Moment of inertia about an axis AB for a rod of mass 40 kg and length 3 m is same as that of a solid sphere of mass 10 kg and radius R about an axis parallel to AB axis with separation of 3 m as shown in figure. The value of R is given as √(α/2) . The value of α is .

Physics · Rotational Dynamics

JEE Main 2026 — 2 April, Evening Shift — Question 22

Moment of inertia about an axis AB for a rod of mass $40\mathrm{kg}$ and length $3\mathrm{m}$ is same as that of a solid sphere of mass $10\mathrm{kg}$ and radius R about an axis parallel to AB axis with separation of $3\mathrm{m}$ as shown in figure. The value of R is given as $\sqrt{\frac{\alpha}{2}}$ . The value of $\alpha$ is ______.

Answer: 15

Numerical answer — enter this value.

Step-by-step solution

$I_{rod} = \frac{ML^2}{3} = \frac{40\times9}{3}=120$ . For sphere: $I = \frac{2}{5}mR^2 + m d^2 = \frac{2}{5}\times10 R^2 + 10\times9 = 4R^2+90$ . Equate: $4R^2+90=120$ → $4R^2=30$ → $R^2=15/2$ → $R=\sqrt{15/2}$ → $\alpha=15$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 2 April, Evening Shift
Subject: Physics
Chapter: Rotational Dynamics
Topic: Moment of Inertia