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Physics · Mechanical Properties of Matter

JEE Main 2025 — 4 April, Evening Shift — Question 58

A cylindrical rod of length 1 m and radius 4 cm is mounted vertically. It is subjected to a shear force of

105 N10^{5} \mathrm{~N} at the top. Considering infinitesimally small displacement in the upper edge,

the angular displacement θ\theta of the rod axis from its original position would be

(shear moduli, G=1010 N/m2G=10^{10} \mathrm{~N} / \mathrm{m}^{2} )

  1. Option A:

    14π\frac{1}{4 \pi}

  2. Option B:

    140π\frac{1}{40 \pi}

  3. Option C:

    12π\frac{1}{2 \pi}

  4. Option D:

    1160π\frac{1}{160 \pi}

    Correct

Answer: D

Step-by-step solution

FA=ηθ\frac{F}{A}=\eta \theta

105π16×10−4×1010=θ\frac{10^{5}}{\pi 16 \times 10^{-4} \times 10^{10}}=\theta

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
4 April, Evening Shift
Subject
Physics
Chapter
Mechanical Properties of Matter
Topic
Longitudinal Strain and Elastic Potential Energy
A cylindrical rod of length 1 m and radius 4 cm is mounted… | JEE Main 2025 PYQ with Solution · DhiX AI