 = 20 cm or –30 cm
 Total distance between the two lines is 20 + 30 = 50 cm.
14. We know,
8
sin i 3 10 sin45
2
sin r v sin30
 
  

 v =
8
3 10
2

m/sec.
Distance travelled by light in the slab is,
x =
1m 2
cos30 3


m
So, time taken = 8
2 2
3 3 10

 
= 0.54  10
–8
= 5.4  10
–9
sec.
15. Shadow length = BA = BD + AD = 0.5 + 0.5 tan r
Now, 1.33 =
sin45
sin r

 sin r = 0.53.
 cos r = 2 2
1 sin r 1 (0.53)   = 0.85
So, tan r = 0.6235
So, shadow length = (0.5) (1 + 0.6235) = 81.2 cm.
16. Height of the lake = 2.5 m
When the sun is just setting,  is approximately = 90°
 2
1
sin i
sin r




1 4/3 3
sinr
sin r 1 4
    r = 49°
As shown in the figure, x/2.5 = tan r = 1.15
 x = 2.5  1.15 = 2.8 m.
17. The thickness of the glass is d = 2.1 cm and  =1.5
Shift due to the glass slab
T =
1 1
1 d 1 2.1
1.5
   
        
= 0.7 CM
So, the microscope should be shifted 0.70 cm to focus the object again.
18. Shift due to water tw =
1 1
1 d 1 20
1.33
   
        
= 5 cm
Shift due to oil, to =
1
1 20
1.3
 
 
 
= 4.6 cm
Total shift t = 5 + 4.6 = 9.6 cm
Apparent depth = 40 – (9.6) = 30.4 cm below the surface.
19. The presence of air medium in between the sheets does not affect the shift.
The shift will be due to 3 sheets of different refractive index other than air.
=
1 1 1
1 (0.2) 1 (0.3) 1 (0.4)
1.2 13 14
     
         
     
= 0.2 cm above point P.
20. Total no. of slabs = k, thickness = t1, t2, t3 … tk
Refractive index = 1,2,3,4,…k
 The shift t = 1 2 k
1 2 k
1 1 1
1 t 1 t ...... 1 t
     
          
       
…(1)
If,   refractive index of combination of slabs and image is formed at same place,
t = 1 2 k
1
1 (t t ... t )
 
     
…(2)
45°
x 30° 1m
45°
A 
r 
0.5m
0.5m
0.5m
A B  D 
45°
x O
i=90°
2.5m
w 
w
Oil 20cm
Water 20cm
t = 0.4 cm
P
= 1.4
1 cm 
= 1.2
= 1.3 t = 0.3 cm
t = 0.2 cm
1 cm 


Chapter 18
18.5
Equation (1) and (2), we get,
 1 2 k
1
1 (t t ... t )
 
     
 1 2 k
1 2 k
1 1 1
1 t 1 t ...... 1 t
     
          
       
= 1 2 k
1 2 k
1 2 k
t t t
(t t ... t ) ...
 
          
=
k
ik k
1 i 1
1 k
1i 1 i 1
1 1
i 1
t
t1
t
(t / )

 

 
         

 

.
21. Given r = 6 cm, r1 = 4 cm, h1 = 8 cm
Let, h = final height of water column.
The volume of the cylindrical water column after the glass piece is put will be,
r
2
h = 800  + r1
2
h1
or r2
h = 800 + r1
2
h1
or 62
h = 800 + 42
 8 = 25.7 cm
There are two shifts due to glass block as well as water.
So, t1 = 0
0
1
1 t
 
  
=
1
1 8
3/ 2
 
 
 
= 2.26 cm
And, t2 = w
w
1 1
1 t 1 (25.7 8)
4/3
   
        
= 4.44 cm.
Total shift = (2.66 + 4.44) cm = 7.1 cm above the bottom.
22. a) Let x = distance of the image of the eye formed above the surface as seen by the fish
So,
H Real depth 1
x Apparent depth
 

or x = H
So, distance of the direct image =
H 1
H H( )
2 2
    
Similarly, image through mirror =
H 3H 3
(H x) H H( )
2 2 2
       
b) Here,
H/ 2
y
= , so, y =
H
2
Where, y = distance of the image of fish below the surface as seen by eye.
So, Direct image = H + y =
H 1
H H 1
2 2
 
     
Again another image of fish will be formed H/2 below the mirror.
So, the real depth for that image of fish becomes H + H/2 = 3H/2
So, Apparent depth from the surface of water = 3H/2
So, distance of the image from the eye =
3H 3
H H(1 )
2 2
  
 
.
23. According to the figure, x/3 = cot r …(1)
Again,
sini 1 3
sinr 1.33 4
 
 sin r =
4 4 3 4
sini
3 3 5 5
   (because sin i =
BC 3
AC 5
 )
 cot r = 3/4 …(2)
From (1) and (2)  x/3 = ¾
 x = 9/4 = 2.25 cm.
 Ratio of real and apparent depth = 4 : (2.25) = 1.78.
h 
h–h1
Glassh1
8cm
Water
8cm
12cm 

y
x
S 
H



H/2
H/2
90°
r 
4cm
A

r 
B
D
3cm
i
C

Chapter 18
18.6
24. For the given cylindrical vessel, dimetre = 30 cm
 r = 15 cm and h = 30 cm
Now, w
sini 3 4
1.33
sinr 4 3
 
     
 sin i = 3/ 4 2 [because r = 45°]
The point P will be visible when the refracted ray makes angle 45° at point of
refraction.
Let x = distance of point P from X.
Now, tan 45° =
x 10
d

 d = x + 10 …(1)
Again, tan i = x/d

3 d 10
d23


3 3
since, sini tani
4 2 23
 
   
 

3 10 23 10
1 d
d23 23 3

    

= 26.7 cm.
sin45 2 sin45 1
sinr r 21
sinr 1 2 2 2
 
      
Therefore,  = (45° – 21°) = 24°
Here, BD = shift in path = AB sin 24°
=
AE
0.406 AB 0.406
cos21
  

= 0.62 cm.
26. For calculation of critical angle,
2
1
sini
sinr




sinC 15 75
sin90 1.72 86
 
 C = 1 75
sin
26
  
 
 
.
27. Let c be the critical angle for the glass
1c
c c
sin 1 1 2 2
sin sin
sin90 x 1.5 3 3
  
         
  
From figure, for total internal reflection, 90° –  > c
  < 90° – c   < cos–1
(2/3)
So, the largest angle for which light is totally reflected at the surface is cos–1
(2/3).
28. From the definition of critical angle, if refracted angle is more than 90°, then reflection occurs, which is
known as total internal reflection.
So, maximum angle of refraction is 90°.
29. Refractive index of glass g = 1.5
Given, 0° < i < 90°
Let, C  Critical angle.
a
g
sinC sinC 1
sinr sin90 15

  
 
= 0.66
 C = 40°48
The angle of deviation due to refraction from glass to air increases as the angle of incidence increases
from 0° to 40°48. The angle of deviation due to total internal reflection further increases for 40°48 to
45° and then it decreases.
30. g = 1.5 = 3/2 ; w = 1.33 = 4/3
15cm
P

r 
d
i
5cm
C
P
x
10cm
45°
r 
A
2cm=AE
shift 
r 
B 
D
E
45°
90°–
A 


90°–
B  C 
90° i 

45°40°48

Chapter 18
18.7
For two angles of incidence,
1) When light passes straight through normal,
 Angle of incidence = 0°, angle of refraction = 0°, angle of deviation = 0
2) When light is incident at critical angle,
w
g
sinC
sinr



(since light passing from glass to water)
 sin C = 8/9  C = sin
–1
(8/9) = 62.73°.
 Angle of deviation = 90° – C = 90° – sin
–1
(8/9) = cos
–1
(8/9) = 37.27°
Here, if the angle of incidence is increased beyond critical angle, total internal
reflection occurs and deviation decreases. So, the range of deviation is 0 to
cos
–1
(8/9).
31. Since,  = 1.5, Critial angle = sin
–1
(1/) = sin
–1
(1/1.5) = 41.8°
We know, the maximum attainable deviation in refraction is (90° – 41.8°) = 47.2°
So, in this case, total internal reflection must have taken place.
In reflection,
Deviation = 180° – 2i = 90°  2i = 90°  i = 45°.
32. a) Let, x = radius of the circular area
x
tanC
h
 (where C is the critical angle)

2
2
x sinC 1/
h 11 sin C 1

 
 

(because sin C = 1/)

2
x 1
h 1

 
or x =
2
h
1 
So, light escapes through a circular area on the water surface directly above the point source.
b) Angle subtained by a radius of the area on the source, C =  1
sin 1/
 .
33. a) As shown in the figure, sin i = 15/25
So,
sini 1 3
sinr 4
 

 sin r = 4/5
Again, x/2 = tan r (from figure)
So, sin r =
2 2
tanr x / 2
1 tan r 1 x / 4

 

2
x 4
54 x


 25x2
= 16(4 + x2
)  9x2
= 64  x = 8/3 m
 Total radius of shadow = 8/3 + 0.15 = 2.81 m
b) For maximum size of the ring, i = critical angle = C
Let, R = maximum radius
 sin C =
2 2
sinC R 3
sinr 420 R
 

(since, sin r = 1)
 16R
2
= 9R
2
+ 9  400
 7R2
= 9  400
 R = 22.67 cm.
i =0
T=0
glass
i =0
T=0 water
T=90°
r 
glass
water
c  h 
S
x  x 
c 
i
r
xm
ceiling
2m 15
20cm

Chapter 18
18.8
34. Given, A = 60°,  = 1.732
Since, angle of minimum deviation is given by,
 =
A m
sin
2
sinA / 2
  
 
   1.732  ½ = sin(30 + m/2)
 sin
–1
(0.866) = 30 + m/2  60° = 30 m/2  m = 60°
Now, m = i + i – A
 60° = i + i – 60° ( = 60° minimum deviation)
 i = 60°. So, the angle of incidence must be 60°.
35. Given  = 1.5
And angle of prism = 4°
  =
m
m
A
sin
(A )/ 22
sinA / 2 (A / 2)
  
      (for small angle sin  = )
  = mA
2
 
 1.5 = m4
4
  

 m = 4°  (1.5) – 4° = 2°.
36. Given A = 60° and  = 30°
We know that,
 =
m m
m
A 60
sin sin
602 2 2sin
sinA / 2 sin30 2
     
       

Since, one ray has been found out which has deviated by 30°, the angle of minimum deviation should
be either equal or less than 30°. (It can not be more than 30°).
So,   m60
2sin
2
  
(because  will be more if m will be more)
or,   2  1/ 2 or,   2 .
37. 1 = 1, 2 = 1.5, R = 20 cm (Radius of curvature), u = –25 cm
 2 1 2 1
v u R
    
  
1.5 0.5 1 1 1 3
v 20 25 40 25 200

    
 v = –200  0.5 = –100 cm.
So, the image is 100 cm from (P) the surface on the side of S.
38. Since, paraxial rays become parallel after refraction i.e. image is formed at .
v = , 1 = 1.33, u = ?, 2 = 1.48, R = 30 cm
2 1 2 1
v u R
    
  
1.48 1.33 1.48 1.33 1.33 0.15
u 30 u 30

    

 u = –266.0 cm
 Object should be placed at a distance of 266 cm from surface (convex) on side A.
39. Given, 2 = 2.0
So, critical angle = 1 1
2
1 1
sin sin
2
    
      
= 30°
a) As angle of incidence is greater than the critical angle, the rays are totally
reflected internally.
b) Here, 2 1 2 1
v u R
    
 

2 1 2 1
v 3
 
   
  
[For parallel rays, u = ]

2 1
v 3
  v = 6 cm
 If the sphere is completed, image is formed diametrically opposite of A.
c) Image is formed at the mirror in front of A by internal reflection.
A
45°
A
C
B
3cm
m
r i
60°
4°
C S 
25cm
=1.0
–sign convertion
+ve
 
=1.5
20cm
30cm
P CO
=1.33
–sign convertion
+ve
 
=1.48 

Chapter 18
18.9
40. a) Image seen from left :
u = (5 – 15) = –3.5 cm
R = –5 cm
 2 1 2 1
v u R
    
  
1 1.5 1 1.5
v 3.5 5

  

1 1 3
v 10 7
   v =
70
23

= –3 cm (inside the sphere).
 Image will be formed, 2 cm left to centre.
b) Image seen from right :
u = –(5 + 1.5) = –6.5 cm
R = –5 cm
 2 1 2 1
v u R
    
  
1 1.5 1 1.5
v 6.5 5

 


1 1 3
v 10 13
   v = –
130
17
= –7.65 cm (inside the sphere).
 Image will be formed, 2.65 cm left to centre.
41. R1 = R2 = 10 cm, t = 5 cm, u = –
For the first refraction, (at A)
g g aa
1v u R
   
  or
1.5 1.5
0
v 10
 
 v = 30 cm.
Again, for 2
nd
surface, u = (30 – 5) = 25 cm (virtual object)
R2 = –10 cm
So,
1 15 0.5
v 25 10

 

 v = 9.1 cm.
So, the image is formed 9.1 cm further from the 2
nd
surface of the lens.
42. For the refraction at convex surface A.
 = –, 1 = 1, 2 = ?
a) When focused on the surface, v = 2r, R = r
So, 2 1 2 1
v u R
    
 
 2 2 1
2r r
  
  2 = 22 – 2  2 = 2
b) When focused at centre, u = r1, R = r
So, 2 1 2 1
v u R
    
 
 2 2 1
R r
  
  2 = 2 – 1.
This is not possible.
So, it cannot focus at the centre.
43. Radius of the cylindrical glass tube = 1 cm
We know, 2 1 2 1
v u R
    
 
Here, u = –8 cm, 2 = 3/2, 1 = 4/3, R = +1 cm
So,
3 4
2v 3 8



3 1 1
2v 6 6
  v = 
 The image will be formed at infinity.
C B
+ve

6.5cm
1.5cm
A CO
3.5cm
+ve

15cm
A B
+ve

–Sign convention
for both surfaces
v=2r
Image
v=r
Image
4/3
Glass rod
water 8cm

3/2

Chapter 18
18.10
44. In the first refraction at A.
2 = 3/2, 1 = 1, u = 0, R = 
So, 2 1 2 1
v u R
    
 
 v = 0 since (R   and u = 0)
The image will be formed at the point, Now for the second refraction at B,
u = –3 cm, R = –3 cm, 1 = 3/2, 2 = 1
So,
1 3 1 1.5 1
v 2 3 3 6

  
 

1 1 1 1
v 6 2 3
   
 v = –3 cm,  There will be no shift in the final image.
45. Thickness of glass = 3 cm, g = 1.5
Image shift =
1
3 1
1.5
 
 
 
[Treating it as a simple refraction problem because the upper surface is flat and
the spherical surface is in contact with the object]
=
0.5
3
1.5
 = 1 cm.
The image will appear 1 cm above the point P.
46. As shown in the figure, OQ = 3r, OP = r
So, PQ = 2r
For refraction at APB
We know, 2 1 2 1
v u R
    
 

1.5 1 0.5 1
v 2r r 2r
  

[because u = –2r]
 v = 
For the reflection in concave mirror
u = 
So, v = focal length of mirror = r/2
For the refraction of APB of the reflected image.
Here, u = –3r/2
1 1.5 0.5
v 3r / 2 r

 
 
[Here, 1 = 1.5 and 2 = 1 and R = –r]
 v = –2r
As, negative sign indicates images are formed inside APB. So, image should be at C.
So, the final image is formed on the reflecting surface of the sphere.
47. a) Let the pin is at a distance of x from the lens.
Then for 1st
refraction, 2 1 2 1
v u R
    
