![1.1. PHYSICAL UNITS 3
1Units
1.1 Physical Units
Physical quantities are measured in comparison of some standard parameters.
These parameters are called units. There are seven fundamental units which
are adapted as International Standards (SI Units) and other units are the
derivative of these fundamental units.
Base Unit Symbol Dimension
Mass kilogram kg [M]
Length metre m [L]
Time second s [T]
Temperature Kelvin K [θ]
Current Ampere A
Molecular Mass mole mol
Illumination Intensity candela cd
Except above seven fundamental units there are two units of angle.
1.1.1 Plane Angle
The unit of plane angle is radian and it is related with degree by
1 radian =
180
π
0
θ 450 π/4
Figure 1.1: Angle notation, angle in degree and angle in radian are shown in
first, second and third part of the figure respectively.](https://image.slidesharecdn.com/units-211010150744/85/Notes-of-Units-Dimensions-Errors-for-IIT-JEE-by-Arun-Umrao-3-320.jpg)
![1.6. DIMENSIONS 11
is written as 1. Zero power of a dimension is one ie L0
= 1 and the inverse
to dimension is dimension−1
ie 1
L
= L−1
. If two similar dimensions are
in product then Lp
× Lq
= Lp+q
, ie the powers of similar dimensions are
additive and if two dimensions are in quotient then powers are subtraction.
Product of unlike dimensions takes place in algebraic form. Dimensions
obey algebraic operations. Addition or subtraction takes place among
like dimensions and multiplication or division takes place among like or unlike
dimensions.
1.6.2 Deriving Physical Relations
Dimensional analysis is used in the finding physical relations by using math-
ematical dimensional properties.
Force Relation
To calculate the force relation F = ma we use the second statement of the
newton’s force law. It states that the force on a body is the product of mass
having power p and acceleration having power q. Where p and q are arbitrary
constants. Now
F = mp
× aq
(1.1)
Now substituting the units of force, mass and accelerations in SI fundamental
units.
kg ·
m
s2
= kgp
m
s2
q
Substituting the dimension value of fundamental units. And
[M][L]
[T]2
= [M]p
[L]
[T]2
q
Using arithmetic operations of dimensions
[MLT−2
] = [Mp
Lq
T−2q
]
Comparing the powers of same fundamental units, and
p = 1 q = 1
Substituting the value of p and q in relation (1.1) the force relation is
F = ma
This is the required relation.](https://image.slidesharecdn.com/units-211010150744/85/Notes-of-Units-Dimensions-Errors-for-IIT-JEE-by-Arun-Umrao-11-320.jpg)
![12 Units
Time Period of Pendulum
Assume that the time period of a pendulum is depends on the mass of bob,
length of pendulum and gravitational acceleration. Hence
T = kma
lb
gc
(1.2)
Substituting the dimensions of time period, mass, length and gravity. Time
period of pendulum becomes
[T] = k[Ma
][Lb
][Lc
T−2c
]
On simplification
[T] = k[Ma
Lb+c
T−2c
]
On comparing of power of same bases in both side
a = 0; −2c = 1; b + c = 0;
On solving these equations the values of a, b and c are
a = 0; c =
−1
2
; b =
1
2
;
Substituting the values of power in equation (1.2), the time period of pen-
dulum is
T = kl
1
2 g
−1
2
T = k
s
l
g
(1.3)
Solved Problem 1.6 The charge on a conductor is defined by q = at2
+ bt,
where a, b are constants. Find the dimension of constants in form of current.
Solution We know that the current in a circuit is given by i = dq
dt
1
I =
d
dt
(at2
+ bt)
= 2at + b
1
Remember that any relative result in physics would be find by using derivative method
in current and next chapters.](https://image.slidesharecdn.com/units-211010150744/85/Notes-of-Units-Dimensions-Errors-for-IIT-JEE-by-Arun-Umrao-12-320.jpg)
![1.6. DIMENSIONS 13
We know that a relation is consists if dimensions in both sides of relation are
same. Again if two are more physical quantities are in addition then each
physical quantity has same dimension as its resultant. Hence the dimensions
of two physical quantities given above in right hand side are
2at = [I] b = [I]
Now substituting the dimension of time in left hand side of first physical
quantity and leaving 2 as constant without dimension. So
a[T] = [I] b = [I]
a =
[I]
[T]
b = [I]
a = [IT−1
] b = [I]
These are the dimensions of constants a and b.
Solved Problem 1.7 A physical quantity A is linearly depend on two other
physical quantities, i.e. C and D. The units of C and D are N/m2
and m/s2
.
If unit of A is s then find the relation between A, C and D.
