Physics · Electrostatics

JEE Main 2025 — 28 January, Morning Shift — Question 63

A particle of mass ' $m$ ' and charge ' $q$ ' is fastened to one end ' $A$ ' of a massless string having equilibrium length $\ell$ , whose other end is fixed at point ' O '. The whole system is placed on a frictionless horizontal plane and is initially at rest. If uniform electric field is switched on along the direction as shown in figure, then the speed of the particle when it crosses the x -axis is

Question figure

Option A:
$\sqrt{\frac{2 q E \ell}{m}}$
Option B:
$\sqrt{\frac{q E \ell}{4 m}}$
Option C:
$\sqrt{\frac{q E \ell}{m}}$
Correct
Option D:
$\sqrt{\frac{q E \ell}{2 m}}$

Answer: C

Step-by-step solution

$\mathrm{W}_{\mathrm{all}}=\Delta \mathrm{k}$

$\mathrm{W}_{\mathrm{e}}=\mathrm{k}_{\mathrm{f}}-\mathrm{k}_{\mathrm{i}}$

$\mathrm{qE} \frac{\ell}{2}=\frac{1}{2} \mathrm{mv}^{2}-0$

$v=\sqrt{\frac{\mathrm{qE} \ell}{\mathrm{m}}}$

IMAGES

Solution figure

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 28 January, Morning Shift
Subject: Physics
Chapter: Electrostatics
Topic: Electrostatic Field & motion of charge

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