 
Here 2 = 1.5, 1 = 1, u = –x, R = –60 cm

1.5 1 0.5
v x 60
 
 
 120(1.5x + v) = –vx …(1)
 v(120 + x) = –180x
 v =
180x
120 x


This image distance is again object distance for the concave mirror.
3cm
B
A
object
3cm
O
I
2r
Q
P
O
r
C
A B
Pin 
x

Chapter 18
18.11
u =
180x
120 x


, f = –10 cm ( f = R/2)

1
1 1 1 1 1 (120 x)
v u f v 10 180x
 
    


1
1 120 x 18x
v 180x
 
  v1 =
180x
120 17x
Again the image formed is refracted through the lens so that the image is formed on the object taken in
the 1
st
refraction. So, for 2
nd
refraction.
According to sign conversion v = –x, 2 = 1, 1 = 1.5, R = –60
Now, 2 1 2 1
v u R
    
  [u =
180x
120 17x
]

1 1.5 0.5
(120 17x)
x 180x 60

  
 

1 120 17x 1
x 120x 120
 
 
Multiplying both sides with 120 m, we get
120 + 120 – 17x = –x
 16x = 240  x = 15 cm
 Object should be placed at 15 cm from the lens on the axis.
48. For the double convex lens
f = 25 cm, R1 = R and R2 = –2R (sign convention)
1 2
1 1 1
( 1)
f R R
 
    
 

1 1 1
(15 1)
25 R 2R
 
   
  
=
3R
0.5
2
 
 
 

1 3 1
25 4 R
  R = 18.75 cm
R1 = 18.75 cm, R2 = 2R = 37.5 cm.
49. R1 = +20 cm ; R2 = +30 cm ;  = 1.6
a) If placed in air :
g
1 2
1 1 1
( 1)
f R R
 
    
 
=
1.6 1 1
1
1 20 30
  
   
  
 f = 60/6 = 100 cm
b) If placed in water :
w
1 2
1 1 1
( 1)
f R R
 
    
 
=
1.6 1 1
1
1.33 20 30
  
   
  
 f = 300 cm
50. Given  = 1.5
Magnitude of radii of curvatures = 20 cm and 30 cm
The 4types of possible lens are as below.
1 2
1 1 1
( 1)
f R R
 
    
 
Case (1) : (Double convex) [R1 = +ve, R2 = –ve]
1 1 1
(15 1)
f 20 30
 
   
 
 f = 24 cm
Case (2) : (Double concave) [R1 = –ve, R2 = +ve]
1 1 1
(15 1)
f 20 30
 
   
 
 f = –24 cm
R1
C2
+ve

C1
R2
R1
C1 C2
R2

Chapter 18
18.12
Case (3) : (Concave concave) [R1 = –ve, R2 = –ve]
1 1 1
(15 1)
f 20 30
 
   
  
 f = –120 cm
Case (4) : (Concave convex) [R1 = +ve, R2 = +ve]
1 1 1
(15 1)
f 20 30
 
   
 
 f = +120 cm
51. a) When the beam is incident on the lens from medium 1.
Then 2 1 2 1
v u R
    
  or 2 1 2 1
v ( ) R
    
 

or 2 1
2
1
v R
  


or v = 2
2 1
R
  
Again, for 2nd
refraction, 3 3 22
v u R
   
 
or, 3 3 2 2
2 1
2
( )
v R R
     
       
 3 2 2 1
R
       
   
or, v = 3
3 2 1
R
2
 
       
So, the image will be formed at = 3
2 1 3
R
2

    
b) Similarly for the beam from 3 medium the image is formed at 1
2 1 3
R
2

    
.
52. Given that, f = 10 cm
a) When u = –9.5 cm
1 1 1
v u f
  
1 1 1 0.2
v 10 9.8 98

  
 v = – 490 cm
So,  m =
v 490
u 9.8



= 50 cm
So, the image is erect and virtual.
b) When u = –10.2 cm
1 1 1
v u f
  
1 1 1 102
v 10 10.2 0.2
  

 v = 510 cm
So, m =
v 510
u 9.8


The image is real and inverted.
53. For the projector the magnification required is given by
m =
v 200
u 3.5
  u = 17.5 cm
[35 mm > 23 mm, so the magnification is calculated taking object size 35 mm]
Now, from lens formula,

1 1 1
v u f
 

1 1 1 1 1 1
v u f 1000 17.5 f
    

 f = 17.19 cm.
+ve

1
2
3
1
2
3
F
A
B F
A
B
9.8cm
10cm
(Virtual image)
F
A
B
F
A
B
10cm
10.2cm
(Real image)

Chapter 18
18.13
54. When the object is at 19 cm from the lens, let the image will be at, v1.

1 1 1
1 1 1 1 1 1
v u f v 19 12
    

 v1 = 32.57 cm
Again, when the object is at 21 cm from the lens, let the image will
be at, v2

2 2 2
1 1 1 1 1 1
v u f v 21 12
    
 v2 = 28 cm
 Amplitude of vibration of the image is A = 1 2v vA B
2 2
  

 A =
32.57 28
2

= 2.285 cm.
55. Given, u = –5 cm, f = 8 cm
So,
1 1 1
v u f
  
1 1 3
8 5 40

 
 v = –13.3 cm (virtual image).
56. Given that,
(–u) + v = 40 cm = distance between object and image
ho = 2 cm, hi = 1 cm
Since i
o
h v
h u


= magnification

1 v
2 u


 u = –2v …(1)
Now,
1 1 1
v u f
  
1 1 1
v 2v f
 

3 1 2v
f
2v f 3
   …(2)
Again, (–u) + v = 40
 3v = 40  v = 40/3 cm
 f =
2 40
3 3


= 8.89 cm = focal length
From eqn. (1) and (2)
u = –2v = –3f = –3(8.89) = 26.7 cm = object distance.
57. A real image is formed. So, magnification m = –2 (inverted image)

v
u
= –2  v = –2u = (–2) (–18) =36
From lens formula,
1 1 1
v u f
  
1 1 1
36 18 f
 

 f = 12 cm
Now, for triple sized image m = –3 = (v/u)

1 1 1
v u f
  
1 1 1
3u u 12
 

 3u = –48  u = –16 cm
So, object should be placed 16 cm from lens.
58. Now we have to calculate the image of A and B. Let the images be A, B. So, length of A B = size of
image.
For A, u = –10 cm, f = 6 cm
MA B
v2
MA B
V1
19cm
21cm
20cm
F
A
B F
A
B
A
B
h2
A
Bh0
40cm
u
v

Chapter 18
18.14
Since,
1 1 1
v u f
  
1 1 1
v 10 6
 

 v = 15 cm = OA
For B, u = –12 cm, f = 6 cm
Again,
1 1 1
v u f
  
1 1 1
v 6 12
 
 v = 12 cm = OB
AB = OA – OB = 15 – 12 = 3 cm.
So, size of image = 3 cm.
59. u = –1.5  1011
m ; f = +20  10–2
m
Since, f is very small compared to u, distance is taken as . So, image will be formed at focus.
 v = +20  10–2
m
 We know, m =
image
object
hv
u h


2
image
11 9
D20 10
1.5 10 1.4 10



 
 Dimage = 1.86 mm
So, radius =
imageD
2
= 0.93 mm.
60. Given, P = 5 diopter (convex lens)
 f = 1/5 m = 20 cm
Since, a virtual image is formed, u and v both are negative.
Given, v/u = 4
 v = 4u …(1)
From lens formula,
1 1 1
v u f
 

1 1 1 1 1 4 3
f 4u u 20 4u 4u

     
 u = –15 cm
 Object is placed 15 cm away from the lens.
61. Let the object to placed at a distance x from the lens further away from the mirror.
For the concave lens (1st
refraction)
u = –x, f = –20 cm
From lens formula,
1 1 1
v u f
  
1 1 1
v 20 x
 
 
 v =
20x
x 20
 
 
 
