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 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 . The value of is ________.

Correct answer
0
Step-by-step solution
Without cavity 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}$$$