Question

During 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

Answer 1

$Y =\frac{ FL }{\ell A }$ since the experiment measures only change in the length of wire
$\therefore \frac{\Delta Y }{ Y } \times 100=\frac{\Delta \ell}{\ell} \times 100$
From the observation $\ell_{1}= MSR +20$ (LC)
$\ell_{2}= MSR +45 $ (LC)
$\Rightarrow $ change in lengths $=25$( LC )
and the maximum permissible error in elongation is one $LC$
$\therefore \frac{\Delta Y }{ Y } \times 100=\frac{( LC )}{25( LC )} \times 100=4$