So, the virtual image due to fist refraction lies on the same side as that of object. (AB)
This image becomes the object for the concave mirror.
For the mirror,
u =
20x 25x 100
5
x 20 x 20
   
      
    
f = –10 cm
1 1 1 1 1 x 20
v u f v 10 25x 100

    
 
OAB AB
2cm
11cm
F
u
v
f = 20 cm
O
A
B
x
B
A
A
B
5cm

Chapter 18
18.15
 v =
50(x 4)
3x 20


So, this image is formed towards left of the mirror.
Again for second refraction in concave lens,
u = –
50(x 4)
5
3x 20
 
  
(assuming that image of mirror is formed between the lens and mirro)
v = +x (Since, the final image is produced on the object)
Using lens formula,
1 1 1
v u f
  
1 1 1
50(x 4x 20
5
3x 20
 
 


 x = 60 cm
The object should be placed at a distance 60 cm from the lens further away from the mirror.
So that the final image is formed on itself.
62. It can be solved in a similar manner like question no.61, by using the sign conversions properly. Left as
an exercise for the student.
63. If the image in the mirror will form at the focus of the converging lens, then after transmission through
the lens the rays of light will go parallel.
Let the object is at a distance x cm from the mirror
 u = –x cm ; v = 25 – 15 = 10 cm (because focal length of lens = 25 cm)
f = 40 cm

1 1 1 1 1 1
v u f x 10 40
    
 x = 400/30 = 40/3
 The object is at distance
40 5
15
3 3
 
  
 
= 1.67 cm from the lens.
64. The object is placed in the focus of the converging mirror.
There will be two images.
a) One due to direct transmission of light through lens.
b) One due to reflection and then transmission of the rays through lens.
Case I : (S) For the image by direct transmission,
u = –40 cm, f = 15 cm

1 1 1
v u f
  
1 1 1
v 15 40
 

 v = 24 cm (left of lens)
Case II : (S) Since, the object is placed on the focus of mirror, after
reflection the rays become parallel for the lens.
So, u = 
 f = 15 cm

1 1 1
v u f
   v = 15 cm (left of lens)
65. Let the source be placed at a distance ‘x’ from the lens as shown, so that images formed by both
coincide.
For the lens,
1 1 1 15x
v
v x 15 x 15
   
 


…(1)
Fro the mirror, u = –(50 – x), f = –10 cm
So,
m
1 1 1
v (50 x) 10
  
 
O FQ

x
15cm
P
O S
x
50cm
S
S
S
40cm
50cm
S

Chapter 18
18.16

m
1 1 1
v (50 x) 10
 
 
So, vm =
10(50 x)
x 40


…(2)
Since the lens and mirror are 50 cm apart,
m
15x 10(50 x)
v v 50 50
x 15 (x 40)

    
 

 x = 30 cm.
So, the source should be placed 30 cm from the lens.
66. Given that, f1 = 15 cm, Fm = 10 cm, ho = 2 cm
The object is placed 30 cm from lens
1 1 1
v u f
  .
 v =
uf
u f
Since, u = –30 cm and f = 15 cm
So, v = 30 cm
So, real and inverted image (AB) will be formed at 30 cm from the lens and it will be of same size as
the object. Now, this real image is at a distance 20 cm from the concave mirror. Since, fm = 10 cm, this
real image is at the centre of curvature of the mirror. So, the mirror will form an inverted image AB at
the same place of same size.
Again, due to refraction in the lens the final image will be formed at AB and will be of same size as that
of object. (AB)
67. For the lens, f = 15 cm, u = –30 cm
From lens formula,
1 1 1
v u f
 

1 1 1 1
v 15 30 30
    v = 30 cm
The image is formed at 30 cm of right side due to lens only.
Again, shift due to glass slab is,
= t =
1
1 1
15
 
 
 
[since, g = 1.5 and t = 1 cm]
= 1 – (2/3) = 0.33 cm
 The image will be formed at 30 + 0.33 = 30.33 cm from the lens on right side.
68. Let, the parallel beam is first incident on convex lens.
d = diameter of the beam = 5 mm
Now, the image due to the convex lens should be formed on its
focus (point B)
So, for the concave lens,
u = +10 cm (since, the virtual object is on the right of concave lens)
f = –10 cm
So,
1 1 1
v u f
  
1 1 1
0
v 10 10
  

 v = 
So, the emergent beam becomes parallel after refraction in concave lens.
As shown from the triangles XYB and PQB,
PQ RB 10 1
XY ZB 20 2
  
So, PQ = ½  5 = 25 mm
So, the beam diameter becomes 2.5 mm.
Similarly, it can be proved that if the light is incident of the concave side, the beam diameter will be 1cm.
A
B
B
A
2cm
30cm
50cm
B
A
B
A
=1.5
1cm30cm
=1.5
A
1cm
B
Z 
Q 
P 
R
10cm 10cm

Chapter 18
18.17
69. Given that, f1 = focal length of converging lens = 30 cm
f2 = focal length of diverging lens = –20 cm
and d = distance between them = 15 cm
Let, F = equivalent focal length
So, 
1 2 1 2
1 1 1 d
F f f f f
   
1 1 15 1
30 20 30( 200 120
   
         
 F = 120 cm
 The equivalent lens is a converging one.
Distance from diverging lens so that emergent beam is parallel (image at infinity),
d1 =
1
dF 15 120
f 30

 = 60 cm
It should be placed 60 cm left to diverging lens
 Object should be placed (120 – 60) = 60 cm from diverging lens.
Similarly, d2 =
2
dF 15 120
f 20

 = 90 cm
So, it should be placed 90 cm right to converging lens.
 Object should be placed (120 + 90) = 210 cm right to converging lens.
70. a) First lens :
u = –15 cm, f = 10 cm
1 1 1
v u f
  
1 1 1
v 15 10
 
    
 
 v = 30 cm
So, the final image is formed 10 cm right of second lens.
b) m for 1st
lens :
image image
object
h hv 30
u h 15 5mm
 
   
 
 himage = –10 mm (inverted)
Second lens :
u = –(40 – 30) = –10 cm ; f = 5 cm
[since, the image of 1st
lens becomes the object for the second lens].
1 1 1
v u f
  
1 1 1
v 10 5
 
   
 
 v = 10 cm
m for 2
nd
lens :
image image
object
h hv 10
u h 10 10
 
   
 
 himage = 10 mm (erect, real).
c) So, size of final image = 10 mm
71. Let u = object distance from convex lens = –15 cm
v1 = image distance from convex lens when alone = 30 cm
f1 = focal length of convex lens
Now, 
1 1
1 1 1
v u f
 
or,
1
1 1 1 1 1
f 30 15 30 15
   

or f1 = 10 cm
Again, Let v = image (final) distance from concave lens = +(30 + 30) = 60 cm
v1 = object distance from concave lens = +30 m
1st 
40cm
15cm
2nd
60cm
30cm
15cm

Chapter 18
18.18
f2 = focal length of concave lens
Now, 
1 1
1 1 1
v v f
 
or,
1
1 1 1
f 60 30
   f2 = –60 cm.
So, the focal length of convex lens is 10 cm and that of concave lens is 60 cm.
72. a) The beam will diverge after coming out of the two convex lens system because, the image formed by
the first lens lies within the focal length of the second lens.
b) For 1
st
convex lens,
1 1 1
v u f
  
1 1
v 10
 (since, u = –)
or, v = 10 cm
for 2
nd
convex lens,
1 1 1
v f u
 

or,
1 1 1 1
v 10 (15 10) 10

  
  
or, v = –10 cm
So, the virtual image will be at 5 cm from 1
st
convex lens.
c) If, F be the focal length of equivalent lens,
Then,
1 2 1 2
1 1 1 d 1 1 15
F f f f f 10 10 100
      =
1
20
 F = 20 cm.
73. Let us assume that it has taken time ‘t’ from A to B.
 AB = 21
gt
2
 BC = h – 21
gt
2
This is the distance of the object from the lens at any time ‘t’.
Here, u = – ( h – 21
gt
2
)
2 = (given) and 1 = i (air)
So, 
2
1 1
1v R
(h gf )
2
  
