Q 11-08-009NEETJEE MainMedium
A metal cube of side $10$ cm has its bottom face fixed. A tangential force of $10^4$ N is applied to its top face. The shear modulus of the metal is $2.5 \times 10^{10}$ N m$^{-2}$. The top face moves sideways by
Answer: (A) $4\ \mu$m
Concept. Shear modulus $\eta = \dfrac{F/A}{\Delta x/L}$, so the sideways shift of the top face is $\Delta x = \dfrac{F}{A\eta}L$.
Step 1. Shear stress. $A = (0.1)^2 = 0.01\ \text{m}^2$:
$$\frac{F}{A} = \frac{10^4}{0.01} = 10^6\ \text{N m}^{-2}$$
Step 2. Shear strain.
$$\frac{10^6}{2.5 \times 10^{10}} = 4 \times 10^{-5}$$
Step 3. Displacement.
$$\Delta x = 4 \times 10^{-5} \times 0.1 = 4 \times 10^{-6}\ \text{m} = 4\ \mu\text{m}$$
Answer. $4\ \mu$m (Option A).
Solution by Sreeraj P, M.Sc Physics