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Question

In Searle's experiment to find Young's modulus the diameter of wire is measured as $$d=0.05cm$$, length of wire is $$l=125cm$$ and when a weight,$$m=20.0kg$$ is put, extension in wire was found to be $$0.100cm$$. Find the maximum permissible error in Young's modulus $$(Y)$$. Use:$$Y=\displaystyle\frac{mgl}{(\pi/4)d^2x}$$.


A
6.3%
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B
5.3%
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C
2.3%
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D
1%
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Solution

The correct option is A $$6.3\%$$
$$Y=\displaystyle\frac{mgl}{(\pi/4)d^2x}$$ .....$$(1)$$
$$\displaystyle\left[\frac{dY}{Y}\right]_{max}$$ $$=\displaystyle \frac{\Delta m}{m}+\frac{\Delta l}{l}+2\frac{\Delta d}{d}+\frac{\Delta x}{x}$$
$$m=20.0kg\Rightarrow \Delta m=0.1kg$$
$$l=125cm\Rightarrow \Delta l=1cm$$
$$d=0.050cm\Rightarrow \Delta d=0.001cm$$
$$x=0.100cm\Rightarrow \Delta x=0.001cm$$
$$\displaystyle\left[\frac{dY}{Y}\right]_{max}=\displaystyle\left[\frac{0.1kg}{20.0kg}+\frac{1cm}{125cm}+2\times \frac{0.001cm}{0.05cm}+\frac{0.001cm}{0.100cm}\right]\times 100\%=6.3\%$$

Physics

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