 
 

2
2 2
1
( 1)(h gt ) R
1 1 2
1 1v R
(h gt ) R(h gt )
2 2
   
  
  
 
So, v = image distance at any time ‘t’ =
2
2
1
R(h gt )
2
1
( 1)(h gt ) R
2
 
   
So, velocity of the image = V =
2
2
2 2
1
R(h gt )
dv d R gt2
1 1dt dt
( 1)(h gt ) R ( 1)(h gt ) R
2 2
 
   
  
        
  
(can be found out).
74. Given that, u = distance of the object = –x
f = focal length = –R/2
and, V = velocity of object = dx/dt
f1 =10cm
15cm
f2 =10cm
B
R 
A
C
h 
½ gt2

Chapter 18
18.19
1 1 2
x v R
  

1 2 1 R 2x
v R x Rx

     v =
Rx
R 2x
= Image distance
So, velocity of the image is given by,
V1 = 2
d d
[ (xR)(R 2x)] [ (R 2x)][xR]
dv dt dt
dt (R 2x)
  


= 2 2
dx dx
R[ (R 2x)] [ 2 x]
R[v(R 2x) 2vx0dt dt
(R 2x) (R 2x)
  
 

 
=
2
2 2
VR R[VR 2xV 2xV
(2x R) (R 2x)
 

 
.
75. a) When t < d/V, the object is approaching the mirror
As derived in the previous question,
Vimage =
2
2
Velocity of object R
[2 distance between them R]

 
 Vimage =
2
2
VR
[2(d Vt) R] 
[At any time, x = d – Vt]
b) After a time t > d/V, there will be a collision between the mirror and the mass.
As the collision is perfectly elastic, the object (mass) will come to rest and the mirror starts to move
away with same velocity V.
At any time t > d/V, the distance of the mirror from the mass will be
x =
d
V t Vt d
V
 
   
 
Here, u = –(Vt – d) = d – Vt ; f = –R/2
So,
1 1 1
v u f
  
1 1 1 R 2(d Vt)
v d Vt ( R / 2) R(d Vt)
  
         
 v =
R(d Vt)
R 2(d Vt)
 
    
= Image distance
So, Velocity of the image will be,
Vimage =
d d R(d Vt)
(Image distance) =
dt dt R 2(d Vt)
 
   
Let, y = (d – Vt)

dy
V
dt
 
So, Vimage = 2
d Ry (R 2y)R( V) Ry( 2)( V)
dt R 2y (R 2y)
     
   
=
2
2 2
R 2y 2y VR
Vr
(R 2y) (R 2y)
   
    
Since, the mirror itself moving with velocity V,
Absolute velocity of image =
2
2
R
V 1
(R 2y)
 
 
 
(since, V = Vmirror + Vimage)
=
2
2
R
V 1
[2(Vt d) R
 
 
  
.
v
ImageObject
x
V 
d
vm = 0
t < (d/v)
M 
V 
x
VB = 0
t < (d/v)


Chapter 18
18.20
76. Recoil velocity of gun = Vg =
mV
M
.
At any time ‘t’, position of the bullet w.r.t. mirror =
mV m
Vt t 1 Vt
M M
 
   
 
For the mirror, u =
m
1 Vt kVt
M
 
   
 
v = position of the image
From lens formula,
1 1 1 1 1 1 1 1 f kVt
v f u v f kVt kVt f kVtf

       

Let
m
1 k
M
 
  
 
,
So, v =
kVft kVtf
kVt f f kVt
 
  
   
So, velocity of the image with respect to mirror will be,
v1 =
2
2 2
dv d kVtf (f kVt)kVf kVtf( kV) kVt
dt dt f kVt (f kVt) (f kVt)
   
      
Since, the mirror itself is moving at a speed of mV/M and the object is moving at ‘V’, the velocity of
separation between the image and object at any time ‘t’ will be,
vs = V +
2
2
mV kVf
M (f kVt)


When, t = 0 (just after the gun is fired),
vs =
mV m m m
V kV V V 1 V 2 1 V
M M M M
   
          
   
77. Due to weight of the body suppose the spring is compressed by which is the mean position of
oscillation.
m = 50  10
–3
kg, g = 10 ms
–2
, k = 500 Nm
–2
, h = 10 cm = 0.1 m
For equilibrium, mg = kx  x = mg/k = 10–3
m = 0.1 cm
So, the mean position is at 30 + 0.1 = 30.1 cm from P (mirror).
Suppose, maximum compression in spring is .
Since, E.K.E. – I.K.E. = Work done
 0 – 0 = mg(h + ) – ½ k
2
(work energy principle)
 mg(h + ) = ½ k
2
 50  10
–3
 10(0.1 + ) = ½ 500 
2
So,  =
0.5 0.25 50
2 250
 

= 0.015 m = 1.5 cm.
From figure B,
Position of B is 30 + 1.5 = 31.5 cm from pole.
Amplitude of the vibration = 31.5 – 30.1 – 1.4.
Position A is 30.1 – 1.4 = 28.7 cm from pole.
For A u = –31.5, f = –12 cm

1 1 1 1 1
v f u 12 31.5
    
 vA = –19.38 cm
For B f = –12 cm, u = –28.7 cm
1 1 1 1 1
v f u 12 28.7
    
 vB = –20.62 cm
The image vibrates in length (20.62 – 19.38) = 1.24 cm.
mirror
v
vg = (mV)/m
30 cm
Fig-A

30.1 cmA 
1.4cm
B 
Fig-B
mean
mirror

1.4cm

Chapter 18
18.21
78. a) In time, t = R/V the mass B must have moved (v  R/v) = R closer to the mirror stand
So, For the block B :
u = –R, f = –R/2

1 1 1 2 1 1
v f u R R R
      
 v = –R at the same place.
For the block A : u = –2R, f = –R/2

1 1 1 2 1 3
v f u R 2R 2R
 
    
 v =
2R
3

image of A at
2R
3
from PQ in the x-direction.
So, with respect to the given coordinate system,
 Position of A and B are
2R
3

, R respectively from origin.
b) When t = 3R/v, the block B after colliding with mirror stand must have come to rest (elastic collision)
and the mirror have travelled a distance R towards left form its initial position.
So, at this point of time,
For block A :
u = –R, f = –R/2
Using lens formula, v = –R (from the mirror),
So, position xA = –2R (from origin of coordinate system)
For block B :
Image is at the same place as it is R distance from mirror. Hence,
position of image is ‘0’.
Distance from PQ (coordinate system)
 positions of images of A and B are = –2R, 0 from origin.
c) Similarly, it can be proved that at time t = 5R/v,
the position of the blocks will be –3R and –4R/3 respectively.
79. Let a = acceleration of the masses A and B (w.r.t. elevator). From the freebody diagrams,
T – mg + ma – 2m = 0 …(1)
Similarly, T – ma = 0 …(2)
From (1) and (2), 2ma – mg – 2m = 0
 2ma = m(g + 2)
 a =
10 2 12
2 2

 = 6 ms–2
so, distance travelled by B in t = 0.2 sec is,
s = 2 21 1
at 6 (0.2)
2 2
   = 0.12 m = 12 cm.
So, Distance from mirror, u = –(42 – 12) = –30 cm ; f = +12 cm
1 1 1 1 1 1
v u f v 30 12
 
      
 
 v = 8.57 cm
Distance between image of block B and mirror = 8.57 cm.
  
m
A  B origin
v
R
m
R
m
2R
m
R
v
mB
v
R
m
2R
a
m
m
A
B
a
2m/s2
FBD-A
ma T 
FBD-B
mg
ma 
T 
m(2) 