Solution From the given problem, A is linearly depend C and D. So,
A = αC + βD (1.4)
Substituting the dimensions of A, C and D as given in the problem.
[T] = α[ML−1
T−2
] + β[LT−2
]
If this relation is possible, then each term at right hand side should have
dimension equal to [T]. So,
[T] = α[ML−1
T−2
] [T] = β[LT−2
]
α = [M−1
LT3
] β = [L−1
T3
]
These are dimensions of α and β. On substituting these values in equation
1.4, we shall get the exact relation between A, C and D.](https://image.slidesharecdn.com/units-211010150744/85/Notes-of-Units-Dimensions-Errors-for-IIT-JEE-by-Arun-Umrao-13-320.jpg)
![14 Units
1.6.3 Dimension in Trigonometric Functions
Consider a trigonometric function, say tan θ, that has two parts. One an
argument and other an operator. Output of trigonometric functions is always
a dimensionless pure numeric value. For example,
tan 45◦
= 1
θ
p
b
Again, trigonometric function with its argument is equal to ratio of ap-
propriate sides of a right angle triangle, i.e.
tan θ =
p
b
Therefore,
θ = tan−1
p
b
As p/b is dimensionless quantity, hence θ shall also be a dimensionless quan-
tity.
Solved Problem 1.8 Find the dimension of physical quantity C if it is related
to velocity as C
v
= sin C
v
.
Solution As output of trigonometric function is a pure numeric value,
hence right hand side is dimensionless quantity. Therefore left side term
should also be a dimensionless quantity. So, dimension of C should be equal
to dimension of v. Now, the dimension of C is [LT−1
].
1.6.4 Dimension in Logarithmic Functions
1.7 Measurement Instruments
Here vernier caliper and Screw Gauge will be discussed as the measuring in-
struments. Important definitions related to vernier caliper and Screw Gauge
are given below:](https://image.slidesharecdn.com/units-211010150744/85/Notes-of-Units-Dimensions-Errors-for-IIT-JEE-by-Arun-Umrao-14-320.jpg)
Notes of Units, Dimensions & Errors for IIT JEE by Arun Umrao
- 1. 1 UNITS A SHORT NOTES Arun Umrao https://sites.google.com/view/arunumrao DRAFT COPY - GPL LICENSING
- 2. 2 Units Contents 1 Units 3 1.1 Physical Units . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.1.1 Plane Angle . . . . . . . . . . . . . . . . . . . . . . . . 3 1.1.2 Solid Angle . . . . . . . . . . . . . . . . . . . . . . . . 4 1.2 Significant Figures . . . . . . . . . . . . . . . . . . . . . . . . 4 1.3 Scientific Notation . . . . . . . . . . . . . . . . . . . . . . . . 5 1.3.1 Significant Figure . . . . . . . . . . . . . . . . . . . . . 5 1.3.2 Identification of Significant Figures . . . . . . . . . . . 5 1.4 Significant Figures in Measurement . . . . . . . . . . . . . . . 6 1.4.1 Significant Figures in Sum & Difference . . . . . . . . . 6 1.4.2 Significant Figures in Multiplication & Division . . . . 7 1.5 Rounding Off . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.5.1 Effect of Rounding Off . . . . . . . . . . . . . . . . . . 8 1.6 Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.6.1 Mathematical properties . . . . . . . . . . . . . . . . . 10 1.6.2 Deriving Physical Relations . . . . . . . . . . . . . . . 11 1.6.3 Dimension in Trigonometric Functions . . . . . . . . . 14 1.6.4 Dimension in Logarithmic Functions . . . . . . . . . . 14 1.7 Measurement Instruments . . . . . . . . . . . . . . . . . . . . 14 1.7.1 Vernier Caliper . . . . . . . . . . . . . . . . . . . . . . 15 1.7.2 Screw Gauge . . . . . . . . . . . . . . . . . . . . . . . 21 1.8 Land Area Units . . . . . . . . . . . . . . . . . . . . . . . . . 25
- 3. 1.1. PHYSICAL UNITS 3 1Units 1.1 Physical Units Physical quantities are measured in comparison of some standard parameters. These parameters are called units. There are seven fundamental units which are adapted as International Standards (SI Units) and other units are the derivative of these fundamental units. Base Unit Symbol Dimension Mass kilogram kg [M] Length metre m [L] Time second s [T] Temperature Kelvin K [θ] Current Ampere A Molecular Mass mole mol Illumination Intensity candela cd Except above seven fundamental units there are two units of angle. 1.1.1 Plane Angle The unit of plane angle is radian and it is related with degree by 1 radian = 180 π 0 θ 450 π/4 Figure 1.1: Angle notation, angle in degree and angle in radian are shown in first, second and third part of the figure respectively.
