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In the determination of Young's modulus $\left(Y=\frac{4 M L g}{\pi l d^{2}}\right)$ 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 $0.5 \mathrm{~mm}$. The number of divisions on their circular scale is 100 . The contributions to the maximum probable error of the $Y$ measurement
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due to the errors in the measurements of $d$ and $l$ are the same.
The maximum possible error in $Y$ due to $l$ and $d$ $\frac{\Delta Y}{Y}=\frac{\Delta l}{l}+\frac{2 \Delta d}{d}$ Least count $=\frac{\text { Pitch }}{\text { No. of division on circular scale }}$ $=\frac{0.5}{100} \mathrm{~mm}=0.005 \mathrm{~mm}$
Here, $\Delta d=\Delta l=0.005 \mathrm{~mm}$
Error contribution of $l=\frac{\Delta l}{l}=\frac{0.005 \mathrm{~mm}}{0.25 \mathrm{~mm}}=\frac{1}{50}$
Error contribution of
$d=\frac{2 \Delta d}{d}=\frac{2 \times 0.005 \mathrm{~mm}}{0.5 \mathrm{~mm}}=\frac{1}{50}$
Hence contribution to the maximum possible error in the measurement of $y$ due to $l$ and $d$ is the same.
Here, $\Delta d=\Delta l=0.005 \mathrm{~mm}$
Error contribution of $l=\frac{\Delta l}{l}=\frac{0.005 \mathrm{~mm}}{0.25 \mathrm{~mm}}=\frac{1}{50}$
Error contribution of
$d=\frac{2 \Delta d}{d}=\frac{2 \times 0.005 \mathrm{~mm}}{0.5 \mathrm{~mm}}=\frac{1}{50}$
Hence contribution to the maximum possible error in the measurement of $y$ due to $l$ and $d$ is the same.
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