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JEE MAIN 2024
31.01.24 S2
Question
Two blocks of mass 2 kg and 4 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 $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )
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Solution
Sol. $$ \begin{aligned} & \mathrm{T}=\left(\frac{2 \mathrm{~m}_1 \mathrm{~m}_2}{\mathrm{~m}_1+\mathrm{m}_2}\right) \mathrm{g}=\frac{80}{3} \mathrm{~N} \\ & \mathrm{~A}=\pi \mathrm{r}^2=16 \pi \times 10^{-10} \mathrm{~m}^2 \\ & \text { Strain }=\frac{\Delta \ell}{\ell}=\frac{\mathrm{F}}{\mathrm{AY}}=\frac{\mathrm{T}}{\mathrm{AY}} \\ & =\frac{80 / 3}{16 \pi \times 10^{-10} \times 2 \times 10^{11}}=\frac{1}{12 \pi} \\ & \alpha=12 \end{aligned} $$
Question Tags
JEE Main
Physics
Medium
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