Chapter 2.14 - Combination of Errors - Solved Examples

In the previous section, we saw the combination of errors. In this section, we will see some solved examples

Solved example 2.22
The resistance R = ^V⁄_I where V = (100 ± 5)V and I = (10 ± 0.2)A. Find the percentage error in R
Solution:
1. We have the rule for division:
If Z = ^A⁄_B, then the 𝚫Z is given by the relation: 
$\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta B}{B}}$
2. Applying this rule, we can write:
Relative error in R = Relative error in V + Relative error in I
3. Relative error in V = $\mathbf\small{\frac{\Delta V}{V}=\frac{5}{100}=0.05}$  
• Relative error in I = $\mathbf\small{\frac{\Delta I}{I}=\frac{0.2}{10}=0.02}$
4. Substituting these values in (2), we get:
Relative error in R = 0.05 + 0.02 = 0.07
5. So percentage error in R = Relative error in R × 100 = (0.07 × 100) = 7%

Solved example 2.23
Two resistors of resistances R₁ = 100 ±3 ohm and R₂ = 200 ± 4 ohm are connected (a) in series, (b) in parallel. Find the equivalent resistance of the (a) series combination, (b) parallel combination
Solution:
Part (a):
1. When two resistors R₁ and R₂ are connected in series, the equivalent resistance R is given by the relation: R = R₁ + R₂ (Details here)
2. We have the rule for addition:
If Z = A + B, then the error 𝚫Z in Z is (𝚫A + 𝚫B)
3. Applying this rule, we can write:
Absolute error in R = ±𝚫R = ±(𝚫R₁ + 𝚫R₂) = ±(3 + 4) = ±7 ohm
4. So the equivalent resistance of the series combination = (100 + 200) ± 7 = 300 ± 7 ohm  
Part (b):
1. When two resistors R₁ and R₂ are connected in parallel, the equivalent resistance R' is given by the relation: $\mathbf\small{\frac{1}{R'}=\frac{1}{R_1}+\frac{1}{R_2}}$ (Details here)
2. Substituting the values, we get:
$\mathbf\small{\frac{1}{R'}=\frac{1}{100}+\frac{1}{200}}$
$\mathbf\small{\Rightarrow R'=\frac{200}{3}\; \text{ohm}}$
3. From the relation in (1), we get: (R')^-1 = (R₁)^-1 + (R₂)^-1
(i) Let Z' = (R')^-1
• Then we can write: $\mathbf\small{\frac{\Delta Z'}{Z'}=-1\left(\frac{\Delta R'}{R'} \right)}$
• Substituting for Z', we get: $\mathbf\small{\frac{\Delta Z'}{(R')^{-1}}=-1\left(\frac{\Delta R'}{R'} \right)}$
$\mathbf\small{\Rightarrow \Delta Z'=-1\left(\frac{\Delta R'}{R'} \right)(R')^{-1}}$
$\mathbf\small{\Rightarrow \Delta Z'=-1\left(\frac{\Delta R'}{R'} \right)\left(\frac{1}{R'} \right)}$
$\mathbf\small{\Rightarrow \Delta Z'=-1\left(\frac{\Delta R'}{(R')^2} \right)}$
(ii) Let Z₁ = (R₁)^-1
• Similar to the above, we will get:
$\mathbf\small{\Delta Z_1=-1\left(\frac{\Delta R_1}{(R_1)^2} \right)}$
(iii) Let Z₂ = (R₂)^-1
• Similar to the above, we will get:
$\mathbf\small{\Delta Z_2=-1\left(\frac{\Delta R_2}{(R_2)^2} \right)}$
4. We succeeded in finding 𝚫Z'
    ♦ It is the absolute error in (R')^-1  
• We succeeded in finding 𝚫Z₁
    ♦ It is the absolute error in (R₁)^-1 
• We succeeded in finding 𝚫Z₂
    ♦ It is the absolute error in (R₂)^-1
5. Now consider the relation that we wrote in (3):
(R')^-1 = (R₁)^-1 + (R₂)^-1
• On the right side, we have an addition of two quantities. So the absolute errors can be added
We get:
$\mathbf\small{-1\left(\frac{\Delta R'}{(R')^2} \right)=-1\left(\frac{\Delta R_1}{(R_1)^2} \right)+-1\left(\frac{\Delta R_2}{(R_2)^2} \right)}$
• Multiplying both sides by '-1', we get:
$\mathbf\small{\left(\frac{\Delta R'}{(R')^2} \right)=\left(\frac{\Delta R_1}{(R_1)^2} \right)+\left(\frac{\Delta R_2}{(R_2)^2} \right)}$
6. Substituting the values, we get:
$\mathbf\small{\left(\frac{\Delta R'}{(200/3)^2} \right)=\left(\frac{3}{(100)^2} \right)+\left(\frac{4}{(200)^2} \right)}$
• Thus we get: 𝚫R' = 1.777 = 1.8 ohm 
7. This 𝚫R' is the absolute error in R'
• We can report the effective resistance in parallel connection as:
(R' ± 𝚫R') = (200/3 ± 1.78) = 66.7 ± 1.8 ohm

