In an experiment of determine the Young's | vedclass

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Question

$L =1\,m$

$\Delta L =0.4 	imes 10^{-3} m$

$m =1\,kg$

$d =0.4 	imes 10^{-3}\,m$

$\frac{ F }{ A }= Y \frac{\Delta L }{ L }$

$Y =\frac{

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Vedclass · Answer

$L =1\,m$

$\Delta L =0.4 	imes 10^{-3} m$

$m =1\,kg$

$d =0.4 	imes 10^{-3}\,m$

$\frac{ F }{ A }= Y \frac{\Delta L }{ L }$

$Y =\frac{ FL }{ A \Delta L }=\frac{( mg ) \cdot(1)}{\left(\frac{\pi d ^{2}}{4}ight) 0.4 	imes 10^{-3}}$

$\Rightarrow \frac{10 	imes 4}{\pi\left(0.4 	imes 10^{-3}ight)^{2} 	imes 0.4 	imes 10^{-3}}$

$Y =\frac{40}{\pi\left(0.4 	imes 10^{-3}ight)^{3}}$

$Y =\frac{40 	imes 7}{22 	imes 64 	imes 10^{-3} 	imes 10^{-9}}$

$Y =0.199 	imes 10^{-12} N / m ^{2}$

$\frac{\Delta Y }{ Y }=\frac{\Delta F }{ F }+\frac{\Delta L }{ L }+\frac{\Delta A }{ A }+\frac{\Delta(\Delta L )}{(\Delta L )}$

$=\frac{0.02}{0.4}+2 \frac{\Delta d }{ d }=\frac{0.2}{4}+2 	imes \frac{0.01}{0.4}$

$=\frac{0.1}{2}+\frac{0.1}{2}=0.1$

$\Rightarrow \Delta Y =0.1 	imes Y$

$=0.199 	imes 10^{11}=1.99 	imes 10^{10}$

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