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QJEEM24Ph08a7: JEE Main 2024 Physics - Class 11 JEEM24Ph08a7 Solution: Two blocks of mass and are connected by a metal wire going over …

Two blocks of mass 2 kg2 \mathrm{~kg} and 4 kg4 \mathrm{~kg} are connected by a metal wire going over a smooth pulley as shown in figure. The radius of wire is 4.0×10−5 m4.0 \times 10^{-5} \mathrm{~m} and Young's modulus of the metal is 2.0×1011 N/m22.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^2 . The longitudinal strain developed in the wire is 1απ\frac{1}{\alpha \pi} . The value of α\alpha is _________. [Use g=10 m/s2g=10 \mathrm{~m} / \mathrm{s}^2 ]

JEE Main 2024 (Online) 31st January Evening Shift Physics - Properties of Matter Question 49 English

Step-by-Step Verified Solution

amp;T=(2 m1 m2 m1+m2)g=803 Namp; A=πr2=16π×10−10 m2amp; Strain =Δℓℓ=FAY=TAYamp;=80/316π×10−10×2×1011=112πamp;α=12\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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