JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Rotational Mechanics

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) $\frac{λ L ^{3}}{8 π ^{2}}$ ✓ Correct

Answer: (D)

Solution

Bending the rod into a ring makes its circumference equal to the rod length, so

L = 2 π R

and

R = \frac{L}{2 π}

. The mass of the ring is

M = λ L

. A ring about any diameter has

I = \frac{M R ^{2}}{2}

. Substituting,

I = \frac{λ L}{2} (\frac{L}{2 π})^{2} = \frac{λ L ^{3}}{8 π ^{2}}

. The moment of inertia of the ring about a diameter is

\frac{λ L ^{3}}{8 π ^{2}}

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Single correct option

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