In the determination of Young's modulus $\left(Y=\frac{4 M L g}{\pi l d^{2}}\right)$ by using Searle

Question

In the determination of Young's modulus $\left(Y=\frac{4 M L g}{\pi l d^{2}}ight)$ by using Searle's method, a wire of length $L=2 \mathrm{~m}$ and diameter $d=0.5 \mathrm{~mm}$ is used. For a load $M=2.5 \mathrm{~kg}$, an extension $l=0.25 \mathrm{~mm}$ 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\

Professor PC Thomas Classes · Answer

(a) : Least count of screw gauge\ $=\frac{	ext { Pitch }}{	ext { Total no. of circular divisions }}$\ on circular scale\ $=\frac{0.5 \mathrm{~mm}}{100}=0.005 \mathrm{~mm}$\ Least count of micrometer\ $=\frac{	ext { Pitch }}{	ext { Total no. of circular divisions }}$ on circular scale\ $=\frac{0.5 \mathrm{~mm}}{100}=0.005 \mathrm{~mm}$\ The errors in both $l$ and $d$ are least count errors.\ $	herefore \quad \Delta l=\Delta d=0.005 \mathrm{~mm}$\ $Y=\frac{4 M L g}{\pi l d^{2}}$\ The maximum probable error in $Y$ is,\ $\frac{\Delta Y}{Y}=\frac{\Delta l}{l}+\frac{2 \Delta d}{d}$\ Error due to $l$ measurement,\ $\frac{\Delta l}{l}=\frac{0.005 \mathrm{~mm}}{0.25 \mathrm{~mm}}=0.02$\ Error due to $d$ measurement,\ $\frac{2 \Delta d}{d}=\frac{2 	imes 0.005 \mathrm{~m}}{0.5 \mathrm{~mm}}=0.02$\ So error in $Y$ due to measurement of $l$ is same as that of due to measurement of $d$.\