JEE Physics 2026 Question with Solution

Given: Two identical metallic spheres, each of mass $m$ and charge $Q$, are initially at a separation $r_i = 4R$ and move towards each other with speed $u$.

Find: The minimum value of $u$ such that they just touch, i.e. reach separation $r_f = 2R$ with zero final speed.

The spheres interact through electrostatic repulsion and gravitational attraction, so the total potential energy at separation $r$ is

$$U = \frac{kQ^2}{r} - \frac{Gm^2}{r} = \frac{kQ^2 - Gm^2}{r}$$

Since only conservative forces act, mechanical energy is conserved.

Initial kinetic energy of the two spheres is

$$K_i = \frac{1}{2}mu^2 + \frac{1}{2}mu^2 = mu^2$$

Initial potential energy is

$$U_i = \frac{kQ^2 - Gm^2}{4R}$$

For the minimum speed required to just touch, the final speed becomes zero at separation $2R$. Thus,

$$K_f = 0$$

and the final potential energy is

$$U_f = \frac{kQ^2 - Gm^2}{2R}$$

Applying conservation of mechanical energy,

$$K_i + U_i = K_f + U_f$$

so,

$$mu^2 + \frac{kQ^2 - Gm^2}{4R} = \frac{kQ^2 - Gm^2}{2R}$$

Therefore,

$$mu^2 = \frac{kQ^2 - Gm^2}{2R} - \frac{kQ^2 - Gm^2}{4R}$$ $$mu^2 = \frac{kQ^2 - Gm^2}{4R}$$

Hence,

$$u^2 = \frac{kQ^2}{4mR}\left(1 - \frac{Gm^2}{kQ^2}\right)$$

and

$$u = \sqrt{\frac{kQ^2}{4mR}\left(1 - \frac{Gm^2}{kQ^2}\right)}$$

Therefore, the minimum speed required is $\sqrt{\frac{kQ^2}{4mR}\left(1 - \frac{Gm^2}{kQ^2}\right)}$, so the correct option is A.

Given: The spheres start from separation $4R$ and must just reach separation $2R$.

Find: The least initial speed needed.

At the instant of just touching, all initial kinetic energy supplied is used to increase the net potential energy of the system against the dominant repulsive interaction.

The increase in potential energy from $4R$ to $2R$ is

$$\Delta U = \frac{kQ^2 - Gm^2}{2R} - \frac{kQ^2 - Gm^2}{4R}$$ $$\Delta U = \frac{kQ^2 - Gm^2}{4R}$$

Since both spheres have equal speed $u$ initially, total initial kinetic energy is

$$K_i = mu^2$$

For the just-touch condition,

$$K_i = \Delta U$$

Thus,

$$mu^2 = \frac{kQ^2 - Gm^2}{4R}$$

which gives

$$u = \sqrt{\frac{kQ^2}{4mR}\left(1 - \frac{Gm^2}{kQ^2}\right)}$$

Therefore, the correct option is A.