General Physics Ques 27
- 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 is
(2012)
(a) due to the errors in the measurements 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 $l$
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Answer:
Correct Answer: 27.( c )
Solution:
- $\begin{aligned} & \Delta d=\Delta l=\frac{0.5}{100} \mathrm{~mm}=0.005 \mathrm{~mm} \\ & Y=\frac{4 M L g}{\pi l d^2} \Rightarrow\left(\frac{\Delta Y}{Y}\right)_{\max }=\left(\frac{\Delta l}{l}\right)+2\left(\frac{\Delta d}{d}\right) \\ & \left(\frac{\Delta l}{l}\right)=\frac{0.5 / 100}{0.25}=0.02 \text { and } \frac{2 \Delta d}{d}=\frac{(2)(0.5 / 100)}{0.5}=0.02 \\ & \text { or } \quad \frac{\Delta l}{l}=2 \cdot \frac{\Delta d}{d}\end{aligned}$
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