Solved example 2.24
The values of two resistors are (5.0 ± 0.2) ㏀ and (10.0 ± 0.1) ㏀. What is the percentage error in the equivalent resistance when they are connected in parallel
Solution:
1. When two resistors R₁ and R₂ are connected in parallel, the effective resistance is given by:
$\mathbf\small{\frac{1}{R'}=\frac{1}{R_1}+\frac{1}{R_2}}$
2. Substituting the values, we get:
$\mathbf\small{\frac{1}{R'}=\frac{1}{5}+\frac{1}{10}}$
$\mathbf\small{\Rightarrow R'=\frac{10}{3}}$ ㏀
3. Now we want the error when they are connected in parallel
• We can use the expression that we derived in the previous solved example 2.23
• We have: $\mathbf\small{\Rightarrow \left(\frac{\Delta R'}{(R')^2} \right)=\left(\frac{\Delta R_1}{(R_1)^2} \right)+\left(\frac{\Delta R_2}{(R_2)^2} \right)}$
4. Substituting the values, we get:
$\mathbf\small{\left(\frac{\Delta R'}{(10/3)^2} \right)=\left(\frac{0.2}{(5)^2} \right)+\left(\frac{0.1}{(10)^2} \right)}$
• Thus we get: 𝚫R' = 0.1 ㏀
5. So the relative error = $\mathbf\small{\frac{\Delta R'}{R'}=\frac{0.1}{10/3}=0.03}$
• Thus we get:
Percentage error = $\mathbf\small{\frac{\Delta R'}{R'}\times 100=0.03 \times 100 = 3 \text{%}}$

Solved example 2.25
Find the relative error in Z, if $\mathbf\small{Z=\frac{A^4 B^{\frac{1}{3}}}{C\,D^{\frac{3}{2}}}}$
Solution:
The relative error $\mathbf\small{\frac{\Delta Z}{Z}}$ is given by:
$\mathbf\small{\frac{\Delta Z}{Z}=4 \left(\frac{\Delta A}{A}\right)+\frac{1}{3}\left(\frac{\Delta B}{B}\right)+\left(\frac{\Delta C}{C}\right)+\frac{3}{2}\left(\frac{\Delta D}{D}\right)}$

Solved example 2.26
The error in measuring the diameter of a circle is 2%. What would be the error in measuring the radius of the circle?
Solution:
1. We have: r = 0.5d
• Here. 0.5 is a factor. It will not contain any error
2. Power of 'd' is '1'. So we get:
$\mathbf\small{\frac{\Delta r}{r}=1 \times\frac{\Delta d}{d}}$
3. That means, the relative error is the same for both r and d
• So percentage error will also be same for both r and d
• Thus the answer is 2%

Solved example 2.27
A force F is applied on a square plate of side L. The percentage error in the measurement of L is 2%. The percentage error in the measurement of F is 4%. What is the percentage error in the pressure?
Solution:
1. We have: $\mathbf\small{\text{Pressure}=\frac{\text{Force}}{\text{Area}}}$
• That is., $\mathbf\small{\text{P}=\frac{\text{F}}{\text{A}}}$
2. Area (A) of a square plate = L²
• So we get: $\mathbf\small{\text{P}=\frac{\text{F}}{{\text{L}}^2}}$
• So $\mathbf\small{\frac{\Delta P}{P}=\frac{\Delta F}{F}+2\left(\frac{\Delta L}{L}\right)}$
3. Given that percentage error in L = 2%. So $\mathbf\small{\frac{\Delta L}{L}}$ = 0.02
• Given that percentage error in F = 4%. So $\mathbf\small{\frac{\Delta L}{L}}$ = 0.04
4. Substituting the values in (2), we get: $\mathbf\small{\frac{\Delta P}{P}=0.04+2 \times 0.02=0.08}$
• So percentage error = $\mathbf\small{\frac{\Delta P}{P}\times 100=0.08 \times 100=8\text{%}}$

Solved example 2.28
The radius of a ball is (5.2 ± 0.2) cm. What is the percentage error in the volume of the ball?
Solution:
1. Relative error in the radius = $\mathbf\small{\frac{\Delta r}{r}=\frac{0.2}{5.2}=0.0385}$
2. We have: Volume of the ball = $\mathbf\small{\frac{4}{3}\pi r^3}$
$\mathbf\small{\frac{4}{3}\pi}$ is a factor. It will not contain any error
• So we get: $\mathbf\small{\frac{\Delta V}{V}=3\left(\frac{\Delta r}{r}\right)=3\times 0.0385=0.1154}$
3. So percentage error = 0.1154 × 100 = 11.54%

