Physics · Rotational Dynamics

JEE Main 2025 — 28 January, Morning Shift — Question 70

Two iron solid discs of negligible thickness have radii R1R_{1} and R2R_{2} and

moment of intertia I1I_{1} and I2I_{2}, respectively. For R2=2R1R_{2}=2 R_{1}, the

ratio of I1I_{1} and I2I_{2} would be 1/x1 / \mathrm{x}, where x=\mathrm{x}= \qquad

Answer: 16

Numerical answer — enter this value.

Step-by-step solution

Given R2=2R1\mathrm{R}_{2}=2 \mathrm{R}_{1}

M1=σ×πR12=MoM_{1}=\sigma \times \pi R_{1}^{2}=M_{o}

M2=σ×πR22=Mo\mathrm{M}_{2}=\sigma \times \pi \mathrm{R}_{2}^{2}=\mathrm{M}_{\mathrm{o}}

M2=σ×πR22=σ×π[2R1]2=σ×4πR12=4Mo\mathrm{M}_{2}=\sigma \times \pi \mathrm{R}_{2}^{2}=\sigma \times \pi\left[2 \mathrm{R}_{1}\right]^{2}=\sigma \times 4 \pi \mathrm{R}_{1}^{2}=4 \mathrm{M}_{\mathrm{o}}

I1I2=M1R122M2R222=M1R12M2R22=14×14=116\frac{I_{1}}{I_{2}}=\frac{\frac{\mathrm{M}_{1} \mathrm{R}_{1}^{2}}{2}}{\frac{\mathrm{M}_{2} \mathrm{R}_{2}^{2}}{2}}=\frac{\mathrm{M}_{1} \mathrm{R}_{1}^{2}}{\mathrm{M}_{2} \mathrm{R}_{2}^{2}}=\frac{1}{4} \times \frac{1}{4}=\frac{1}{16}

IMAGES

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Physics
Chapter
Rotational Dynamics
Topic
Moment of Inertia