- 4. 4 Units 1.1.2 Solid Angle This is the angle swept by curved surface at the center. Maximum solid angle swept by any surface is 4π. The solid angle is given by Solid Angle = Surface Area Square of distance of surface from center r r̂ ˆ dS dS dΩ Figure 1.2: Solid Angle. dΩ = dS r2 1.2 Significant Figures The significant figures of a number are the digits necessary to take in cal- culations. The significant figures must be carefully chosen in calculations to find the accurate result. A digit in a figure is significant if presence of it rejects the claim that the measured value is equal to the true value. This is the probability of mistakenly rejecting equal values. For example, while measuring length of pen by a scale precise upto millimeter, we find that it is 8.6 centimeter long. The actual length of pen is 8.6 centimeter long. If a student measures it 8.65 centimeter then digit 5 has no significance as we do not reject the claim that pen is 8.65 centimeter long by using the same scale. Due to precision upto millimeter, digit 5 is removed. But if another student claim that length of pen is 8.7 centimeter then it shall be rejected as 8.6 centimeter (true value) is not equal to 8.7 centimeter (measured value). So, a significant figure is that digit, whose presence rejects our claim that measured value is equal to actual/true value.
- 5. 1.3. SCIENTIFIC NOTATION 5 1.3 Scientific Notation A number n is written in scientific notation as n = ±y.yy × 10±xxx Here x, y are two arbitrary numbers. y.yy is called mantissa and xxx is exponent. Mantissa is always written in decimal form and its modulus is always greater than 1 and less than 10. For example 12.4533 = 1.24533×101 is correct format where 12.4533 = 0.124533 × 102 is wrong format. 1.3.1 Significant Figure The digits of mantissa of a number expressed in scientific notation are called significant figures or significant digits. A zero is always significant if it is right of the non zero digits or right of the decimal. For example Number Scientific Notation Significant Figures 132.00 1.3200 × 102 5 0.0123540 1.23540 × 10−2 6 -0.125 −1.25 × 10−1 3 25250 2.5250 × 104 5 A point is to be remember that 2.5250×104 and 2.525×104 are same values but first has five significant figures while second has four significant figures. In calculation first number is more accurate because it give the precision value upto fourth place of decimal while second number gives precision upto third place of decimal. 1.3.2 Identification of Significant Figures There are no strict rules for identification of significant figures but some basic rules should be followed during identification significant figures. 1. All non-zero digits are considered significant. For example 23.221 has five significant figures and 2001.22 has six significant figures. 2. Zeroes between two non-zero digits are significant. For example 20010001 has eight significant figures and 400001 has six significant figures.
- 6. 6 Units 3. Leading zeros are not significant. For example 0014114 has five signif- icant figures and 000014114 has five significant figures. 4. Trailing zeros in a number containing decimal point are significant. For example 2.300 has four significant digits and 2.30 has three significant digits. Both numbers are same in arithmetic but first one is more accurate than second. 5. A bar may be placed at the last significant digit; any trailing zeros following bar digit are not significant and can leave in calculation. 12.001̄0000 has only five significant digits. 6. All values betweenn zero and one, and values larger than one and with- out decimal are first transform into scientific notations before finding significant figure. 7. All values larger than one and with decimal, decimal part is also sig- nificant. A number ends with decimal has only significant figures equal to the digits which are before decimal. 1.4 Significant Figures in Measurement There are two similar mathematical operations in unit. (i) Adding and sub- traction and (ii) Multiplication and Division. 1.4.1 Significant Figures in Sum & Difference If two physical quantities are in addition or in subtraction then the significant figures in result are taken upto the decimal place which is equal to the least decimal place in the physical quantities. For example, two physical quantities are 1.230m and 2.30m and their addition is 3.530m. But 2.30m has least decimal places (2 decimal places). Hence the result is 3.53m. Solved Problem 1.1 Two physical quantities 2.31 and 4.591 are added to get the result. Find the result with maximum precision. Solution The sum of two given quantities is 2.31 + 4.591 = 6.901. The final result should have same number of decimal digits which are in the addend which have minimum decimal digits. So, the final answer will be 6.90.