Solved example 2.28
The time period (T) of a pendulum is given by $\mathbf\small{T=2\pi\sqrt{\frac{l}{g}}}$. If l and g are measured with an error of 1% and 2% respectively, what is the percentage error in T?
Solution:
1. We have: $\mathbf\small{T=2\pi\sqrt{\frac{l}{g}}=2\pi\frac{l^{\frac{1}{2}}}{g^{\frac{1}{2}}}}$
$\mathbf\small{2\pi}$ is a factor. It will not contain any error
• So we get: $\mathbf\small{\frac{\Delta T}{T}=\frac{1}{2}\left(\frac{\Delta l}{l}\right)+\frac{1}{2}\left(\frac{\Delta g}{g}\right)}$
2. Substituting the values, we get: $\mathbf\small{\frac{\Delta T}{T}=\frac{1}{2}\left(0.01\right)+\frac{1}{2}\left(0.02\right)=0.015}$
• So percentage error = 0.015 × 100 = 1.5%

Solved example 2.29
The period of oscillation of a simple pendulum is $\mathbf\small{T=2\pi \sqrt{\frac{L}{g}}}$. Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. What is the accuracy in the determination of g ?
Solution:
1. Given that: Measured value of L is 20.0 cm known to 1 mm accuracy  
• This is a least count error. The true value of L may be higher by 1 mm or lower by 1 mm
• 1 mm = 0.1 cm
• So we can write: L = (20 ± 0.1) cm
2. Given that: Time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution
• Time for 100 oscillations = 90 s
• The resolution of the watch is 1 s. That watch cannot measure time durations which are less than 1 s 
• That means least count of that watch is 1 s
• The 90 s may be 91 s or 89 s
• So we can write: The time for 100 oscillations is (90 ± 1) s 
• So time for 1 oscillation (T) = $\mathbf\small{\frac{90\pm 1}{100}=(0.9\pm0.01)\;\text{s}}$
3. Thus we get:
• Absolute error (𝚫L) in L = 0.1 cm
• Absolute error (𝚫T) in T = 0.01 s
4. Given that: The period of oscillation of a simple pendulum is $\mathbf\small{T=2\pi \sqrt{\frac{L}{g}}}$
• Squaring both side, we get: $\mathbf\small{T^2=4\pi^2\left(\frac{l}{g}\right)}$
• Rearranging the expression, we get: $\mathbf\small{g=4\pi^2\left(\frac{l}{T^2}\right)}$
5. $\mathbf\small{4\pi^2}$ is a factor. It will not contain any error
• So we get: $\mathbf\small{\frac{\Delta g}{g}=\left(\frac{\Delta l}{l}\right)+2\left(\frac{\Delta T}{T}\right)}$
6. Substituting the values, we get:
$\mathbf\small{\frac{\Delta g}{g}=\left(\frac{0.1}{20}\right)+2\left(\frac{0.01}{0.9}\right)}$ = 0.02722
7. So percentage error = $\mathbf\small{\frac{\Delta g}{g}\times 100}$ = 0.02722 × 100 = 2.722% = 3%

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In the next section, we will see dimensions of physical quantities

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Chapter 2.13 - Combination of Errors

In the previous section, we saw the details about absolute error and relative error. In this section, we will see combination of errors.

We will write the basics in steps:
1. Consider an experiment to determine the density of the material of an object
• We first find the mass of that object
• For that, we weigh that object in a balance
• We make several observations so as to get the most accurate mass
2. Then we find the volume of that object
• For finding the volume, we may have to measure it’s length, width and thickness
Or, we may have to measure the diameter
• Here also, we make several observations so as to get the most accurate dimension/dimensions
3. Even after taking all the necessary precautions, there will be some error in the measured mass
4. Also there will be some error in the measured dimensions
• So there will be some error in the calculated volume
5. When we apply the equation density = ^mass⁄_volume, we are combining two quantities (mass and volume) which have errors in them
■ So we want to know 'how much error will be there in the density'
6. To make such estimates, we must learn how errors combine when we do the mathematical operations
• We can derive rules for each of the mathematical operations: addition/subtraction, multiplication/division

