JEE Main 2025 Apr 8 Shift 2, Physics Q2: Moment of Inertia

JEE Main2025Apr 8, Shift 2Physics

A rod of linear mass density $λ$ and length $L$ is bent to form a ring of radius $R$ . Moment of inertia of ring about any of its diameter is:

A
$\frac{λ L ^{3}}{16 π ^{2}}$
B
$\frac{λ L ^{3}}{12}$
C
$\frac{λ L ^{3}}{4 π ^{2}}$
D
$\frac{λ L ^{3}}{8 π ^{2}}$

Solution

Bending the rod into a ring gives circumference $L = 2 π R$ , so $R = \frac{L}{2 π}$ .

The ring's mass is $M = λ L$ .

For a thin ring, the moment of inertia about an axis through its centre perpendicular to the plane is $M R^{2}$ ; by the perpendicular axis theorem, the moment of inertia about any diameter is half of this: $I = \frac{M R ^{2}}{2}$ .

Substituting: $I = \frac{λ L}{2} \cdot (\frac{L}{2 π})^{2} = \frac{λ L ^{3}}{8 π ^{2}}$ , matching option (4).

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JEE Main 2025 Apr 8 Shift 2, Physics Q2: Moment of Inertia

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