Question 272: Units , Dimensions and Error Analysis

Question

In the determination of Young's modulus $$\left( {Y = {{4MLg} \over {\pi l{d^2}}}} \right)$$
by using Searle's method, a wire of length L = 2 m
and diameter d = 0.5 mm is used. For a load M = 2.5 kg, an extension l = 0.25 mm in the length of the wire
is observed. Quantities d and l are measured using a screw gauge and a micrometer, respectively. They
have the same pitch of 0.5 mm. The number of divisions on their circular scale is 100. The contributions to
the maximum probable error of the Y measurement

Accepted Answer

To determine Young's modulus using Searle's method, we use the formula:
$$ Y = \frac{4MLg}{\pi l d^2} $$
Given:
Length of wire, $L = 2 	ext{ m}$
Diameter of wire, $d = 0.5 	ext{ mm}$
Load, $M = 2.5 	ext{ kg}$
Extension in length, $l = 0.25 	ext{ mm}$
The quantities $d$ and $l$ are measured using a screw gauge and a micrometer, respectively, both having a pitch of 0.5 mm and 100 divisions on their circular scale. The least count, which is the smallest measurement that can be accurately read, is:
$$ 	ext{Least count} = \frac{	ext{Pitch}}{	ext{Total number of divisions on the circular scale}} = \frac{0.5 	ext{ mm}}{100} = 0.005 	ext{ mm} $$
Hence, the least count, $\Delta d$ and $\Delta l$, are both 0.005 mm.
Error in measurement of $l$ (extension length) is:
$$ \frac{\Delta l}{l} = \frac{0.005}{0.25} = \frac{1}{50} $$
Error in measurement of $d$ (diameter) is:
$$ \frac{\Delta d}{d} = \frac{0.005}{0.5} = \frac{1}{100} $$
Using the formula for Young's modulus, the combined error is:
$$ Y = \frac{4MLg}{\pi ld^2} $$
The relative error in $Y$ is given by:
$$ \frac{\Delta Y}{Y} = \frac{\Delta l}{l} + 2 \frac{\Delta d}{d} $$
As can be seen:
$$ \frac{\Delta l}{l} = 2 \frac{\Delta d}{d} = \frac{1}{50} $$
Therefore, the contributions to the error in the measurement of $Y$ from errors in measuring $l$ and $d$ are the same.

Answer

due to the errors in the measurements of d and l are the same.

Answer

due to the error in the measurement of d is twice that due to the error in the measurement of l.

Answer

due to the error in the measurement of l is twice that due to the error in the measurement of d.

Answer

due to the error in the measurement of d is four times that due to the error in the measurement of l.

Units , Dimensions and Error Analysis

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