Error of a sum

We will write the derivation by demonstrating an example in steps:
1. In an experiment, lengths of two rods are to be determined
■ The total length of the two rods is to be reported
2. Length of first rod is denoted as A
• Length of second rod is denoted as B
• Total length is denoted as Z
• So Z = A + B
3. But while measuring the lengths, errors (even though small), are unavoidable
• The length of the first rod can be written only as: (A ± 𝚫A) cm
    ♦ Where 𝚫A is the absolute error in the measurement of the length of rod 
• The length of the second rod can be written only as: (B ± 𝚫B) cm
    ♦ Where 𝚫B is the absolute error in the measurement of the length of second rod
4. So the total length Z will contain some error. Let us denote it as 𝚫Z
• We want to find this 𝚫Z
5. We have: Z ± 𝚫Z = (A ± 𝚫A) + (B ± 𝚫B)
= A + B + (± 𝚫A ± 𝚫B)
6. So the maximum possible value that (Z ± 𝚫Z) can have is: A + B + (+ 𝚫A + 𝚫B)  
= A + B + ( 𝚫A + 𝚫B)
• The least possible value that (Z ± 𝚫Z) can have is: A + B + (- 𝚫A - 𝚫B)      
= A + B - ( 𝚫A + 𝚫B)
7. So the total length has the probability to be anywhere between 
[A + B + ( 𝚫A + 𝚫B)] and [A + B - ( 𝚫A + 𝚫B)]
• This can be represented as: [A+ B ± (𝚫A + 𝚫B)]
■ So the error 𝚫Z in Z is (𝚫A + 𝚫B)

Error of a difference

1. In an experiment, lengths of two rods are to be determined
■ The difference in the lengths of the two rods is to be reported
2. Length of first rod is denoted as A
• Length of second rod is denoted as B
• Difference in length is denoted as Z'
• So Z' = A - B
3. But while measuring the lengths, errors (even though small), are unavoidable
• The length of the first rod can be written only as: (A ± 𝚫A) cm
    ♦ Where 𝚫A is the absolute error in the measurement of the length of rod 
• The length of the second rod can be written only as: (B ± 𝚫B) cm
    ♦ Where 𝚫B is the absolute error in the measurement of the length of second rod
4. So the difference Z' will contain some error. Let us denote it as 𝚫Z'
• We want to find this 𝚫Z'
5. We have: Z' ± 𝚫Z' = (A ± 𝚫A) - (B ± 𝚫B)
= A - B + (± 𝚫A) - (± 𝚫B)
= A - B + (± 𝚫A) + (± 𝚫B)
= A - B + (± 𝚫A ± 𝚫B) 
6. So the maximum possible value that (Z' ± 𝚫Z') can have is: A - B + (+ 𝚫A + 𝚫B)  
= A - B + ( 𝚫A + 𝚫B)
• The least possible value that (Z' ± 𝚫Z') can have is: A - B + (- 𝚫A - 𝚫B)      
= A - B - ( 𝚫A + 𝚫B)
7. So the difference has the probability to be any where between 
[A - B + ( 𝚫A + 𝚫B)] and [A - B - ( 𝚫A + 𝚫B)]
• This can be represented as: [A- B ± (𝚫A + 𝚫B)]
■ So the error 𝚫Z' in Z' is (𝚫A + 𝚫B)
• This is the same expression that we obtained for 𝚫Z

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■ So we can write the rule for addition/subtraction:

When two quantities are added or subtracted, the absolute error in the final result is the sum of the absolute errors in the individual quantities

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Solved example 2.21

The volumes of two bodies are (8.4 ± 0.03) cm³ and (3.7 ± 0.01) cm³. Find the sum and difference in volumes with error limits
Solution:
1. We have: V₁ = (8.4 ± 0.03) cm³ and V₂ = (3.7 ± 0.01) cm³
2. Sum of the absolute errors = (0.03 + 0.01) = 0.04
3. Sum of the volumes (V) = (8.4 + 3.7) ± 0.04 cm³ = 12.1 ± 0.04 cm³  
4. Difference of the volumes (V') = (8.4 - 3.7) ± 0.04 cm³ = 4.7 ± 0.04 cm³

