Physics · Gravitation

JEE Main 2026 — 21 January, Evening Shift — Question 26

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 $k=\frac{1}{4 \pi \epsilon_{0}}$ and assume $k Q^{2}>G m^{2}$ where G is the Gravitational constant)

Option A:
$\sqrt{\dfrac{kQ^{2}}{4mR}\left(1 - \dfrac{Gm^{2}}{kQ^{2}}\right)}$
Correct
Option B:
$\sqrt{\dfrac{kQ^{2}}{4mR}\left(1 + \dfrac{Gm^{2}}{kQ^{2}}\right)}$
Option C:
$\sqrt{\dfrac{kQ^{2}}{2mR}\left(1 - \dfrac{Gm^{2}}{kQ^{2}}\right)}$
Option D:
$\sqrt{\dfrac{kQ^{2}}{2mR}\left(1 - \dfrac{Gm^{2}}{2kQ^{2}}\right)}$

Answer: A

Step-by-step solution

Using energy conservation (2) $\left(\frac{1}{2} m u^{2}\right)-\frac{G m^{2}}{4 r}+\frac{K Q^{2}}{4 r}=-\frac{G m^{2}}{2 r}+\frac{K Q^{2}}{2 r}$ $u=\sqrt{\frac{1}{4 m r}\left(K Q^{2}-G m^{2}\right)}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 21 January, Evening Shift
Subject: Physics
Chapter: Gravitation
Topic: Newton's Law of Gravitation