MHTCET 13 Apr 2026 Shift 2 Paper

Concept:When two conducting spheres are connected by a wire, charge flows between them until their electric potentials become equal.Explanation:For a conducting sphere, surface charge density is given by$$\sigma = \frac{Q}{4\pi r^2}$$Initially, both spheres have the same charge density $\sigma$.For the smaller sphere of radius $R$,$$Q_1 = \sigma \cdot 4\pi R^2$$For the larger sphere of radius $2R$,$$Q_2 = \sigma \cdot 4\pi (2R)^2 = 16\pi R^2 \sigma$$Total initial charge,$$Q = Q_1 + Q_2 = 4\pi R^2 \sigma + 16\pi R^2 \sigma = 20\pi R^2 \sigma$$After connecting, let the final charges be $Q_1'$ and $Q_2'$.Potentials become equal:$$\frac{kQ_1'}{R} = \frac{kQ_2'}{2R}$$Cancelling $k$ and $R$,$$Q_2' = 2Q_1'$$Charge is conserved:$$Q_1' + Q_2' = 20\pi R^2 \sigma$$Substituting $Q_2' = 2Q_1'$,$$3Q_1' = 20\pi R^2 \sigma$$So,$$Q_2' = \frac{40}{3}\pi R^2 \sigma$$New charge density of the larger sphere,$$\sigma_1 = \frac{Q_2'}{4\pi (2R)^2} = \frac{\frac{40}{3}\pi R^2 \sigma}{16\pi R^2} = \frac{5}{6}\sigma$$Hence,$$\frac{\sigma_1}{\sigma} = \frac{5}{6}$$Answer:$\sigma_1 : \sigma = 5 : 6$, so the correct option is D.