Error of a product

1. In an experiment, length and width of a rectangle is to be determined
■ The product of the length and width is to be reported as the area 
2. Length is denoted as A
• Width is denoted as B
• Area is denoted as Z
• So Z = AB
3. But while measuring the lengths, errors (even though small), are unavoidable
• The length can be written only as: (A ± 𝚫A) cm
    ♦ Where 𝚫A is the absolute error in the measurement of the length
• The width can be written only as: (B ± 𝚫B) cm
    ♦ Where 𝚫B is the absolute error in the measurement of the width
4. So the area Z will contain some error. Let us denote it as 𝚫Z  
• We want to find this 𝚫Z
5. We have: Z ± 𝚫Z = (A ± 𝚫A) × (B ± 𝚫B) = AB ± B𝚫A ± A𝚫B ± 𝚫A𝚫B
⇒ Z ± 𝚫Z = AB ± B𝚫A ± A𝚫B ± 𝚫A𝚫B
6. Let us divide both sides by Z
• But from (2), we have: Z = AB
• So we will divide left side by Z and right side by AB. We get:
$\mathbf\small{1 \pm \frac{\Delta Z}{Z}=1 \pm \frac{\Delta A}{A} \pm \frac{\Delta B}{B} \pm \frac{\Delta A \Delta B}{AB}}$
$\mathbf\small{\Rightarrow \frac{\Delta Z}{Z}=\frac{\Delta A}{A} \pm \frac{\Delta B}{B} \pm \frac{\Delta A \Delta B}{AB}}$
• 𝚫A and 𝚫B are small. So their product (𝚫A𝚫B) will be very small
    ♦ Example: (0.01× 0.02) = 0.0002
    ♦ We can ignore the last term containing (𝚫A𝚫B)
7. We get: $\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A} \pm \frac{\Delta B}{B}}$
• The error 𝚫Z will be maximum when we add the two terms on the right
• So we can discard the '-' sign from '±'
• We can write:
Maximum 𝚫Z is given by the relation: $\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta B}{B}}$
• Note that, each of the three terms are 'relative errors'
8. What if we want to multiply 3 measurements?
An example:
If (U ± 𝚫U) = (A ± 𝚫A) × (B ± 𝚫B) × (C ± 𝚫C), find 𝚫U
Solution:
(i) (A ± 𝚫A) × (B ± 𝚫B) × (C ± 𝚫C)
= [(A ± 𝚫A) × (B ± 𝚫B)] × (C ± 𝚫C)  
= [(Z ± 𝚫Z)] × (C ± 𝚫C)
⇒ (U ± 𝚫U) = (Z ± 𝚫Z) × (C ± 𝚫C)
(ii) Applying the relation in (7), we can write:
$\mathbf\small{\frac{\Delta U}{U}=\frac{\Delta Z}{Z}+\frac{\Delta C}{C}}$
(iii) But $\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta B}{B}}$ 
• So we get 𝚫U from the relation:
$\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta B}{B}+\frac{\Delta C}{C}}$
• In this way, we can multiply any number of measurements we want
• This method can be used to find the volume of a rectangular prism, where we have to multiply length, width and height together 

Error of a quotient

1. In an experiment, mass and volume of a rectangular prism is to be determined
■ The quotient ^mass⁄_volume is to be reported as the density
2. Mass is denoted as A
• Volume is denoted as B
• Density is denoted as Z
• So Z = ^A⁄_B
3. But while measuring the mass, errors (even though small), are unavoidable
• The mass can be written only as: (A ± 𝚫A) kg
    ♦ Where 𝚫A is the absolute error in the measurement of the mass
4. The volume is calculated as the product of length, width and height
• All those 3 measurements will contain errors. So the volume will contain errors
• We have seen how to report the product of three 'measurements with errors'
• The volume that we have in our present experiment is B
• It will be written as: (B ± 𝚫B) cm³
    ♦ Where 𝚫B is the absolute error in the calculation of volume
5. So the density Z will contain some error. Let us denote it as 𝚫Z  
• We want to find this 𝚫Z
6. We have: $\mathbf\small{Z\pm \Delta Z=\frac{(A \pm\Delta A)}{(B \pm\Delta B)}}$
$\mathbf\small{\Rightarrow Z\left(1\pm\frac{\Delta Z}{Z} \right)=\frac{A\left(1 \pm\frac{\Delta A}{A} \right)}{B\left(1 \pm\frac{\Delta B}{B} \right)}}$
7. But from (2), we have:
The Z on the left side is equal to ^A⁄_B on the right side
• So they will cancel each other
• We will get:
$\mathbf\small{\left(1\pm\frac{\Delta Z}{Z} \right)=\frac{\left(1 \pm\frac{\Delta A}{A} \right)}{\left(1 \pm\frac{\Delta B}{B} \right)}}$
$\mathbf\small{\Rightarrow \left(1 \pm\frac{\Delta Z}{Z} \right)=\left(1 \pm\frac{\Delta A}{A} \right)\left(1 \pm \frac{\Delta B}{B} \right)^{-1}}$
8. The term $\mathbf\small{\left(1 \pm \frac{\Delta B}{B} \right)^{-1}}$ can be expanded by applying binomial theorem
• After simplification, the result in (7) will become:
$\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A} \pm \frac{\Delta B}{B}}$
9. The error 𝚫Z will be maximum when we add the two terms on the right
• So we can discard the '-' sign from '±'
• We can write:
Maximum 𝚫Z is given by the relation: $\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta B}{B}}$
• Note that, each of the three terms are 'relative errors'
• This is the same equation that we obtained for multiplication. In both cases, the relative errors are being added

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■ So we can write the rules:

