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JEE Main 202528 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Morning Shift
Subject
Physics
Chapter
Rotational Motion

JEE Main 2025 · 28 January · Morning Shift · Physics · Rotational Motion

Two iron solid discs of negligible thickness have radii R_1 and R_2...

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 / x, where x=\mathrm{x}= ________ -Two iron solid discs of negligible thickness have radii R_1 and R_2 and moment of intertia I_1 and I_2 , respectively.

Correct answer

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Step-by-step solution

Two iron solid discs of negligible thickness have radii R_1 and R_2 and moment of intertia I_1 and I_2 , respectively.
Given R2=2R1R_2=2 R_1M1=σ×πR12=MoM2=σ×πR22=MoM2=σ×πR22=σ×π[2R1]2=σ×4πR12=4Mo\begin{aligned} & M_1=\sigma \times \pi R_1^2=M_o \\ & M_2=\sigma \times \pi R_2^2=M_o \\ & M_2=\sigma \times \pi R_2^2=\sigma \times \pi\left[2 R_1\right]^2=\sigma \times 4 \pi R_1^2=4 M_o \end{aligned}I1I2=M1R122M2R222=M1R12M2R22=14×14=116\frac{I_1}{I_2}=\frac{\frac{M_1 R_1^2}{2}}{\frac{M_2 R_2^2}{2}}=\frac{M_1 R_1^2}{M_2 R_2^2}=\frac{1}{4} \times \frac{1}{4}=\frac{1}{16}
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