- 7. 1.5. ROUNDING OFF 7 Solved Problem 1.2 Find the best result of the sum of of two numbers 2.01 and 2.0120. Solution The sum of two given quantities is 2.01 + 2.0120 = 4.0220. The best final result should have same number of decimal digits which are in the addend which have minimum decimal digits. So, the best result will be 2.02. Solved Problem 1.3 Find 2.45/0.2935. Solution The result of division 2.45/0.293 is 8.3475. In significant rules, in division, the result will have same number of significant digits that is in the physical quantity having least number of significant digits. In this problem least number of significant digits is 3. So, the result is 8.34. 1.4.2 Significant Figures in Multiplication & Division If two physical quantities are in product or in division then the result has the same number of significant digits that is in the physical quantity having least number of significant digits. For example, if quantity 1.2482m is divided by 2.2s then the result is 0.5673m/s. This result is rounded off for two significant digits as the least significant digits are two in the quantity 2.2s, one of the both physical quantities. The rounded off result is 0.56m/s. Solved Problem 1.4 Find the division of 2.459 by 1.2. Solution The division of 2.459 by 1.2 is 2.459/1.2 = 2.0492. In significant rules, in division, the result will have same number of significant digits that is in the physical quantity having least number of significant digits. In this problem least number of significant digits is 2. So, the result is 2.0 1.5 Rounding Off Let a number is to be round off upto n significant digits as 1. If non-significant digit is less than 5 and is to be rounded off, the previ- ous digit does not change and non-significant digit is simply dropped. For example 2.301 is significant upto the second place of the decimal then digit 1 is simply dropped and number is rounded off to 2.30. Sim- ilarly 2.304 is rounded off upto two place of decimal as 2.30.
- 8. 8 Units 2. If non-significant digit is greater than 5 and is to be rounded off, the previous digit (left side digit) is raised by 1 and non-significant digit is dropped. For example 2.306 is significant upto the second place of the decimal then digit 6 is simply dropped and previous digit (left side digit) 0 is raised by 1 and number is rounded off to 2.31. 3. If the rounding off digit is greater than 5, rounding off digit is simply dropped when the previous digit (left side digit) is even. For example, when 2.3546 is rounded off, result is 2.354. 4. If the rounding off digit is greater than 5, rounding off digit is dropped and previous digit (left side digit) is raised by 1 when the previous digit (left side digit) is odd. For example, when 2.3576 is rounded off, result is 2.358. 5. In mathematics and specially in pure science, the large numbers are mostly written in scientific notation form. Hence these numbers are rounded-off upto significant digits. To round-off a large number upto n digits, the rounding-off process is always started from leftmost non-zero digit. For example start process from 2 in 2500 and from 3 in 0.0325. i Keep n significant digits and replace rest of digits with zeros. ii Write number in scientific notation, i.e. a × 10±b form, where a and b are real and significant numbers. iii Round off mantissa upto the last significant digit. For example 0.2336 can be written as 2.336 ×10−1 in scientific notation and if there are only three significant digits then third significant digit becomes 4 and number becomes 2.34 × 10−1 . Rest of digits are dropped. 1.5.1 Effect of Rounding Off Rounding off a fraction number is required to limit the observed value to the nearest actual value. For example 2.505 rupees is rounded to 2.50 to represent rupees in standard form. Sometime rounding off creates havoc in calculations if rounding a number off is not taken properly, specially in inverse functions. For example, take the function f(t) = 1 n1 − n2
- 9. 1.6. DIMENSIONS 9 Assuming n1 = 300, n2 and f(t) = 20 are finite numbers. Comparatively f(t) and n1 are not negligible. Now the value of n2 is 20 = 1 300 − n2 On simplification, 6000 − 20n2 = 1 Here comparatively 1 is negligible to 6000. If 1 is neglected then n2 becomes 300 and again function will tends to infinity. Hence 1 can not be neglected here. Now value of n2 from above relation is ‘299.955’. If the same question is modified as; n1 = 300, n2 = 299.955 and f(t) are finite numbers then what will be the value of function f(t). Here n2 is slightly different to n1. If n2 is upper rounded off then value of function f(t) tends to infinity and if n2 is taken as it is then function value is f(t) = 1 n1 − n2 = 1 300.000 − 299.955 Or, f(t) = 22.222. Hence rounding off the number is not permitted in these type of cases. 1.6 Dimensions Dimensional analysis is a tool to find or check relations among physical quan- tities by using their dimensions. The dimension of a physical quantity is the combination of the basic physical dimensions (usually mass, length, time, electric charge, and temperature) which describe it. Dimensional analysis is based on the fact that a physical law must be independent of the units used to measure the physical variables. A straightforward practical consequence is that any meaningful equation (and any inequality and in-equation) must have the same dimensions in the left and right sides. Checking this is the basic way of performing dimensional analysis. Dimension method is used for unification of various units of a physical quantity.