• Rule for multiplication:
When two quantities are multiplied, the relative error in the product is the sum of the relative errors in the multipliers
• Rule for division:
When a quantity (dividend) is divided by another quantity (divisor), the relative error in the quotient is the sum of the relative errors in the dividend and divisor

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Error in the case of a measured quantity raised to a power

1. In an experiment, side of a square is to be determined

■ The square (raising to the power 2) of the side is to be reported as the area 
2. Side is denoted as A
• Area is denoted as Z
• So Z = A²
3. But while measuring the side, errors (even though small), are unavoidable
• The side can be written only as: (A ± 𝚫A) cm
    ♦ Where 𝚫A is the absolute error in the measurement of the side
4. So the area Z will contain some error. Let us denote it as 𝚫Z  
• We want to find this 𝚫Z
5. We have: Z ± 𝚫Z = (A ± 𝚫A)² 
⇒ Z ± 𝚫Z = (A ± 𝚫A) × (A ± 𝚫A)
6. So now we have a product of two measurements. We know how to find 𝚫Z in such a case
• Applying the rule for multiplication, we can write:
Maximum 𝚫Z is given by the relation: $\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta A}{A}}$
• That means:
Maximum 𝚫Z is given by the relation: $\mathbf\small{\frac{\Delta Z}{Z}=2 \left(\frac{\Delta A}{A} \right)}$
7. What if the exponent is '3'?
We will see an example:
• In an experiment, side of a cube is to be determined
■ The cube (raising to the power 3) of the side is to be reported as the volume
8. Side is denoted as A
• Volume is denoted as Z
• So Z = A³
9. But while measuring the side, errors (even though small), are unavoidable
• The side can be written only as: (A ± 𝚫A) cm
    ♦ Where 𝚫A is the absolute error in the measurement of the side
10. So the volume Z will contain some error. Let us denote it as 𝚫Z  
• We want to find this 𝚫Z
11. We have: Z ± 𝚫Z = (A ± 𝚫A)³ 
⇒ Z ± 𝚫Z = (A ± 𝚫A) × (A ± 𝚫A) × (A ± 𝚫A)
12. So now we have a product of three measurements. We know how to find 𝚫Z in such a case
• Applying the rule for multiplication, we can write:
Maximum 𝚫Z is given by the relation: $\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta A}{A}+\frac{\Delta A}{A}}$
• That means:
Maximum 𝚫Z is given by the relation: $\mathbf\small{\frac{\Delta Z}{Z}=3 \left(\frac{\Delta A}{A} \right)}$
13. In general, we can write:
• If Z = A^p, then $\mathbf\small{\frac{\Delta Z}{Z}=p \left(\frac{\Delta A}{A} \right)}$

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14. Based on the above, we can derive two more results:
(A) If Z = (A^p × B^q), then $\mathbf\small{\frac{\Delta Z}{Z}=p \left(\frac{\Delta A}{A} \right)+q \left(\frac{\Delta B}{B} \right)}$
Proof:
(i) We have:
Z ± 𝚫Z = [(A ± 𝚫A) × (A ± 𝚫A) × . . . p terms] × [(B ± 𝚫B) × (B ± 𝚫B) × . . . q terms]
(ii) Then:
$\mathbf\small{\frac{\Delta Z}{Z}=\left(\frac{\Delta A}{A}+\frac{\Delta A}{A}+\;.\;.\;.\;\text{p terms} \right)+\left(\frac{\Delta B}{B}+\frac{\Delta B}{B}+\;.\;.\;.\;\text{q terms} \right)}$
$\mathbf\small{\Rightarrow\frac{\Delta Z}{Z}=p \left(\frac{\Delta A}{A} \right)+q \left(\frac{\Delta B}{B} \right)}$
(B) If $\mathbf\small{Z=\frac{A^p B^q}{C^r}}$, then $\mathbf\small{\frac{\Delta Z}{Z}=p \left(\frac{\Delta A}{A} \right)+q \left(\frac{\Delta B}{B} \right)+r \left(\frac{\Delta C}{C} \right)}$
• Proof can be written using binomial theorem and the quotient rule

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■ So we can write the rule:

The relative error in a physical quantity raised to the power k is k times the relative error in the individual quantity

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In the next section, we will see some solved examples

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Chapter 2.12 - Relative Error

In the previous section, we saw the different types of errors that can occur while taking measurements. In this section, we will see how the 'quantity of errors' can be calculated. In other words, we will see the method to calculate 'how much error occurred', when the experiment was conducted

