Physics · Rotational Dynamics

JEE Main 2025 — 7 April, Evening Shift — Question 66

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

Answer: 9

Numerical answer — enter this value.

Step-by-step solution

Mass of removed disc:

m = M\left(\frac{r^2}{R^2}\right)=\frac{M}{9}

Moment of inertia of full disc about axis $AB$ :

I_{\text{full}}=\frac{1}{2}MR^2

Distance of centre of removed disc from $O$ :

d=R-\frac{R}{3}=\frac{2R}{3}

MOI of removed disc about axis $AB$ :

I_{\text{removed}} =\frac{1}{2}m r^2 + m d^2 =\frac{1}{2}\cdot\frac{M}{9}\cdot\frac{R^2}{9} +\frac{M}{9}\cdot\frac{4R^2}{9} =\frac{MR^2}{162}+\frac{4MR^2}{81} =\frac{MR^2}{18}

MOI of remaining part:

I = I_{\text{full}}-I_{\text{removed}} =\frac{1}{2}MR^2-\frac{1}{18}MR^2 =\frac{4}{9}MR^2

Given:

I=\frac{4}{x}MR^2 \Rightarrow x=9

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 7 April, Evening Shift
Subject: Physics
Chapter: Rotational Dynamics
Topic: Moment of Inertia