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 × 10 -5 m and Young's modulus of the metal is 2.0 × 10 11 N / m 2 . The longitudinal strain developed in the wire is 1/α π . The value of α is \ \ \ \ \ \ \ . [Use . g =10 m / s 2 )

Physics · Mechanical Properties of Matter

JEE Main 2024 — 31 January, Shift 2 — Question 51

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 $4.0 \times 10^{-5}$ 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 $\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$

Answer: 12

Numerical answer — enter this value.

Step-by-step solution

$T=\left(\frac{2 \mathrm{~m}_{1} \mathrm{~m}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\right) \mathrm{g}=\frac{80}{3} \mathrm{~N}$

$\mathrm{A}=\pi \mathrm{r}^{2}=16 \pi \times 10^{-10} \mathrm{~m}^{2}$

$\operatorname{Strain}=\frac{\Delta \ell}{\ell}=\frac{\mathrm{F}}{\mathrm{AY}}=\frac{\mathrm{T}}{\mathrm{AY}}$

$=\frac{80 / 3}{16 \pi \times 10^{-10} \times 2 \times 10^{11}}=\frac{1}{12 \pi}$ $\alpha=12$

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Exam: JEE Main 2024
Paper: 31 January, Shift 2
Subject: Physics
Chapter: Mechanical Properties of Matter
Topic: Stress,Strain and Modulus of Elasticity

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Answer: 12

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