1. Suppose that, the length of a cylinder is being measured
• Let n trials be done
    ♦ In the first trial, length obtained is a₁ cm 
    ♦ In the second trial, length obtained is a₂ cm 
    ♦ In the third trial, length obtained is a₃ cm
    ♦ . . . .
    ♦ . . . .
    ♦ In the n^th trial, length obtained is a_n cm
2. Each of the above measurements may be different from one another
• We cannot say which among the n values, is the true length
3. In such a situation, the ‘mean of the n values’ will be very close to the true value
• But how can we prove that 'mean is indeed very close to the true value'?
4. For proving it, let us recall the 'basics about mean (average)' that we learned in our previous classes:
• Consider a group of seven students. It is decided to make uniforms for them
• Their heights are: 142 cm, 141 cm, 138 cm , 145 cm, 139 cm, 143 cm and 144 cm
• How much clothes should be bought?
Solution:
(i) Sum of the heights works out to 992 cm
(ii) If we divide this sum by 7, we will get the average height
• So average height = ⁹⁹²⁄₇ = 142 cm
(iii) Assume that all 7 students have a height of 142 m
• Let 2 m of cloth be required for a student of 142 cm height
(iv) Then for the 7 students, (2 × 7) = 14 m of cloth will be required
• This is an easy method for calculating the total quantity of cloth, especially if the number of students in the group or class is large
(v) But doubts arise:
• The student of 138 cm height do not need 2 m of cloth
    ♦ Some cloth will go as waste
    ♦ Same is the case with students of heights less than 142 cm
• Similarly, the student of 145 cm will need more than 2 m
    ♦ His uniform will be incomplete
    ♦ Same is the case with students of heights more than 142 cm
(vi) But in reality, such a wastage or insufficiency do not occur. The reason can be explained using the schematic diagram in fig.2.25 below:

Fig.2.25

• In fig.a, all students stand in ascending order of their heights
• The fourth student have the average height of 142 cm. He is shown in yellow color in fig.b
• The average height is marked by a cyan line in fig.c
• The white rectangles represents the wastage caused by the first 3 students
    ♦ They are below the cyan line
• The magenta rectangles represent the insufficiency suffered by the last 3 students
    ♦ They are above the cyan line
• If the mean is calculated accurately, the wastage will be very nearly equal to the insufficiency
• That is., the total heights of the white rectangles will be equal to the total heights of the magenta rectangles
• In short, the negatives will be cancelled by the positives. This is the advantage of using the mean value
5. When we calculate the mean of the measured values, a similar cancellation occurs. Let us see how:
(i) We have n number of readings
(ii) Half of those readings will be greater than the true value
    ♦ The errors in this case will be all positive
    ♦ Because, Error = Measured value - True value
    ♦ The errors of such over estimated readings are represented by the magenta rectangles in fig.2.25 above 
(iii) The other half will be lesser than the true value
    ♦ The errors in this case will be all negative
    ♦ Because, Error = Measured value - True value
    ♦ The errors of such under estimated readings are represented by the white rectangles in fig.2.25 above
(iv) How can we be sure that exactly half readings are over estimates and the other half are under estimates?
• The answer is that, it is an assumption
• It is a reasonable assumption. Because, 'likely hood of over estimating' is same as the 'likely hood of under estimating'
• Nobody would make all readings greater than the true value  
• Also, nobody would make all readings lesser than the true value  
• Every body wants the readings to be as accurate as possible
• The 'possibility for making positive errors' is same as the 'possibility for making negative errors'
6. So we can confirm that, the mean value is very close to the true value
• We know the formula for calculating the mean value: $\mathbf\small{a_{mean}=\frac{a_1+a_2+a_3+\;.\;.\;.\;+\;a_n}{n}}$
• In short form, this formula is written as: $\mathbf\small{a_{mean}=\frac{\sum\limits_{i=1}^{i=n}{a_i} }{n}}$
7. So we have obtained the mean value. The mean value will be very close to the true value
• So we can now find 'how much error was made in each reading'
• Let 𝚫a₁ be the error made in the first reading
• We have: Error = Measured value - True value
    ♦ So 𝚫a₁ = a₁ - a_mean 
    ♦ 𝚫a₂ = a₂ - a_mean 
    ♦ 𝚫a₃ = a₃ - a_mean
    ♦ . . . .
    ♦ . . . .
    ♦ 𝚫a_n = a_n - a_mean.
8. The above 𝚫a values will be +ve in some cases and -ve in the rest of the cases
• For our present discussion, we do not need the sign
• We need only 'how much error' occurred in each reading 
• So we take absolute values:
    ♦ |𝚫a₁| = |(a₁ - a_mean)|
    ♦ |𝚫a₂| = |(a₂ - a_mean)|
    ♦ |𝚫a₃| = |(a₃ - a_mean)|
    ♦ . . . .
    ♦ . . . .
    ♦ |𝚫a_n| = |(a_n - a_mean)|
9. The mean of the above absolute error values will give us the final absolute error or the mean absolute error
• It is denoted as 𝚫a_mean
• This 𝚫a_mean can be calculated using the same method: $\mathbf\small{\Delta a_{mean}=\frac{|\Delta a_1|+|\Delta a_2|+|\Delta a_3|+\;.\;.\;.\;+\;|\Delta a_n|}{n}}$
• In short form, this formula is written as: $\mathbf\small{\Delta a_{mean}=\frac{\sum\limits_{i=1}^{i=n}{|\Delta a_i|} }{n}}$
10. So now we have two quantities: a_mean and 𝚫a_mean.
■ We must clearly understand the difference between the two
• a_mean is the value which is very close to the true value
• 𝚫a_mean is an indication of 'how much error' occurred when the experiment was conducted
11. So the true value will be within the range: $\mathbf\small{a_{\rm mean}\pm \Delta a_{\rm mean}}$
• That is., (a_mean - 𝚫a_mean) ≤ true value ≤ (a_mean + 𝚫a_mean)

