Physics · Rotational Dynamics
JEE Main 2026 — 4 April, Morning Shift — Question 5
A solid sphere of mass M and radius R is divided into two unequal parts. The smaller part having mass M/8 is converted into a sphere of radius r and the larger part is converted into a circular disc of thickness t and radius 2R. If I₁ is moment of inertia of the sphere about its centre and I₂ is moment of inertia of the disc about its diameter, the ratio I₂/I₁ is
- Option A:
35
- Option B:Correct
70
- Option C:
140
- Option D:
210
Answer: B
Step-by-step solution
Mass density constant ⇒ r = R/2. I₁ = (2/5)(M/8)(R/2)² = MR²/80. I₂ = (1/4)(7M/8)(2R)² = (7M/8)×R² = 7MR²/8? Wait recalc: Disc mass = 7M/8, radius 2R, moment about diameter = (1/4)MR² → I₂ = (1/4)(7M/8)(4R²) = (7M/8)R². Then I₂/I₁ = (7/8)/(1/80) = 70. Correct
Answer key and solution verified before publishing.
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- Exam
- JEE Main 2026
- Subject
- Physics
- Chapter
- Rotational Dynamics
- Topic
- Moment of Inertia