In the determination of Young's modulus left( | vedclass

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Question

(A) The formula for Young's modulus is $Y = \frac{4 MLg}{\pi \ell d^2}$.The maximum relative error in $Y$ is given by $\left(\frac{\Delta Y}{Y}\right)

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due to the errors in the measurements of $d$ and $\ell$ are the same.

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(A) The formula for Young's modulus is $Y = \frac{4 MLg}{\pi \ell d^2}$.The maximum relative error in $Y$ is given by $\left(\frac{\Delta Y}{Y}ight)_{\max} = \frac{\Delta \ell}{\ell} + 2 \frac{\Delta d}{d}$.The least count of both instruments is $	ext{LC} = \frac{	ext{pitch}}{	ext{number of divisions}} = \frac{0.5 \ mm}{100} = 0.005 \ mm$. Thus,$\Delta d = \Delta \ell = 0.005 \ mm$.The relative error contribution due to $\ell$ is $\frac{\Delta \ell}{\ell} = \frac{0.005 \ mm}{0.25 \ mm} = 0.02$.The relative error contribution due to $d$ is $2 \frac{\Delta d}{d} = 2 	imes \frac{0.005 \ mm}{0.5 \ mm} = 2 	imes 0.01 = 0.02$.Since both contributions are equal to $0.02$,the errors in the measurements of $d$ and $\ell$ contribute equally to the maximum probable error of $Y$.

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