JEE Main 2026 — Physics Electromagnetism

2026mcqmedium

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)

Official previous-year question

Held on 21 Jan 2026 · Verified 6 Jul 2026.

Options

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

Did you get this right?

Solution

Using energy conservation, the system's kinetic energy must overcome the increase in electrostatic potential energy as the spheres approach.

Initial state: separation 4R, speed u each.

Final state: touching (separation 2R), speed zero.

Change in potential energy: $\Delta U = k\frac{Q^2}{2R} - k\frac{Q^2}{4R} = k\frac{Q^2}{4R}$ .

Initial kinetic energy: $KE_i = mu^2$ .

Energy equation: $mu^2 = k\frac{Q^2}{4R}\left(1 - \frac{Gm^2}{kQ^2}\right)$ .

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

Your note

JEE Main Electromagnetism in other years

Electromagnetism 2025 Electromagnetism 2024 Electromagnetism 2023 Electromagnetism 2022 Electromagnetism 2021 Electromagnetism 2020 Electromagnetism 2019 Electromagnetism 2018 Electromagnetism 2017 Electromagnetism 2016 Electromagnetism 2015 Electromagnetism 2014

Keep practicing Electromagnetism

Work through every JEE Main Electromagnetism PYQ, year by year.

Browse all Physics topics

Options

$\sqrt{\frac{k Q^{2}}{2 m R}\left(1-\frac{G m^{2}}{k Q^{2}}\right)}$

$\sqrt{\frac{k Q^{2}}{4 m R}\left(1-\frac{G m^{2}}{k Q^{2}}\right)}$

$\sqrt{\frac{k Q^{2}}{2 m R}\left(1-\frac{G m^{2}}{2 k Q^{2}}\right)}$

$\sqrt{\frac{k Q^{2}}{4 m R}\left(1+\frac{G m^{2}}{k Q^{2}}\right)}$

Solution

Using energy conservation, the system's kinetic energy must overcome the increase in electrostatic potential energy as the spheres approach.

Initial state: separation 4R, speed u each.

Final state: touching (separation 2R), speed zero.

Change in potential energy: $\Delta U = k\frac{Q^2}{2R} - k\frac{Q^2}{4R} = k\frac{Q^2}{4R}$ .

Initial kinetic energy: $KE_i = mu^2$ .

Energy equation: $mu^2 = k\frac{Q^2}{4R}\left(1 - \frac{Gm^2}{kQ^2}\right)$ .

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

JEE Main 2026 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism

JEE Main 2026 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism