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Rotational Motion question

2025 · 8 Apr · Shift 2 · Q61

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Rotational Motion
2025 · 8 Apr · Shift 2 · Q61

Rotational Motion question

2025 · 8 Apr · Shift 2 · Q61

JEE MainPhysicsRotational MotionMCQ+4 / −1

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
$\frac{\lambda L^3}{8 \pi^2}$
B
$\frac{\lambda L^3}{16 \pi^2}$
C
$\frac{\lambda L^3}{4 \pi^2}$
D
$\frac{\lambda L^3}{12}$

View written solutionFree

Correct answer: A

Find the total mass of the rod/ring

The rod has linear mass density $\lambda$ and length $L$ . So its total mass is

M = \lambda L

Relate the length of the rod to the radius of the ring

When the rod is bent into a ring, its length becomes the circumference:

L = 2\pi R

Hence,

R = \frac{L}{2\pi}

Moment of inertia of a ring about a diameter

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

I_z = MR^2

By the perpendicular axis theorem, for two perpendicular diameters in the plane,

I_x + I_y = I_z

Since the ring is symmetric,

I_x = I_y

Therefore,

2I_{\text{diameter}} = MR^2

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

Substitute $M$ and $R$

I = \frac{1}{2}(\lambda L)\left(\frac{L}{2\pi}\right)^2

I = \frac{1}{2}(\lambda L)\cdot \frac{L^2}{4\pi^2}

I = \frac{\lambda L^3}{8\pi^2}

Match with the options

\boxed{I = \frac{\lambda L^3}{8\pi^2}}

So the correct option is A.

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