19.1
CHAPTER 19
1. The visual angles made by the tree with the eyes can be
calculated be below.
 = A
Height of the tree AB 2
0.04
Distance from the eye OB 50
    
similarly, B = 2.5 / 80 = 0.03125
C = 1.8 / 70 = 0.02571
D = 2.8 / 100 = 0.028
Since, A >B >D >C, the arrangement in decreasing order is given by A, B, D and C.
2. For the given simple microscope,
f = 12 cm and D = 25 cm
For maximum angular magnification, the image should be produced
at least distance of clear vision.
So, v = – D = –25 cm
Now,
1 1 1
v u f
 

1 1 1 1 1 37
u v f 25 12 300
     

 u = –8.1 cm
So, the object should be placed 8.1 cm away from the lens.
3. The simple microscope has, m = 3, when image is formed at D = 25 cm
a) m =
D
1
f
 
25
3 1
f
 
 f = 25/2 = 12.5 cm
b) When the image is formed at infinity (normal adjustment)
Magnifying power =
D 25
f 12.5
 = 2.0
4. The child has D = 10 cm and f = 10 cm
The maximum angular magnification is obtained when the image is formed at near point.
m =
D 10
1 1
f 10
   = 1 + 1 = 2
5. The simple microscope has magnification of 5 for normal relaxed eye (D = 25 cm).
Because, the eye is relaxed the image is formed at infinity (normal adjustment)
So, m = 5 =
D 25
f f
  f = 5 cm
For the relaxed farsighted eye, D = 40 cm
So, m =
D 40
f 5
 = 8
So, its magnifying power is 8X.
A
A
+ve
B
B
D=25cm
(Simple Microscope)
Distance
A
height
B


Chapter 19
19.2
6. For the given compound microscope
f0 =
1
25 diopter
= 0.04 m = 4 cm, fe =
1
5 diopter
= 0.2 m = 20 cm
D = 25 cm, separation between objective and eyepiece = 30 cm
The magnifying power is maximum when the image is formed by
the eye piece at least distance of clear vision i.e. D = 25 cm
for the eye piece, ve = –25 cm, fe = 20 cm
For lens formula,
e e e
1 1 1
v u f
 

e e e
1 1 1 1 1
u v f 25 20
   

 ue = 11.11 cm
So, for the objective lens, the image distance should be
v0 = 30 – (11.11) = 18.89 cm
Now, for the objective lens,
v0 = +18.89 cm (because real image is produced)
f0 = 4 cm
So,
o o o
1 1 1 1 1
u v f 18.89 4
    = 0.053 – 0.25 = –0.197
 uo = –5.07 cm
So, the maximum magnificent power is given by
m = o
o e
v D 18.89 25
1 1
u f 5.07 20
   
          
= 3.7225  2.25 = 8.376
7. For the given compound microscope
fo = 1 cm, fe = 6 cm, D = 24 cm
For the eye piece, ve = –24 cm, fe = 6 cm
Now,
e e e
1 1 1
v u f
 

e e e
1 1 1 1 1 5
u v f 24 6 24
 
        
 ue = –4.8 cm
a) When the separation between objective and eye piece is 9.8 cm, the image distance for the objective
lens must be (9.8) – (4.8) = 5.0 cm
Now,
00 u
1
v
1
 =
0f
1

0u
1
=
00 f
1
v
1
 =
1
1
5
1
 = –
5
4
 u0 = –
4
5
= – 1.25 cm
So, the magnifying power is given by,
m = 






f
D
1
u
v
o
0
= 








6
24
1
25.1
5
= 4 × 5 = 20
(b) When the separation is 11.8 cm,
v0 = 11.8 – 4.8 = 7.0 cm, f0 = 1 cm

0u
1
=
00 f
1
v
1
 =
1
1
7
1
 =
7
6

fo = 0.04m
objective
A
B
A
BB
11.11
cm
25cm
30cm
A
fe = 0.2m
objective
Fig-A
5 cm
9.8cm
4.8cm
24 cm
Fig-B
7 cm
11.8cm
4.8cm
24 cm

Chapter 19
19.3
So, m = – 






f
D
1
u
v
o
0
= 














6
24
1
6
7
7
= 6 × 5 = 30
So, the range of magnifying power will be 20 to 30.
8. For the given compound microscope.
f0 =
D20
1
= 0.05 m = 5 cm, fe =
D10
1
= 0.1 m = 10 cm.
D = 25 cm, separation between objective & eyepiece= 20 cm
For the minimum separation between two points which can be distinguished by eye using the
microscope, the magnifying power should be maximum.
For the eyepiece, v0 = –25 cm, fe = 10 cm
So,
eu
1
=
ee f
1
v
1
 =
10
1
25
1


= – 




 
50
52
 ue = –
7
50
cm
So, the image distance for the objective lens should be,
V0 = 20 –
7
50
=
7
90
cm
Now, for the objective lens,
000 f
1
v
1
u
1
 =
5
1
90
7
 = –
90
11
 u0 = –
11
90
cm
So, the maximum magnifying power is given by,
m = 







e0
0
f
D
1
u
v
= 



















10
25
1
11
90
7
90
= 5.3
7
11
 = 5.5
Thus, minimum separation eye can distinguish =
5.5
22.0
mm = 0.04 mm
9. For the give compound microscope,
f0 = 0.5cm, tube length = 6.5cm
magnifying power = 100 (normal adjustment)
Since, the image is formed at infinity, the real image produced by the objective lens should lie on the
focus of the eye piece.
So, v0 + fe = 6.5 cm …(1)
Again, magnifying power=
e0
0
f
D
u
v
 [for normal adjustment]
 m = –
e0
0
f
D
f
v
1 





 






0
0
0
0
f
v
1
u
v

 100 = –
e
0
f
25
5.0
v
1 





 [Taking D = 25 cm]
 100 fe = –(1 – 2v0) × 25
 2v0 – 4fe = 1 …(2)

v0 Eye piece
fe
Fe
Objective 

Chapter 19
19.4
Solving equation (1) and (2) we can get,
V0 = 4.5 cm and fe = 2 cm
So, the focal length of the eye piece is 2cm.
10. Given that,
fo = = 1 cm, fe = 5 cm, u0 = 0.5 cm, ve = 30 cm
For the objective lens, u0 = – 0.5 cm, f0 = 1 cm.
From lens formula,
000 f
1
u
1
v
1
 
000 f
1
u
1
v
1
 =
1
1
5.0
1


= – 1
 v0 = – 1 cm
So, a virtual image is formed by the objective on the same side as that of the object at a distance of 1
cm from the objective lens. This image acts as a virtual object for the eyepiece.
For the eyepiece,
000 f
1
u
1
v
1
 
000 f
1
v
1
u
1
 =
5
1
30
1
 =
30
5
=
6
1
 u0 = – 6 cm
So, as shown in figure,
Separation between the lenses = u0 – v0 = 6 – 1 = 5 cm
11. The optical instrument has
f0 =
D25
1
= 0.04 m = 4 cm
fe =
D20
1
= 0.05 m = 5 cm
tube length = 25 cm (normal adjustment)
(a) The instrument must be a microscope as f0 < fe
(b) Since the final image is formed at infinity, the image produced
by the objective should lie on the focal plane of the eye piece.
So, image distance for objective = v0 = 25 – 5 = 20 cm
Now, using lens formula.
000 f
1
u
1
v
1
 
000 f
1
v
1
u
1
 =
4
1
20
1
 =
20
4
=
5
1
 u0 = – 5 cm
So, angular magnification = m = –
e0
0
f
D
u
v
 [Taking D = 25 cm]
= –
5
25
5
20


= 20
12. For the astronomical telescope in normal adjustment.
Magnifying power = m = 50, length of the tube = L = 102 cm
Let f0 and fe be the focal length of objective and eye piece respectively.
m =
e
0
f
f
= 50  f0 = 50 fe …(1)
and, L = f0 + fe = 102 cm …(2)
Putting the value of f0 from equation (1) in (2), we get,
f0 + fe = 102  51fe = 102  fe = 2 cm = 0.02 m
So, f0 = 100 cm = 1 m
 Power of the objective lens =
0f
1
= 1D
And Power of the eye piece lens =
ef
1
=
02.0
1
= 50D
B 
A 

