Q 11-08-101JEE MainJEE Main 2018 (15 Apr, Shift 2)Easy
Forces of $10^5\ \text{N}$ each are applied in opposite directions, on the upper and lower faces of a cube of side $10\ \text{cm}$, shifting the upper face parallel to itself by $0.5\ \text{cm}$. If the side of another cube of the same material is $20\ \text{cm}$ then under similar conditions as above, the displacement will be:
Answer: (B) $0.25\ \text{cm}$
Modulus of rigidity:
$$\eta = \frac{F/A}{\Delta x/L} = \frac{FL}{A\,\Delta x} \quad\Rightarrow\quad \Delta x = \frac{FL}{\eta L^2} = \frac{F}{\eta L}$$
For the same force and material, $\Delta x \propto \dfrac1L$:
$$\Delta x_2 = 0.5\times\frac{10}{20} = 0.25\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics