MHTCET 11 Apr 2026 Shift 1 Paper

Concept:When two conducting spheres are brought into contact, charges redistribute until their electric potentials become equal.Explanation:For a metal sphere, $Q = \sigma \cdot 4\pi r^{2}$.Initially, both spheres have the same surface charge density $\sigma$.For the smaller sphere of radius $R$: $Q_{1} = \sigma \cdot 4\pi R^{2}$.For the bigger sphere of radius $3R$: $Q_{2} = \sigma \cdot 4\pi (3R)^{2} = 9\sigma \cdot 4\pi R^{2}$.On contact, charge flows from one sphere to the other until potentials become equal.For a conducting sphere, $V = \frac{kQ}{r}$.After contact, equate the potentials: $\frac{kQ_{1}'}{R} = \frac{kQ_{2}'}{3R}$.Cancelling $k$ and $R$ gives $Q_{2}' = 3Q_{1}'$.After separation, the new surface charge densities are:$\sigma_{1} = \frac{Q_{1}'}{4\pi R^{2}}$ and $\sigma_{2} = \frac{Q_{2}'}{4\pi (3R)^{2}}$.Substitute $Q_{2}' = 3Q_{1}'$: $\sigma_{2} = \frac{3Q_{1}'}{36\pi R^{2}} = \frac{Q_{1}'}{12\pi R^{2}}$.Thus, $\frac{\sigma_{1}}{\sigma_{2}} = \frac{\frac{Q_{1}'}{4\pi R^{2}}}{\frac{Q_{1}'}{12\pi R^{2}}} = 3$.Answer:The ratio $\frac{\sigma_{1}}{\sigma_{2}}$ is $3$, which is option D.