QJEEAdv12Phe947: JEE Advanced 2012 Physics - Class 11 JEEAdv12Phe947 Solution: In the determination of Young's modulus by using Searle's…

To determine Young's modulus using Searle's method, we use the formula:

$Y = \frac{4MLg}{\pi l d^2}$

Given:
Length of wire, $L = 2 \text{ m}$
Diameter of wire, $d = 0.5 \text{ mm}$
Load, $M = 2.5 \text{ kg}$
Extension in length, $l = 0.25 \text{ mm}$

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:

$\text{Least count} = \frac{\text{Pitch}}{\text{Total number of divisions on the circular scale}} = \frac{0.5 \text{ mm}}{100} = 0.005 \text{ mm}$

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.