20cm
5cm
Fe

f0 Eye piece
fe
Fe
Objective 
A A
0.5cm
Eye piece1cm
Objective 
30cm
60cm
B
A
B
B 

Chapter 19
19.5
13. For the given astronomical telescope in normal adjustment,
Fe = 10 cm, L = 1 m = 100cm
S0, f0 = L – fe = 100 – 10 = 90 cm
and, magnifying power =
e
0
f
f
=
10
90
= 9
14. For the given Galilean telescope, (When the image is formed at infinity)
f0 = 30 cm, L = 27 cm
Since L = f0 – ef
[Since, concave eyepiece lens is used in Galilean Telescope]
 fe = f0 – L = 30 – 27 = 3 cm
15. For the far sighted person,
u = – 20 cm, v = – 50 cm
from lens formula
f
1
u
1
v
1

f
1
=
20
1
50
1



=
50
1
20
1
 =
100
3
 f =
3
100
cm =
3
1
m
So, power of the lens =
f
1
= 3 Diopter
16. For the near sighted person,
u =  and v = – 200 cm = – 2m
So,
u
1
v
1
f
1
 =



1
2
1
= –
2
1
= – 0.5
So, power of the lens is –0.5D
17. The person wears glasses of power –2.5D
So, the person must be near sighted.
u = , v = far point, f=
5.2
1

= – 0.4m = – 40 cm
Now,
f
1
u
1
v
1


f
1
u
1
v
1
 =
40
1
0

  v = – 40 cm
So, the far point of the person is 40 cm
18. On the 50
th
birthday, he reads the card at a distance 25cm using a glass of +2.5D.
Ten years later, his near point must have changed.
So after ten years,
u = – 50 cm, f =
D5.2
1
= 0.4m = 40 cm v = near point
Now,
f
1
u
1
v
1
 
f
1
u
1
v
1
 =
40
1
50
1


=
200
1
So, near point = v = 200cm
To read the farewell letter at a distance of 25 cm,
U = – 25 cm
For lens formula,
f
1
u
1
v
1
 
f
1
=
25200
1


 =
25
1
200
1
 =
200
9
 f =
9
200
cm =
9
2
m
 Power of the lens =
f
1
=
2
9
= 4.5D
He has to use a lens of power +4.5D.

Chapter 19
19.6
19. Since, the retina is 2 cm behind the eye-lens
v = 2cm
(a) When the eye-lens is fully relaxed
u = , v = 2cm = 0.02 m

u
1
v
1
f
1
 =


1
02.0
1
= 50D
So, in this condition power of the eye-lens is 50D
(b) When the eye-lens is most strained,
u = – 25 cm = – 0.25 m, v = +2 cm = +0.02 m

u
1
v
1
f
1
 =
25.0
1
02.0
1

 = 50 + 4 = 54D
In this condition power of the eye lens is 54D.
20. The child has near point and far point 10 cm and 100 cm respectively.
Since, the retina is 2 cm behind the eye-lens, v = 2cm
For near point u = – 10 cm = – 0.1 m, v = 2 cm = 0.02 m
So,
u
1
v
1
f
1
near
 =
1.0
1
02.0
1

 = 50 + 10 = 60D
For far point, u = – 100 cm = – 1 m, v = 2 cm = 0.02 m
So,
u
1
v
1
f
1
far
 =
1
1
02.0
1

 = 50 + 1 = 51D
So, the rage of power of the eye-lens is +60D to +51D
21. For the near sighted person,
v = distance of image from glass
= distance of image from eye – separation between glass and eye
= 25 cm – 1cm = 24 cm = 0.24m
So, for the glass, u =  and v = – 24 cm = –0.24m
So,
u
1
v
1
f
1
 =



1
24.0
1
= – 4.2 D
22. The person has near point 100 cm. It is needed to read at a distance of 20cm.
(a) When contact lens is used,
u = – 20 cm = – 0.2m, v = – 100 cm = –1 m
So,
u
1
v
1
f
1
 =
2.0
1
1
1



= – 1 + 5 = + 4D
(b) When spectacles are used,
u = – (20 – 2) = – 18 cm = – 0.18m, v = – 100 cm = –1 m
So,
u
1
v
1
f
1
 =
18.0
1
1
1



= – 1 + 5.55 = + 4.5D
23. The lady uses +1.5D glasses to have normal vision at 25 cm.
So, with the glasses, her least distance of clear vision = D = 25 cm
Focal length of the glasses =
5.1
1
m =
5.1
100
cm
So, without the glasses her least distance of distinct vision should be more
If, u = – 25cm, f =
5.1
100
cm
Now,
f
1
u
1
v
1
 =
25
1
100
5.1
 =
100
45.1 
=
100
5.2
 v = – 40cm = near point without glasses.
Focal length of magnifying glass =
20
1
m = 0.05m = 5 cm = f
Retina
Eye lens
2cm

Chapter 19
19.7
(a) The maximum magnifying power with glasses
m =
f
D
1 =
5
25
1 = 6 [D = 25cm]
(b) Without the glasses, D = 40cm
So, m =
f
D
1 =
5
40
1 = 9
24. The lady can not see objects closer than 40 cm from the left eye and 100 cm from the right eye.
For the left glass lens,
v = – 40 cm, u = – 25 cm

u
1
v
1
f
1
 =
25
1
40
1



=
40
1
25
1
 =
200
3
 f =
3
200
cm
For the right glass lens,
v = – 100 cm, u = – 25 cm
u
1
v
1
f
1
 =
25
1
100
1



=
100
1
25
1
 =
100
3
 f =
3
100
cm
(a) For an astronomical telescope, the eye piece lens should have smaller focal length. So, she should
use the right lens (f =
3
100
cm) as the eye piece lens.
(b) With relaxed eye, (normal adjustment)
f0 =
3
200
cm, fe =
3
100
cm
magnification = m =
e
0
f
f
=
 
 3/100
3/200
= 2
    

20.1
CHAPTER – 20
1. Given that,
Refractive index of flint glass = f = 1.620
Refractive index of crown glass = c = 1.518
Refracting angle of flint prism = Af = 6.0°
For zero net deviation of mean ray
(f – 1)Af = (c – 1) Ac
 Ac = f
c
f
A
1
1


= 


)0.6(
1518.1
1620.1
= 7.2°
2. Given that
r = 1.56,  y = 1.60, and v = 1.68
(a) Dispersive power =  =
1y
rv


=
160.1
56.168.1


= 0.2
(b) Angular dispersion = (v - r)A = 0.12 × 6° = 7.2° 
3. The focal length of a lens is given by
f
1
= ( – 1) 






21 R
1
R
1
 ( – 1) =








21 R
1
R
1
1
f
1
=
f
K
…(1)
So, r – 1 =
100
K
…(2)
y – 1 =
98
K
…(3)
And v – 1 =
96
K
(4)
So, Dispersive power = =
1y
rv


=
)1(
)1()1(
y
rv


=
98
K
100
K
96
K

=
9600
498 
= 0.0408
4. Given that, v – r = 0.014
Again, y =
depthApparent
depthalRe
=
30.1
00.2
= 1.515
So, dispersive power =
1y
rv


=
1515.1
014.0

= 0.027
5. Given that, r = 1.61, v = 1.65,  = 0.07 and y = 4°
Now,  =
1y
rv



 0.07 =
1
61.165.1
y 


 y – 1 =
07.0
04.0
=
7
4
Again,  = ( – 1) A
 A =
1y
y


=
)7/4(
4
= 7°
2cmImage
Object
1.32cm

Chapter 20
20.2
6. Given that, r = 38.4°, y = 38.7° and v = 39.2°
Dispersive power =
1y
rv


