JEE Main 2026 — Jan 21, Shift 2 — Physics solution

JEE Main2026Jan 21, Shift 2Physics

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)

A
$\frac{k Q ^{2}}{4 m R} (1 - \frac{G m ^{2}}{k Q ^{2}})$
B
$\frac{k Q ^{2}}{4 m R} (1 + \frac{G m ^{2}}{k Q ^{2}})$
C
$\frac{k Q ^{2}}{2 m R} (1 - \frac{G m ^{2}}{k Q ^{2}})$
D
$\frac{k Q ^{2}}{2 m R} (1 - \frac{G m ^{2}}{2 k Q ^{2}})$

Solution

Use energy conservation between the initial state (separation $4 R$ , each sphere moving with speed $u$ ) and the just-touching state (separation $2 R$ , both at rest).

Initial KE of the two-sphere system: $2 \times \frac{1}{2} m u^{2} = m u^{2}$ .

Net potential energy at separation $d$ (electrostatic repulsion minus gravitational attraction): $U (d) = \frac{k Q ^{2}}{d} - \frac{G m ^{2}}{d}$ .

Conservation gives: $m u^{2} + \frac{k Q ^{2} - G m ^{2}}{4 R} = \frac{k Q ^{2} - G m ^{2}}{2 R}$ .

Solving: $m u^{2} = \frac{k Q ^{2} - G m ^{2}}{4 R}$ , so $u = \frac{k Q ^{2}}{4 m R} (1 - \frac{G m ^{2}}{k Q ^{2}})$ .

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