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JEE Main 2026 (5 Apr, Shift 1) · Shear Modulus & Poisson's Ratio

Shear Modulus & Poisson's Ratio question from JEE Main 2026 (5 Apr, Shift 1)

Q1JEE Main 2026 · 5 APR · SHIFT 1Easy

Shear Modulus & Poisson's Ratio

A cube has side length 5 cm and modulus of rigidity 105 N/m210^{5}\text{ N/m}^2. The displacement produced by a force of 10 N in the upper face of cube is ______ mm.

Numerical answer type. Work it out, then open the answer.

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Answer: 2

  1. η=F/AΔx/L⇒Δx=FLηA\eta = \frac{F/A}{\Delta x/L} \Rightarrow \Delta x = \frac{FL}{\eta A}
  2. A=(0.05)2=2.5×10−3 m2A = (0.05)^2 = 2.5\times10^{-3}\ \text{m}^2, L=0.05 mL = 0.05\ \text{m}
  3. Δx=10×0.05105×2.5×10−3=2×10−3 m\Delta x = \frac{10\times0.05}{10^{5}\times2.5\times10^{-3}} = 2\times10^{-3}\ \text{m}
  4. Δx=2 mm\Delta x = 2\ \text{mm}
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