=
)1(
)1()1(
y
rv


=





 





 





 
A
AA
v
rv
[  = ( – 1) A]
=
y
rv


=
7.38
4.382.39 
= 0.0204
7. Two prisms of identical geometrical shape are combined.
Let A = Angle of the prisms
v = 1.52 and v = 1.62, v = 1°
v = (v – 1)A – (v – 1) A [since A = A]
 v = (v – v)A
 A =
vv
v


=
52.162.1
1

= 10°
8. Total deviation for yellow ray produced by the prism combination is
y = cy – fy + cy = 2cy – fy = 2(cy – 1)A – (cy – 1)A
Similarly the angular dispersion produced by the combination is
v – r = [(vc – 1)A – (vf – 1)A + (vc – 1)A] – [(rc – 1) A – (rf – 1)A + (r – 1) A)]
= 2(vc – 1)A – (vf – 1)A
(a) For net angular dispersion to be zero,
v – r = 0
 2(vc – 1)A = (vf – 1)A

A
A
=
)(
)(2
rfvf
rccv


=
)(
)(2
rv
rv


(b) For net deviation in the yellow ray to be zero,
y = 0
 2(cy – 1)A = (fy – 1)A

A
A
=
)1(
)1(2
fy
cy


=
)1(
)1(2
y
y



9. Given that, cr = 1.515, cv = 1.525 and fr = 1.612, fv = 1.632 and A = 5°
Since, they are similarly directed, the total deviation produced is given by,
 = c + r = (c – 1)A + (r – 1) A = (c + r – 2)A
So, angular dispersion of the combination is given by,
v – y = (cv + fv – 2)A – (cr + fr – 2)A
= (cv + fv – cr – fr)A = (1.525 + 1.632 – 1.515 – 1.612) 5 = 0.15°
10. Given that, A = 6°,  = 0.07, y = 1.50
A = ?  = 0.08, y = 1.60
The combination produces no deviation in the mean ray.
(a) y = (y – 1)A – (y – 1)A = 0 [Prism must be oppositely directed]
 (1.60 – 1)A = ((1.50 – 1)A
 A =
60.0
650.0 
= 5° 
(b) When a beam of white light passes through it,
Net angular dispersion = (y – 1)A – (y – 1)A
 (1.60 – 1)(0.08)(5°) – (1.50 – 1)(0.07)(6°)
 0.24° – 0.21° = 0.03°
(c) If the prisms are similarly directed,
y = (y – 1)A + (y – 1)A
= (1.60 – 1)5° + (1.50 – 1)6° = 3° + 3° = 6°
(d) Similarly, if the prisms are similarly directed, the net angular dispersion is given by,
v – r = (y – 1)A – (y – 1) A = 0.24° + 0.21° = 0.45°
5°6°
5°6°
5° 5°
Prism2
Prism1
CrownCrown
A A
A
Flint

Chapter 20
20.3
11. Given that, v – r = 0.014 and v – r = 0.024
A = 5.3° and A = 3.7°
(a) When the prisms are oppositely directed,
angular dispersion = (v – r)A – (v – r)A
= 0.024 × 3.7° – 0.014 × 5.3° = 0.0146°
(b) When they are similarly directed,
angular dispersion = (v – r)A + (v – r)A
= 0.024 × 3.7° + 0.014 × 5.3° = 0.163°
♠ ♠ ♠ ♠ ♠
3.7° 5.3°
5.3°3.7°

21.1
CHAPTER 21
1. In the given Fizeau’ apparatus,
D = 12 km = 12 × 10
3
m
n = 180
c = 3 × 10
8
m/sec
We know, c =

Dn2
  =
Dn2
c
rad/sec =


 180
Dn2
c
deg/sec
  =
1801024
103180
3
8


= 1.25 × 10
4
deg/sec 
2. In the given Focault experiment,
R = Distance between fixed and rotating mirror = 16m
 = Angular speed = 356 rev/ = 356 × 2 rad/sec
b = Distance between lens and rotating mirror = 6m
a = Distance between source and lens = 2m
s = shift in image = 0.7 cm = 0.7 × 10
–3
m
So, speed of light is given by,
C =
)bR(s
aR4 2


=
)616(107.0
22356164
3
2



= 2.975 × 10
8
m/s
3. In the given Michelson experiment,
D = 4.8 km = 4.8 × 10
3
m
N = 8
We know, c =


2
ND
  =
DN
c2
rad/sec =
DN
c
rev/sec =
8108.4
103
3
8


= 7.8 × 10
3
rev/sec
* * * * *

22.1
CHAPTER 22
1. Radiant Flux =
Time
emittedenergyTotal
=
s15
45
= 3W
2. To get equally intense lines on the photographic plate, the radiant flux (energy) should be same.
S0, 10W × 12sec = 12W × t
 t =
W12
sec12W10 
= 10 sec.
3. it can be found out from the graph by the student.
4. Relative luminousity =
powersameofnm555ofsourceaoffluxousminLu
wavelengthgivenofsourceaoffluxousminLu
Let the radiant flux needed be P watt.
Ao, 0.6 =
P685
watt'P'sourceoffluxousminLu
 Luminous flux of the source = (685 P)× 0.6 = 120 × 685
 P =
6.0
120
= 200W
5. The luminous flux of the given source of 1W is 450 lumen/watt
 Relative luminosity =
powersameofsourcenm555offluxousminLu
wavelengthgivenofsourcetheoffluxousminLu
=
685
450
= 66%
[ Since, luminous flux of 555nm source of 1W = 685 lumen]
6. The radiant flux of 555nm part is 40W and of the 600nm part is 30W
(a) Total radiant flux = 40W + 30W = 70W
(b) Luminous flux = (L.Fllux)555nm + (L.Flux)600nm
= 1 × 40× 685 + 0.6 × 30 × 685 = 39730 lumen
(c) Luminous efficiency =
fluxradiantTotal
fluxousminluTotal
=
70
39730
= 567.6 lumen/W
7. Overall luminous efficiency =
inputPower
fluxousminluTotal
=
100
68535 
= 239.75 lumen/W
8. Radiant flux = 31.4W, Solid angle = 4
Luminous efficiency = 60 lumen/W
So, Luminous flux = 60 × 31.4 lumen
And luminous intensity =
4
FluxousminLu
=


4
4.3160
= 150 candela
9. I = luminous intensity =
4
628
= 50 Candela
r = 1m,  = 37°
So, illuminance, E = 2
r
cosI 
= 2
1
37cos50 
= 40 lux 
10. Let, I = Luminous intensity of source
EA = 900 lumen/m
2
EB = 400 lumen/m
2
Now, Ea = 2
x
cosI 
and EB = 2
)10x(
cosI


So, I =
cos
xE 2
A
=


cos
)10x(E 2
B
 900x
2
= 400(x + 10)
2

10x
x

=
3
2
 3x = 2x + 20  x = 20 cm
So, The distance between the source and the original position is 20cm.
1m
Source
37°
Area
Normal
10cmx
O
BA

Chapter 22
22.2
11. Given that, Ea = 15 lux = 2
0
60
I
 I0 = 15 × (0.6)
2
= 5.4 candela
So, EB = 2
0
)OB(
cosI 
= 2
1
5
3
4.5 






= 3.24 lux
12. The illuminance will not change.
13. Let the height of the source is ‘h’ and the luminous intensity in the normal direction is I0.
So, illuminance at the book is given by,
E = 2
0
r
cosI 
= 3
0
r
hI
= 2/322
0
)hr(
hI

For maximum E,
dh
dE
= 0  322
)222/322
0
)hR(
h2hR(h
2
3
)hR(I
2/1








 (R
2
+ h
2
)
1/2
[R
2
+ h
2
– 3h
2
] = 0
 R
2
– 2h
2
= 0  h =
2
R
♠ ♠ ♠ ♠ ♠

0.6m
A
1m
O
B0.8m
rh 
R

H c verma part 1 solution

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H c verma part 1 solution