- 10. 10 Units bcb bcb Pressure bcb Tension Figure 1.3: Equality in dimensions. Solved Problem 1.5 Height, h, of water in a leaky tank is time dependent function and the rate of water leakage is directly proportional to the instan- taneous height of the water in the tank. Mathematically, it is given by dh dt ∝ −h Or dh dt = −kh Find the dimension of constant k. Solution The given relation is dh dt = −kh From the property of the dimension, dimension units must be equal in left and right hand side of the above relation. So, left hand unit is m/s and right hand side is m. Including the dimension of constant k, to make the unit both side equal, k must has unit of 1/τ (where τ represents to the time and known as time constant). Now, k = 1 τ This is unit of k. 1.6.1 Mathematical properties The dimensions that can be formed from a given collection of basic physical dimensions, such as M, L, and T, form a group. The identity of dimension
- 11. 1.6. DIMENSIONS 11 is written as 1. Zero power of a dimension is one ie L0 = 1 and the inverse to dimension is dimension−1 ie 1 L = L−1 . If two similar dimensions are in product then Lp × Lq = Lp+q , ie the powers of similar dimensions are additive and if two dimensions are in quotient then powers are subtraction. Product of unlike dimensions takes place in algebraic form. Dimensions obey algebraic operations. Addition or subtraction takes place among like dimensions and multiplication or division takes place among like or unlike dimensions. 1.6.2 Deriving Physical Relations Dimensional analysis is used in the finding physical relations by using math- ematical dimensional properties. Force Relation To calculate the force relation F = ma we use the second statement of the newton’s force law. It states that the force on a body is the product of mass having power p and acceleration having power q. Where p and q are arbitrary constants. Now F = mp × aq (1.1) Now substituting the units of force, mass and accelerations in SI fundamental units. kg · m s2 = kgp m s2 q Substituting the dimension value of fundamental units. And [M][L] [T]2 = [M]p [L] [T]2 q Using arithmetic operations of dimensions [MLT−2 ] = [Mp Lq T−2q ] Comparing the powers of same fundamental units, and p = 1 q = 1 Substituting the value of p and q in relation (1.1) the force relation is F = ma This is the required relation.
- 12. 12 Units Time Period of Pendulum Assume that the time period of a pendulum is depends on the mass of bob, length of pendulum and gravitational acceleration. Hence T = kma lb gc (1.2) Substituting the dimensions of time period, mass, length and gravity. Time period of pendulum becomes [T] = k[Ma ][Lb ][Lc T−2c ] On simplification [T] = k[Ma Lb+c T−2c ] On comparing of power of same bases in both side a = 0; −2c = 1; b + c = 0; On solving these equations the values of a, b and c are a = 0; c = −1 2 ; b = 1 2 ; Substituting the values of power in equation (1.2), the time period of pen- dulum is T = kl 1 2 g −1 2 T = k s l g (1.3) Solved Problem 1.6 The charge on a conductor is defined by q = at2 + bt, where a, b are constants. Find the dimension of constants in form of current. Solution We know that the current in a circuit is given by i = dq dt 1 I = d dt (at2 + bt) = 2at + b 1 Remember that any relative result in physics would be find by using derivative method in current and next chapters.
- 13. 1.6. DIMENSIONS 13 We know that a relation is consists if dimensions in both sides of relation are same. Again if two are more physical quantities are in addition then each physical quantity has same dimension as its resultant. Hence the dimensions of two physical quantities given above in right hand side are 2at = [I] b = [I] Now substituting the dimension of time in left hand side of first physical quantity and leaving 2 as constant without dimension. So a[T] = [I] b = [I] a = [I] [T] b = [I] a = [IT−1 ] b = [I] These are the dimensions of constants a and b. Solved Problem 1.7 A physical quantity A is linearly depend on two other physical quantities, i.e. C and D. The units of C and D are N/m2 and m/s2 . If unit of A is s then find the relation between A, C and D. Solution From the given problem, A is linearly depend C and D. So, A = αC + βD (1.4) Substituting the dimensions of A, C and D as given in the problem. [T] = α[ML−1 T−2 ] + β[LT−2 ] If this relation is possible, then each term at right hand side should have dimension equal to [T]. So, [T] = α[ML−1 T−2 ] [T] = β[LT−2 ] α = [M−1 LT3 ] β = [L−1 T3 ] These are dimensions of α and β. On substituting these values in equation 1.4, we shall get the exact relation between A, C and D.
