JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Work, Power & Energy

Consider two identical metallic spheres of radius

R

each having charge

Q

and mass

m

. Their centers have an initial separation of

4 R

. Both the spheres are given an initial speed of

u

towards each other. The minimum value of

u

, so that they can just touch each other is :
(Take

k = \frac{1}{4 π ϵ _{0}}

and assume

k Q^{2} > G m^{2}

where G is the Gravitational constant)

Answer: (A)

Solution

Apply conservation of energy between the initial state (separation

4 R

) and final state (just touching, separation

2 R

). Initial kinetic energy:

2 \times \frac{1}{2} m u^{2} = m u^{2}

. Initial potential energy (electrostatic + gravitational):

\frac{k Q ^{2}}{4 R} - \frac{G m ^{2}}{4 R}

. Final kinetic energy: 0 (minimum speed condition). Final potential energy:

\frac{k Q ^{2}}{2 R} - \frac{G m ^{2}}{2 R}

. By conservation:

m u^{2} + \frac{k Q ^{2}}{4 R} - \frac{G m ^{2}}{4 R} = \frac{k Q ^{2}}{2 R} - \frac{G m ^{2}}{2 R}

. Solving:

m u^{2} = \frac{k Q ^{2}}{4 R} - \frac{G m ^{2}}{4 R} = \frac{1}{4 R} (k Q^{2} - G m^{2})

. Therefore

u = \frac{k Q ^{2}}{4 m R} (1 - \frac{G m ^{2}}{k Q ^{2}})

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Single correct option