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JEE Main 2018 (15 Apr, Shift 2) · Shear Modulus & Poisson's Ratio

Shear Modulus & Poisson's Ratio question from JEE Main 2018 (15 Apr, Shift 2)

Q9JEE Main 2018 · 15 APR · SHIFT 2Medium

Shear Modulus & Poisson's Ratio

As shown in the figure, forces of 105 N each are applied in opposite directions, on the upper and lower faces of a cube of side 10 cm, shifting the upper face parallel to itself by 0.5 cm. If the side of another cube of the same material is 20 cm, then under similar conditions as above, the displacement will be:

Figure

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Answer: (a)

  1. η=F/L2Δx/L⇒Δx=FηL\eta = \frac{F/L^2}{\Delta x/L} \Rightarrow \Delta x = \frac{F}{\eta L}
  2. Same FF and η\eta: Δx∝1L\Delta x \propto \frac{1}{L}
  3. LL doubles ⇒Δx′=0.52=0.25 cm\Rightarrow \Delta x' = \frac{0.5}{2} = 0.25\ \text{cm}
  4. So the answer is (a).
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