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JEE Main 20257 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
7 April
Shift
Evening Shift
Subject
Physics
Chapter
Rotational Motion

JEE Main 2025 · 7 April · Evening Shift · Physics · Rotational Motion

M and R be the mass and radius of a disc. A small disc of radius R...

M and R be the mass and radius of a disc. A small disc of radius R/3R / 3 is removed from the bigger disc as shown in figure. The moment of inertia of remaining part of bigger disc about an axis AB passing through the centre O and perpendicular to the plane of disc is 4xMR2\frac{4}{x} M R^2. The value of xx is ________.
M and R be the mass and radius of a disc. A small disc of radius R / 3 is removed from the bigger disc as shown in

Correct answer

0

Step-by-step solution

Without cavity I1=MR22\mathrm{I}_1=\frac{\mathrm{MR}^2}{2} Mass of  removed disc =MπR2×(R3)2π=(M9)\begin{aligned}\text{ Mass\text{ of } removed disc }= & \frac{\mathrm{M}}{\pi \mathrm{R}^2} \times\left(\frac{\mathrm{R}}{3}\right)^2 \pi \\ & =\left(\frac{\mathrm{M}}{9}\right)\end{aligned}M.I. of removed disc $I_2=\frac{\frac{M}{9}\left(\frac{R}{3}\right)^2}{2}+\frac{M}{9} \times\left(\frac{2 R}{3}\right)^2\begin{aligned}& =\frac{\mathrm{MR}^2}{18} \\ & \mathrm{I}=\mathrm{I}_1-\mathrm{I}_2=\frac{\mathrm{MR}^2}{2}-\frac{\mathrm{MR}^2}{18}=\frac{4 \mathrm{MR}^2}{9} \\ & (\mathrm{n}=9)\end{aligned}$$$
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