Q 11-08-073JEE MainJEE Main 2022 (27 Jul, Shift 1)Medium
A square aluminium (shear modulus is $25\times10^9\ \text{N m}^{-2}$) slab of side $60$ cm and thickness $15$ cm is subjected to a shearing force (on its narrow face) of $18.0\times10^4$ N. The lower edge is riveted to the floor. The displacement of the upper edge is ______ $\mu$m.
Numerical value type. Enter your answer.
Answer: 48
The force acts on the narrow face of area $A = 0.60\times0.15 = 0.09\ \text{m}^2$; the height of the slab is $L = 0.60$ m.
$$\text{stress} = \frac{18\times10^4}{0.09} = 2\times10^6\ \text{N m}^{-2}$$
$$\Delta x = \frac{\text{stress}}{\eta}L = \frac{2\times10^6}{25\times10^9}\times0.6 = 4.8\times10^{-5}\ \text{m} = 48\ \mu\text{m}$$
Solution by Sreeraj P, M.Sc Physics