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Physics · Rotational Dynamics

JEE Main 2026 — 22 January, Morning Shift — Question 25

A solid sphere of mass 5 kg and radius 10 cm is kept in contact with another solid sphere of mass 10 kg and radius 20 cm . The moment of inertia of this pair of spheres about the tangent passing through the point of contact is ____\_\_\_\_ kg.m2\mathrm{kg} . \mathrm{m}^{2}.

  1. Option A:

    0.36

  2. Option B:

    0.72

  3. Option C:

    0.18

  4. Option D:

    0.63

    Correct

Answer: D

Step-by-step solution

I=75[m1R12+m2R22]I=\frac{7}{5}\left[m_{1} R_{1}^{2}+m_{2} R_{2}^{2}\right] =75[5(10)2+10×(20)2]×10−4=\frac{7}{5}\left[5(10)^{2}+10 \times(20)^{2}\right] \times 10^{-4} I=63×10−2 kg m2\mathrm{I}=63 \times 10^{-2} \mathrm{~kg} \mathrm{~m}^{2} I=0.63 kg m2\mathrm{I}=0.63 \mathrm{~kg} \mathrm{~m}^{2}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
22 January, Morning Shift
Subject
Physics
Chapter
Rotational Dynamics
Topic
Moment of Inertia

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