Solution
When a conducting ring is charged, the mutual repulsion between elements of the charge creates an outward electrostatic pressure, resulting in tension $T$ in the ring. The relationship for tension in a ring of radius $R$ and charge $q$ is proportional to:
$$T \propto \frac{q^2}{R^2}$$Let the breaking strength (maximum tension) of the first ring be $S$. We are given:
$$S \propto \frac{q^2}{R^2}$$For the second ring:
- Radius is $nR$.
- Tensile strength is $kS$.
- Let the maximum charge be $Q$.
The condition for the second ring to just rupture is:
$$kS \propto \frac{Q^2}{(nR)^2}$$Taking the ratio of the two conditions:
$$\frac{kS}{S} = \frac{\frac{Q^2}{n^2 R^2}}{\frac{q^2}{R^2}}$$ $$k = \frac{Q^2}{n^2 q^2}$$ $$Q^2 = k n^2 q^2$$ $$Q = q n \sqrt{k}$$Answer: (b) $< qn\sqrt{k}$