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JEE Main 2025 (4 Apr, Shift 2) · Shear Modulus & Poisson's Ratio

Shear Modulus & Poisson's Ratio question from JEE Main 2025 (4 Apr, Shift 2)

Q3JEE Main 2025 · 4 APR · SHIFT 2Medium

Shear Modulus & Poisson's Ratio

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}\text{ 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}\text{ N/m}^2)

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Answer: (a)

  1. G=F/Aθ⇒θ=FGAG = \frac{F/A}{\theta} \Rightarrow \theta = \frac{F}{GA}, with A=πr2A = \pi r^2 (top face).
  2. A=π(0.04)2=16π×10−4 m2A = \pi(0.04)^2 = 16\pi\times10^{-4}\ \text{m}^2
  3. θ=1051010×16π×10−4=1160π\theta = \frac{10^{5}}{10^{10}\times16\pi\times10^{-4}} = \frac{1}{160\pi}
  4. So the answer is (a).
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