Mechanical Properties of Solids

JEE Main level, single correct, medium

77 of 118

Question 77 of 118

Two blocks of masses m\mathrm{m} and M\mathrm{M} are connected by means of a metal wire of cross-sectional area A passing over a frictionless fixed pulley as shown in the figure. The system is then released. If M=2 m\mathrm{M}=2 \mathrm{~m}, then the stress produced in the wire is:

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Answer: (B) 4mg3 A\frac{4 \mathrm{mg}}{3 \mathrm{~A}}

Tension in the wire, T=(2mMm+M)g\mathrm{T}=\left(\frac{2 \mathrm{mM}}{\mathrm{m}+\mathrm{M}}\right) \mathrm{g}
 Stress = Force / Tension  Area =2mMA(m+M)g=2( m×2 m)gA(m+2 m)(M=2 m given )=4 m23 mA g=4mg3 A\begin{aligned} \text { Stress } &=\frac{\text { Force } / \text { Tension }}{\text { Area }}=\frac{2 \mathrm{mM}}{\mathrm{A}(\mathrm{m}+\mathrm{M})} \mathrm{g} \\ &=\frac{2(\mathrm{~m} \times 2 \mathrm{~m}) \mathrm{g}}{\mathrm{A}(\mathrm{m}+2 \mathrm{~m})}(\mathrm{M}=2 \mathrm{~m} \text { given }) \\ &=\frac{4 \mathrm{~m}^2}{3 \mathrm{~mA}} \mathrm{~g}=\frac{4 \mathrm{mg}}{3 \mathrm{~A}} \end{aligned}

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