QJEEAdv07Ph44b0: JEE Advanced 2007 Physics - Class 11 JEEAdv07Ph44b0 Solution: A student performs an experiment to determine the Young's…

Young's modulus $(Y)=\frac{\text { Stress }}{\text { Strain }}=\frac{4 m g l}{\pi d^{2} \Delta l}$

$\begin{gathered}\mathrm{Y}=\frac{4 m g l}{\pi d^{2} l} \\\\=\frac{4 \times 1 \times 9.8 \times 2}{3.14 \times(0.4)^{2} \times(0.8) \times 10^{-6} \times 10^{-3}} \\\\=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\end{gathered}$

Where, $\quad$ Mass $(m)=1 \mathrm{~kg}$

$\begin{aligned}g & =9.8 \mathrm{~m} / \mathrm{s}^{2} \\\\\text { Length of wire } & =2 \mathrm{~m} \\\\\pi & =3.14 \\\\\text { Diameter }(d) & =0.4 \mathrm{~mm} \times 10^{-3} \mathrm{~m}\end{aligned}$

Extension in the length of wire

$\begin{aligned}(\Delta l) & =0.8 \mathrm{~mm} \\\\& =0.8 \times 10^{-3} \mathrm{~m} \\\\\frac{\Delta y}{y} & =\frac{2 \Delta d}{d}+\frac{\Delta l}{l} \\\\& =\frac{2 \times(0.01)}{0.4}+\frac{(0.05)}{0.8} \\\\& =0.1125\end{aligned}$

$\begin{aligned}\Rightarrow \Delta \mathrm{Y} & =(\mathrm{Y}) \times 0.1125 \\\\& =2 \times 10^{11} \times 0.1125 \\\\& =0.225 \times 10^{11} \\\\& \simeq 0.2 \times 10^{11} \\\\\text { (b) }\left(2 \pm 0.2 \times 10^{11} \mathrm{~N/}\right. & \left.\mathrm{m}^{2}\right)\end{aligned}$