Rigid Body Dynamics | JEE Main Solution

JEE MAIN 2025

22-01-2025 SHIFT-1

Question

A uniform circular disc of radius 'R' and mass 'M' is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius R/2 is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above.

Select the correct option:

$\frac{9}{32} \mathrm{MR}^2$

$\frac{17}{32} \mathrm{MR}^2$

$\frac{13}{32} \mathrm{MR}^2$

$\frac{7}{32} \mathrm{MR}^2$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\begin{aligned} & I=\frac{M R^2}{2}-\left[\frac{\frac{M}{4}\left(\frac{R}{2}\right)^2}{2}+\frac{M}{4}\left(\frac{R}{2}\right)^2\right] \\ & I=\frac{13}{32} M R^2\end{aligned}$

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