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Durring Searle's experiment, zero of the Vernier scale lies between $3.20 \times 10^{-2} m$ and $3.25 \times 10^{-2} m$ of the main scale. The $20^{\text {th }}$ division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of $2 \ kg$ is applied to the wire, the zero of the Vernier scale still lies between $3.20 \times 10^{-2} m$ and $3.25 \times 10^{-2} m$ of the main scale but now the $45^{\text {th }}$ division of Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is $2 m$. and its cross-sectional area is $8 \times 10^{-7} m ^2$. The least count of the Vernier scale is $1.0 \times 10^{-5} m$. The maximum percentage error in the Young's modulus of the wire is
$8$
$7$
$6$
$5$
Solution
Observation $-1$
Let weight used is $W _1$, extension $\ell_1$
$W_1$, $\ell_1$
$y =\frac{ W _1 / A }{\ell_1 / L } \Rightarrow W _1=\frac{ yA \ell_1}{ L } \quad \ell_1=3.2 \times 10^{-2}+20 \times 10^{-5}$
Observation $-2$
Let weight used is $W _2$ extension $\ell_2$
$W _2$, $\ell_2$
$y =\frac{ W _2 / A }{\ell_2 / L } \Rightarrow W _1=\frac{ yA \ell_2}{ L } \quad \ell_1=3.2 \times 10^{-2}+45 \times 10^{-5}$
$W _2- W _1=\frac{y A}{L}\left(\ell_2-\ell_1\right) \Rightarrow y=\frac{\left(W_2-W_1\right) / L}{y A\left(\ell_2-\ell_1\right)} $
$\left(\frac{\Delta y}{y}\right)_{\max }=\frac{\Delta \ell_2+\Delta \ell_1}{\ell_2-\ell_1}=\frac{2 \times 10^{-5}}{25 \times 10^{-5}} $
$\left(\frac{\Delta y}{y}\right)_{\max } \times 100 \% \quad=\frac{2}{25} \times 100 \%=8 \%$