JEE Physics 2025 Question with Solution

Given: Two metal spheres have radii $R$ and $3R$, and both initially have the same surface charge density $\sigma$.

Find: The ratio $\frac{\sigma_1}{\sigma_2}$ after the spheres are brought into contact and then separated.

Initial charges are obtained from surface charge density multiplied by surface area:

$$Q_{\text{small}} = \sigma \times 4\pi R^2$$ $$Q_{\text{large}} = \sigma \times 4\pi (3R)^2 = 36\pi R^2 \sigma$$

So, total charge is

$$Q_{\text{total}} = 4\pi R^2 \sigma + 36\pi R^2 \sigma = 40\pi R^2 \sigma$$

When the spheres are brought into contact, their potentials become equal. For an isolated conducting sphere, potential is proportional to $\frac{Q}{r}$. Hence,

$$\frac{Q_1}{R} = \frac{Q_2}{3R}$$

which gives

$$Q_1 = \frac{Q_2}{3}$$

Using charge conservation,

$$Q_1 + Q_2 = 40\pi R^2 \sigma$$

Substituting $Q_1 = \frac{Q_2}{3}$,

$$\frac{Q_2}{3} + Q_2 = 40\pi R^2 \sigma$$ $$\frac{4Q_2}{3} = 40\pi R^2 \sigma$$ $$Q_2 = 30\pi R^2 \sigma$$

Therefore,

$$Q_1 = 10\pi R^2 \sigma$$

Now the final surface charge densities are

$$\sigma_1 = \frac{Q_1}{4\pi R^2} = \frac{10\pi R^2 \sigma}{4\pi R^2} = \frac{5\sigma}{2}$$ $$\sigma_2 = \frac{Q_2}{36\pi R^2} = \frac{30\pi R^2 \sigma}{36\pi R^2} = \frac{5\sigma}{6}$$

Hence,

$$\frac{\sigma_1}{\sigma_2} = \frac{\frac{5\sigma}{2}}{\frac{5\sigma}{6}} = 3$$

Therefore, the correct option is D.

Given: The initial surface charge density on both spheres is the same, and their radii are $R$ and $3R$.

Find: The ratio $\frac{\sigma_1}{\sigma_2}$ after contact and separation.

Since the initial surface charge density is the same, initial charges are proportional to surface areas:

$$Q_{\text{small}} : Q_{\text{large}} = 4\pi R^2 : 36\pi R^2 = 1 : 9$$

So total charge is distributed over the two spheres after contact in the ratio of their radii, because equal potential implies charge proportional to radius:

$$Q_1 : Q_2 = R : 3R = 1 : 3$$

Now surface charge density is charge divided by surface area. Therefore,

$$\sigma_1 : \sigma_2 = \frac{Q_1}{4\pi R^2} : \frac{Q_2}{36\pi R^2}$$

Using $Q_1 : Q_2 = 1 : 3$,

$$\sigma_1 : \sigma_2 = \frac{1}{4} : \frac{3}{36} = \frac{1}{4} : \frac{1}{12} = 3 : 1$$

Hence,

$$\frac{\sigma_1}{\sigma_2} = 3$$

This shortcut works because after contact, the common potential condition makes the final charges proportional to sphere radii, not surface areas. Therefore, the correct option is D.