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

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Rotational Mechanics

The moment of inertia of semicircular ring about an axis, passing through the center and perpendicular to the plane of ring, is

\frac{1}{x} MR^{2}

, where R is the radius and M is the mass of semicircular ring. The value of x will be

Answer: 1

Solution

For a semicircular ring about an axis through its center perpendicular to its plane, the moment of inertia is

I = M R^{2}

(same as a full circular ring, since all mass elements are at distance R from the axis). Therefore

\frac{1}{x} M R^{2} = M R^{2}

, giving

x = 1

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Integer answer

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