Relative error or Percentage error

1. Instead of mean absolute error, we often use the relative error or percentage error
• Relative error is a ratio. It can be obtained using the relation:
Relative error = $\mathbf\small{\frac{\Delta a_{mean}}{a_{mean}}}$
2. This ratio helps us to compare two quantities:
(i) The error (ii) a_mean (which is close to the true value)
3. If the error is very small (when compared to the true value), we will get a small ratio    
• In that case, the error can be considered as negligible
4. If the error is large (when compared to the true value), we will get a large ratio
• In that case, the error cannot be considered as negligible
An example is given below. It is written in 3 steps:
(i) Consider the experiment to find the distance between earth and a star
(ii) If the error made is a few kilometers, it is negligible. We will get a small relative error
(iii) If the error made is thousands of kilometers, it is not negligible. We will get a large relative error
5. Once we calculate the relative error, we can easily convert it into percentage format
• All we need to do is: multiply by 100
• The result thus obtained is called the percentage error. It is denoted as $\mathbf\small{\delta a}$
• So we get: $\mathbf\small{\delta a=\left(\frac{\Delta a_{mean}}{a_{mean}}\right)\times 100 \text{%}}$

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Now we will see two solved examples

Solved example 2.19 
We measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to be 2.63 s, 2.56 s, 2.42 s, 2.71 s and 2.80 s. Calculate the absolute errors, relative error or percentage error
Solution:
The calculations are done in the table below:

1. We get: a_mean = 2.624 s
• This is written in column 3
• This result is got by addition of various numbers
• So the result must not have more decimal places than the least in the given data
• Thus, the result cannot have more than 2 decimal places
• Rounding off, we get: a_mean = 2.62 s
(After addition, we divide the sum by (n = 5). But n is a factor. It has infinite number of significant numbers)   
2. The absolute errors are calculated in the fourth column
• We get: Mean absolute error 𝚫a_mean = 0.107 s
• This is also obtained by addition. Addition of numbers having two decimal places
• So rounding off, we get: 𝚫a_mean = 0.11 s 
3. Now we can write three statements about that pendulum:
(i) The true value of the time period of that pendulum is very close to 2.62 s
(ii) The mean absolute error is 0.11 s
(iii) There is a probability that, the actual time period lies within the range:
(2.62 - 0.11) to (2.62 + 0.11)
• That is., 2.51 ≤ actual time period ≤ 2.73 s
4. Calculation of relative error
• We have: Relative error = $\mathbf\small{\frac{\Delta a_{mean}}{a_{mean}}}$
• Substituting the values, we get:
Relative error = $\mathbf\small{\frac{0.11}{2.62}=0.04}$
• Percentage error = Relative error ×100 = 0.04 × 100 = 4%

Solved example 2.20
Diameter of a wire as measured by a screw gauge was found to be 2.620, 2.625, 2.630, 2.628 and 2.626 cm. Calculate
(i) Mean value of diameter
(ii) absolute error in each measurement
(iii) Mean absolute error
(iv) relative error
(v) Percentage error
(vi) Express the result in terms of percentage error
Solution:
The calculations are done in the table below:

1. We get: a_mean = 2.626 cm
• This is written in column 3
• This is the answer for part (i) 
2. The absolute error of each measurement is written in column 4
• These are the answers for part (ii)
3. The mean absolute error is written in column 5
• This is the answer for part (iii)
4. Calculation of relative error
• We have: Relative error = $\mathbf\small{\frac{\Delta a_{mean}}{a_{mean}}}$
• Substituting the values, we get:
Relative error = $\mathbf\small{\frac{0.003}{2.626}=0.001}$
• This is the answer for part (iv)
5. Percentage error = Relative error ×100 = 0.001 × 100 = 0.1%
• This is the answer for part (v)
6. Diameter of the wire can be written as:
d = 2.626 cm ± 0.1%
• This is the answer for part (vi)

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In the next section, we will see combination of errors

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