Q. In the determination of Young’s modulus $\left( Y = \frac{4\ MLg}{\pi\mathcal{\ l}d^{2}} \right)$ by using Searle’s method, a wire of length $L = 2m$ and diameter $d = 0.5mm$ is used. For a load $M = 2.5\ kg$, an extension $\mathcal{l =}0.25mm$ in the length of the wire is observed. Quantities $d$ and $\mathcal{l}$ 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 scale is $100$. The contributions to the maximum probable error of the $Y$ measurement
A) Due to the errors in the measurements of $d$ and $\mathcal{l}$ are the same
B) Due to the error in the measurement of $d$ is twice that due to the error in the measurement of $\mathcal{l}$
C) Due to the error in the measurement of $\mathcal{l}$ is twice that due to the error in the measurement of $d$
D) Due to the error in the measurement of $d$ is four time that due to the error in the measurement of $\mathcal{l}$
Explanation
Least count of both instrument $\mathrm{\Delta}d\mathcal{= \mathrm{\Delta}l =}\frac{0.5}{100}\ mm = 5 \times 10^{- 3}\ mm$ $Y = \frac{4MLg}{\pi\mathcal{l}d^{2}}$ $\left( \frac{\mathrm{\Delta}Y}{Y} \right)_{\max} = \frac{\mathcal{\mathrm{\Delta}l}}{\mathcal{l}} + 2\frac{\mathrm{\Delta}d}{d}$ Error due to $\mathcal{l}$ measurement $\frac{\mathcal{\mathrm{\Delta}l}}{\mathcal{l}} = \frac{0.5/100\ mm}{0.25\ mm} = 2\%$ Error due to $d$ measurement $2\frac{\mathrm{\Delta}d}{d} = \frac{2 \times \frac{0.5}{100}}{0.5\ mm}$ $= \frac{0.5/100}{0.25} = 2\%$ Hence due to the errors in the measurements of $d$ and $\mathcal{l}$ are the same
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.