Start free

Physics · Rotational Dynamics

JEE Main 2025 — 2 April, Morning Shift — Question 49

Moment of inertia of a rod of mass ' MM ' and length ' LL ' about an axis passing through its center and normal to its length is ' α\alpha '. Now the rod is cut into two equal parts and these parts are joined symmetrically to form a cross shape. Moment of inertia of cross about an axis passing through its center and normal to plane containing cross is

  1. Option A:

    α/4\alpha / 4

    Correct
  2. Option B:

    α/2\alpha / 2

  3. Option C:

    α/8\alpha / 8

  4. Option D:

    α\alpha

Answer: A

Step-by-step solution

α=ML212\alpha=\frac{M L^{2}}{12} I=(M2)(L2)212×2I=\frac{\left(\frac{M}{2}\right)\left(\frac{L}{2}\right)^{2}}{12} \times 2 I=α4I=\frac{\alpha}{4}

Answer key and solution verified before publishing.

Practise Rotational Dynamics

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 2025
Paper
2 April, Morning Shift
Subject
Physics
Chapter
Rotational Dynamics
Topic
Moment of Inertia

More Rotational Dynamics questions from this paper

Moment of inertia of a rod of mass ' M ' and length ' L ' about an… | JEE Main 2025 PYQ with Solution · DhiX AI