- 14. 14 Units 1.6.3 Dimension in Trigonometric Functions Consider a trigonometric function, say tan θ, that has two parts. One an argument and other an operator. Output of trigonometric functions is always a dimensionless pure numeric value. For example, tan 45◦ = 1 θ p b Again, trigonometric function with its argument is equal to ratio of ap- propriate sides of a right angle triangle, i.e. tan θ = p b Therefore, θ = tan−1 p b As p/b is dimensionless quantity, hence θ shall also be a dimensionless quan- tity. Solved Problem 1.8 Find the dimension of physical quantity C if it is related to velocity as C v = sin C v . Solution As output of trigonometric function is a pure numeric value, hence right hand side is dimensionless quantity. Therefore left side term should also be a dimensionless quantity. So, dimension of C should be equal to dimension of v. Now, the dimension of C is [LT−1 ]. 1.6.4 Dimension in Logarithmic Functions 1.7 Measurement Instruments Here vernier caliper and Screw Gauge will be discussed as the measuring in- struments. Important definitions related to vernier caliper and Screw Gauge are given below:
- 15. 1.7. MEASUREMENT INSTRUMENTS 15 Least Count of Instrument This is the minimum possible measurement by an instrument. It is also called resolution of the instrument. The quantity measured by the instrument is always integer multiple of its least count. Least Count of Main Scale It is the length between two consecutive marks on the main scale of vernier caliper. Least Count of Vernier Scale It is the length between two consecutive marks on the vernier scale of vernier caliper. Divisions These are number of marks between two fixed ends or between two major distant marks. For example, in centimeter and millimeter scale, there are ten divisions in one centimeter length. 1.7.1 Vernier Caliper A vernier scale is a device that lets the user measure more precisely than could be done by reading a uniformly-divided straight or circular measurement scale. It is scale that indicates where the measurement lies in between two of the marks on the main scale. It has two scales, (a) main scale or fixed scale and (b) vernier scale. Main scale and vernier scale has partitions of equal size with labels. Main scale is used to measure the main reading of the scale. It is an estimated value and further precise value is measured with help of vernier scale. Vernier scale has a leas count value. Least count depends on the number of partitions in vernier scale. If there are n partitions in the vernier scale then least count ℓ of the vernier caliper is given as the ratio of difference of length of ‘n’ partitions of main scale dn ms and vernier scale dn vs to the ‘n’. ℓ = dn ms − dn vs n Second method of finding least count is ℓ = d1 ms − d1 vs Where d1 ms is length of one partition in main scale and d1 vs is length of one partition in vernier scale.
- 16. 16 Units 0 1 0 10 ℓ 5ℓ There are two ways to scale vernier scale; (i) by marking more divisions in the vernier scale than the number of divisions in main scale in equal length, for example, within 1cm, there are n divisions in main scale and m n divisions in vernier scale. Or (ii) marking equal number of divisions in vernier scale whose length is less than by one division to the length of main scale for the same number of divisions, for example, 10 divisions are made in 1cm of main scale and 0.9cm of vernier scale each. First case is not acceptable as the precision measurement passes to the least count value of main scale at the minimum count of vernier divisions. For example, in main scale there are ten divisions, hence two consecutive divisions has value 0.1cm. There are 20 divisions in vernier scale in 1cm. Least count of vernier scale is 1/20 = 0.05. When second tick of the vernier scale is perfectly coincide with the main scale tick then precision value is 0.05 × 2 = 0.1. It is equal to the least count of the main scale. In this case other ticks of vernier scale are unused as they give values larger than the least count of main scale. Second case is best suited as highest positioned tick gives value less than the least count of the main scale. 0 1 0 10 5ℓ Thus between two consecutive divisions of main scale distance is 1/10 cm while distance between two consecutive divisions of vernier scale is 0.9/10 cm. Now, the least count of this Vernier caliper (i.e. instrument) is ℓ = 1 10 − 0.9 10 = 0.1 10 = 0.01cm There are restricted number of divisions in vernier scale. It means, For equal number of division of main scale and vernier scale, the difference between
- 17. 1.7. MEASUREMENT INSTRUMENTS 17 distances should be equal to the least count of the main scale. Again, lm is least count of main scale and lv is least count of the vernier scale. There are m and n divisions in main scale and vernier scale respectively. Now, relation between these two least count is lm × m − lv × n = lm 0 1 0 20 In short form, the vernier precision value for any tick of the vernier scale can not be larger than the least count of the main scale. This is main principle of design of the vernier caliper. Solved Problem 1.9 A vernier caliper has 20 divisions in 1 cm of main scale. The vernier scale of the same caliper has 20 divisions in 0.95 cm. Find the least count of the caliper. Solution The least count of the main scale is lm = 1 20 = 0.05cm Similarly least count of the vernier scale is lv = 0.95 20 = 0.0475cm Now, the least count of the vernier caliper is ℓ = lm − lv = 0.05 − 0.0475 = 0.0025cm Solved Problem 1.10 A vernier caliper has main scale of least count 0.05cm. There are 50 divisions in 2.45 cm of vernier scale. Find the least count of the caliper.
