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Rotational Motion

JEE Main level, numerical answer, medium

34 of 280

Question 34 of 280

The moment of inertia of a semicircular ring about an axis, passing through the center and perpendicular to the plane of ring, is $\dfrac{1}{x}{MR}^{2}$ , where $R$ is the radius and $M$ is the mass of the semicircular ring. The value of $x$ will be $_______$ .

Show answer and solution

Answer: 1

The moment of inertia of a semicircular ring about a line perpendicular to the plane of the ring and passing through its centre is given as

I

=M{R}^{2}

, where

m

and

R

are the mass and radius of the ring. Comparing the given value
.

I={MR}^{2}=\dfrac{1}{x}M{R}^{2}

\Rightarrow x=1

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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More Physics practice

JEE Main Physics practice question on Rotational Motion (Moment of inertia of rigid bodies), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.

The moment of inertia of a semicircular ring about an axis, passing through the center and perpendicular to the plane of ring, is1xMR2\dfrac{1}{x}{MR}^{2}x1​MR2, where RRR is the radius and MMM is the mass of the semicircular ring. The value of xxx will be _______.

JEE Main question paper PDF

More Physics practice

The moment of inertia of a semicircular ring about an axis, passing through the center and perpendicular to the plane of ring, is $\dfrac{1}{x}{MR}^{2}$ , where $R$ is the radius and $M$ is the mass of the semicircular ring. The value of $x$ will be $_______$ .