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JEE Main 20246 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Evening Shift
Subject
Physics
Chapter
Mechanical Properties of Solids

JEE Main 2024 · 6 April · Evening Shift · Physics · Mechanical Properties of Solids

A wire of cross sectional area A, modulus of elasticity 2 × 10^ 11...

A wire of cross sectional area A, modulus of elasticity 2×1011Nm−22 \times 10^{11} \mathrm{Nm}^{-2} and length 2 m2 \mathrm{~m} is stretched between two vertical rigid supports. When a mass of 2 kg2 \mathrm{~kg} is suspended at the middle it sags lower from its original position making angle θ=1100\theta=\frac{1}{100} radian on the points of support. The value of A\mathrm{A} is _______ ×10−4 m2\times 10^{-4} \mathrm{~m}^2 (consider x<<Lx< < \mathrm{L} ). (given : g=10 m/s2\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2 )A wire of cross sectional area A, modulus of elasticity 2 × 10^ 11 Nm^ -2 and length 2 ~m is stretched between two

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Step-by-step solution

A wire of cross sectional area A, modulus of elasticity 2 × 10^ 11 Nm^ -2 and length 2 ~m is stretched between two In vertical direction 2 Tsin⁡θ=202 \mathrm{~T} \sin \theta=20 using small angle approximation sin⁡θ=θ\sin \theta=\theta θ=1100∴T=10θT=1000 N\begin{aligned} & \theta=\frac{1}{100} \\ & \therefore \mathrm{T}=\frac{10}{\theta} \\ & \mathrm{T}=1000 \mathrm{~N} \end{aligned} Change in length ΔL=2x2+L2−2 L=2 L[1+x22 L2−1]ΔL=x2 L\begin{array}{ll} \text{Change in length } \Delta \mathrm{L} =2 \sqrt{\mathrm{x}^2+\mathrm{L}^2}-2 \mathrm{~L} \\ =2 \mathrm{~L}\left[1+\frac{\mathrm{x}^2}{2 \mathrm{~L}^2}-1\right] \\ \Delta \mathrm{L} =\frac{\mathrm{x}^2}{\mathrm{~L}} \end{array} ∴\therefore Modulus of elasticity = stress  strain =\frac{\text{ stress }}{\text{ strain }} 2×1011=103 A×x2 L×2 L∴A=1×10−4 m2\begin{aligned} & 2 \times 10^{11}=\frac{10^3}{\mathrm{~A} \times \frac{\mathrm{x}^2}{\mathrm{~L}}} \times 2 \mathrm{~L} \\ \therefore & \mathrm{A}=1 \times 10^{-4} \mathrm{~m}^2 \end{aligned}
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