Start free

Physics · Work, Power & Energy

JEE Main 2026 — 22 January, Morning Shift — Question 41

A simple pendulum has a bob with mass mm and charge q . The pendulum string has negligible mass. When a uniform and horizontal electric field E⃗\vec{E} is applied, the tension in the string changes. The final tension in the string, when pendulum attains an equilibrium position is ____\_\_\_\_ . (g : acceleration due to gravity)

  1. Option A:

    mg−qEm g-q E

  2. Option B:

    mg+qEm g+q E

  3. Option C:

    m2g2+q2E2\sqrt{m^{2} g^{2}+q^{2} E^{2}}

    Correct
  4. Option D:

    m2g2−q2E2\sqrt{m^{2} g^{2}-q^{2} E^{2}}

Answer: C

Step-by-step solution

T=(qE)2+(mg)2\mathrm{T}=\sqrt{(\mathrm{qE})^{2}+(\mathrm{mg})^{2}}

Solution figure

Answer key and solution verified before publishing.

Practise Work, Power & Energy

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2026
Paper
22 January, Morning Shift
Subject
Physics
Chapter
Work, Power & Energy
Topic
Conservative Forces and Potential Energy
A simple pendulum has a bob with mass m and charge q . The pendulum… | JEE Main 2026 PYQ with Solution · DhiX AI