Rotational Motion

JEE Main level, single correct, medium

98 of 280

Question 98 of 280

A circular disc XX of radius RR is made from an iron plate of thickness tt, and another disc YY of radius 4R4 \mathrm{R} is made from an iron plate of thickness t4\frac{\mathrm{t}}{4}. Then the relation between the moment of inertia IX\mathrm{I}_{\mathrm{X}} and IY\mathrm{I}_{\mathrm{Y}} is

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Answer: (D) IY=64IX\mathrm{I}_{\mathrm{Y}}=64 \mathrm{I}_{\mathrm{X}}

I=12mR or M∝t∝R2\mathrm{I}=\frac{1}{2} \mathrm{mR} \quad \text { or } \quad \mathrm{M} \propto \mathrm{t} \propto \mathrm{R}^2
For disc⁡X,Ix=12(m)(R)2=12(πr2t).(R)2\operatorname{disc} X, I_x=\frac{1}{2}(m)(R)^2=\frac{1}{2}\left(\pi r^2 t\right) .(R)^2
For disc Y,Iy=12[π(4R)2⋅t/4][4R]2Y, I_y=\frac{1}{2}\left[\pi(4 R)^2 \cdot t / 4\right][4 R]^2
⇒IxIy=1(4)3⇒Iy=64Ix\Rightarrow \frac{I_x}{I_y}=\frac{1}{(4)^3} \Rightarrow I_y=64 I_x

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