JEE MAIN - Physics (2024 - 31st January Evening Shift - No. 21)
Two blocks of mass $$2 \mathrm{~kg}$$ and $$4 \mathrm{~kg}$$ are connected by a metal wire going over a smooth pulley as shown in figure. The radius of wire is $$4.0 \times 10^{-5} \mathrm{~m}$$ and Young's modulus of the metal is $$2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$$. The longitudinal strain developed in the wire is $$\frac{1}{\alpha \pi}$$. The value of $$\alpha$$ is _________. [Use $$g=10 \mathrm{~m} / \mathrm{s}^2$$]
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Answer
12
Explanation
$$\begin{aligned}
& \mathrm{T}=\left(\frac{2 \mathrm{~m}_1 \mathrm{~m}_2}{\mathrm{~m}_1+\mathrm{m}_2}\right) \mathrm{g}=\frac{80}{3} \mathrm{~N} \\
& \mathrm{~A}=\pi \mathrm{r}^2=16 \pi \times 10^{-10} \mathrm{~m}^2 \\
& \text { Strain }=\frac{\Delta \ell}{\ell}=\frac{\mathrm{F}}{\mathrm{AY}}=\frac{\mathrm{T}}{\mathrm{AY}} \\
& =\frac{80 / 3}{16 \pi \times 10^{-10} \times 2 \times 10^{11}}=\frac{1}{12 \pi} \\
& \alpha=12
\end{aligned}$$
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