- 18. 18 Units Solution The least count of the main scale is 0.05cm. Least count of the vernier scale is lv = 2.45 50 = 0.049cm Now, the least count of the vernier caliper is ℓ = lm − lv = 0.05 − 0.049 = 0.001cm Measurement of Size To measure the length or size of an object by Vernier Caliper, it is placed between two heads of vernier caliper and heads are gently tightened. Now two observations are taken 0 1 2 3 4 0 10 Figure 1.4: Vernier Calipers. Principal Measurement Firstly, observe the length on main scale that is passed by zero marker of the Vernier scale. Generally zero marker lies in between two consecutive equally spaced partition markers in the main scale. It is principal measurement. Assume it is m and denoted by p. Precision Measurement Secondly, observe the partition number from zero marker of Vernier scale which is perfectly coincide with any of the par- tition mark of the main scale. Assume it is nth marker of vernier scale. Now multiply it by least count (ℓ). The value obtained is precision measurement. ∆p = n × ℓ Now the actual dimension of the object is sum of principal measurement (p) and precision measurement (∆p). d = m + n × ℓ
- 19. 1.7. MEASUREMENT INSTRUMENTS 19 Solved Problem 1.11 In a Vernier Caliper, the distance between two consec- utive divisions of main scale is 0.05cm. In vernier scale, there are 40 divisions in a span of 1.95cm. Find the least count of the Vernier Caliper. Solution 0 1 2 3 4 0 20 40 The least count of vernier caliper is difference of distances between two consecutive divisions of main scale and vernier scale. Distance between two consecutive divisions of main scale is 0.05cm. Similarly, distance between two consecutive divisions of vernier scale is given by 1.95/40 = 0.04875cm. Now, least count of the vernier caliper is 0.05 − 0.04875 = 0.00125cm. Solved Problem 1.12 Solution Measurement Errors There are following errors observed in the measurement by using Vernier Calipers. Zero Error When two forks are closed, zero markers of both, main scale and Vernier scale must be coincide to each other. If they are not coincide then there is a deviation in the measured value from actual value. This deviation of measured value from actual value is called zero error. It is sub classified as (i) positive zero error and (ii) negative zero error. Positive Zero Error When two jaws of the Vernier calipers are closed and zero marker of Vernier scale leads to the zero marker of main scale then it adds additional value to the actual value of measure- ment. This error is taken as positive value. For exact measure- ments, this error is subtracted from the final result.
- 20. 20 Units 0 1 2 0 10 Figure 1.5: Vernier Calipers : Positive Error. Considering the positive error, the actual measured value is given by t = d − e+ Negative Zero Error When two jaws of the vernier caliper are closed and zero marker of Vernier scale trails to the zero marker of main scale then it measures less value to the actual value of measure- ment. This error is taken as negative value. For exact measure- ments, this error is subtracted to the final result. 0 1 2 0 10 Figure 1.6: Vernier Calipers : Negative Error. Considering the negative error, the actual measured value is given by t = d − e− Mechanical Error If heads are clamped tightly over the object then mea- sured value is lesser than the actual value. Environmental Errors In hot and cold environment, scales are expanded or contracted and cause the errors in measurements. Design Error If scales are faulty designed then measured value shows de- viation from the actual value. Solved Problem 1.13 The diameter of a cylinder is measured using a Vernier Calipers with no zero errors. It is found that the zero of the Vernier scale lies between 5.10cm and 5.15cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45cm. The 24th division of the Vernier scale exactly coincides with one of the main scale divisions. Calculate the diameter of the cylinder.
- 21. 1.7. MEASUREMENT INSTRUMENTS 21 Solution The zero marker of the Vernier scale lies between 5.10cm and 5.15cm hence the principal measurement is dp = 5.10cm 0 1 2 3 4 5 6 7 8 0 10 20 30 40 50 Now from the least count relation ℓ = dn ms − dn vs n • Number of partitions in Vernier scale = 50 • Length of one partition in main scale = 0.05cm • Length of 50 partitions of main scale = 50 × 0.05 = 2.50cm • Length of 50 partitions of Vernier scale = 2.45cm (given) Now, ℓ = 2.50 − 2.45 50 = 0.001 Now precision measurement is 0.001 × 24 = 0.024cm. Now the diameter of the cylinder is Dcyl = 5.100cm + 0.024cm = 5.124cm This is required result. 1.7.2 Screw Gauge A screw gauge is also known as micro-meter. Its structure is shown in the figure 1.7.2. It composed of: Frame The C-shaped body made of thick material of larger coefficient of thermal expansion that holds the anvil and barrel. Anvil The shiny part fixed at the one end of C-shape body. Spindle moves toward it when it is rotated. Samples are rests against it.
