Question

In the determination of Young’s modulus $(Y=\frac{4MLg}{\pi l{d}^2})$ by using Searle’s method, a wire of length L=2m and diameter d=0.5mm is used. For a load M=2.5kg, an extension l=0.25mm 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 0.5mm. The number of divisions on their circular scale is 100. The contributions to the maximum probable error of the Y measurement is:

• A. Due to the errors in the measurement of d and l are the same
• B. Due to the error in the measurement of d is twice that due to the error in the measurement of l
• C. Due to the error in the measurement of l is twice that due to the error in the measurement of d
• D. Due to the error in the measurement of d is four times that due to the error in the measurement of l

Answer 1

The correct option is A

Due to the errors in the measurement of d and l are the same

From the formula for Young's modulus,
$\frac{\Delta Y}{Y}=\frac{\Delta l}{l}+2\frac{\Delta d}{d}$
Here both screw gauge and micrometer used have same least count and hence the maximum probable error in the measurement of $l$ and $d$ will be same i.e., $\Delta l=\Delta d=\Delta x$
The error contribution to measured $Y$ from error in $d$ is $e_d=2\frac{\Delta d}{d}=\frac{2\Delta x}{0.5}=\frac{\Delta x}{0.25}$
The error contribution to measured $Y$ from error in $l$ is $e_l=\frac{\Delta l}{l}=\frac{\Delta x}{0.25}$