Mechanical Properties of Solids

JEE Main level, numerical answer, medium

26 of 118

Question 26 of 118

Two blocks of mass 2kg2kg and 4kg4kg are connected by a metal wire going over a smooth pulley as shown in figure. The radius of wire is 4.0×10−5m4.0\times {10}^{-5}m and Young's modulus of the metal is 2.0×1011Nm−22.0\times {10}^{11}N{m}^{-2}. The longitudinal strain developed in the wire is 1απ\dfrac{1}{\alpha \pi }. The value of α\alpha is _____. [Use g=10ms−2g=10m{s}^{-2})

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Answer: 12


For the given system,
T=(2m1m2m1+m2)g=803NT=\left(\dfrac{2{m}_{1}{m}_{2}}{{m}_{1}+{m}_{2}}\right)g=\dfrac{80}{3}N
Area of the wire,
A=πr2=16π×10−10m2A=\pi {r}^{2}=16\pi \times {10}^{-10}{m}^{2}
Strain
=Δll=FAY=TAY=\dfrac{\Delta l}{l}=\dfrac{F}{AY}=\dfrac{T}{AY}
=80316π×10−10×2×1011=112π=\dfrac{\dfrac{80}{3}}{16\pi \times {10}^{-10}\times 2\times {10}^{11}}=\dfrac{1}{12\pi }
⇒α=12\Rightarrow \alpha =12

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