- 22. 22 Units Sleeve It is stationary cylindrical component with the linear scale on a base line. Screw It is the heart of the micrometer. It is inside the the barrel. Distance between two consecutive grooves of the screw are called pitch. Spindle This is cylindrical component that is revolved with help of thim- ble. It moves towards the anvil. Thimble It is used to rotate the screw by help of thumb. It has graduated 50-100 markings. Ratchet Stop It is device on the end of handle that limits applied pressure by slipping at a calibrated torque. 0.0 0.5 10 15 20 Measurement of Thickness We can measured the value with help of screw gauge by using following steps. Least Count: First we calculate the least count of the device by using re- lation ℓ = p N Here p is pitch and N is number of marks in round scale. Pitch (p) is the distance moved by the screw leftward or rightward when it is rotated by one complete revolution. Note that, the pitch is always equal to the length between two consecutive division on main scale. Principal Measurement: Further we check the main scale and find which main scale mark is visible correctly. This is main scale value (say m). It is principal value. Precision Measurement: Again we see the round scale to get the division which is perfectly coincide with main scale lien. Let it is the nth di- vision. This value is multiplied by the least count ℓ. This measured value obtained is called precision value (p).
- 23. 1.7. MEASUREMENT INSTRUMENTS 23 Measured Value: The final measured value is the sum of principal value and precision value. d = m + n × ℓ Taking the considerations of the positive or negative error, actual measured value is given by t = d − e± In screw gauge the pitch of the round scale is always equal to the least count of the main scale. If not so, then screw gauge has a design error/flaws. Zero Errors When we put close the two heads of the screw gauge, 0-mark of round scale should be coincide perfectly to the base line of main scale. If it is not, then there is a zero error. There are two types of measurement errors in screw gauge. Positive Zero Error 0.0 0 5 10 Initially, by rotating thimble, two jaws of the screw gauge are placed in contact. If zero line of the round scale is not aligned with the main scale line then it is said that the screw gauge has measurement error. Now if we can see zero of main scale clearly then the error is positive error. Positive error is obtained as e+ = n × ℓ Here n is nth division aligned to the main line and ℓ is least count of the screw gauge. This error is always taken as positive value. It adds extra value to the actual dimension of substrate. If positive error is larger than the least count of the main scale then fixed jaw is adjusted with help of screw driver to align the zero mark of round scale to main scale line. Negative Zero Error
- 24. 24 Units 0.0 30 35 40 Initially, by rotating thimble, two jaws of the screw gauge are brought in contact. If zero line of the round scale is not aligned with the main scale line then it is said that the screw gauge has measurement error. Now if we can hardly saw the zero of main scale clearly then the error is negative error. Negative error is obtained as e− = −(N − n) × ℓ Here n is nth division aligned to the main scale line and ℓ is least count of the screw gauge. N is total numbers of divisions in main scale. It subtracts an amount of value from the actual dimension of substrate. This error is always taken as negative value. If negative error is larger than the least count of the main scale then fixed jaw is adjusted with help of screw driver to align the zero mark of round scale to main scale line. Solved Problem 1.14 A screw gauge with a pitch of 0.5mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5mm and the 25th division coincides with the main scale line? Solution When round scale is rotated anticlockwise direction, it moves away from fixed jaw of screw gauge. Round scale has graduated marks in re- ducing order when it rotates anticlockwise (i.e. when jaws move away). This is why when two jaws are brought constant and main scale line is coincide with 45th division, then it is said negative error. The total divisions on round scale passed to main scale line are (50-45)=5. Now, the least count of the screw gauge is 0.5/50 = 0.01mm. Negative error is −5 × 0.01 = −0.05mm. Using the standard measurements of screw gauge, the measured value of sheet is t = 0.5 + 25 × 0.01 = 0.75mm
- 25. 1.8. LAND AREA UNITS 25 Including the error in measurement, the true thickness of the sheet is 0.75 − (−0.05) = 0.80mm. 1.8 Land Area Units The following units are useful for measuring the land area. 1. One international acre is equal to 4,046.8564224 square meters and 0.40468564224 hectare. 2. One hectare is equal to a square with 100m × 100m area. 3. One acre is equal to 43,560 square feet. 4. One chain is equals to 66 feet or 22 yards or 4 rods or 100 links. 5. A furlong is equals to 220 yards or ten chains. 6. One square mile is equal to 640 acres.