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 $2kg$ 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 $2m$ 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

Answer: 4

$Y=\frac{FL}{\ell A}$ 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 $LC$
$\therefore \frac{\Delta Y}{Y}\times 100=\frac{(LC)}{25(LC)}\times 100=4$