JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Rotational Mechanics

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Consider a uniform wire of mass M and length L . It is bent into a semi-circle. Its moment of inertia about a line perpendicular to the plane of the wire passing through centre is :

(A) $\frac{M L ^{2}}{π ^{2}}$ ✓ Correct
(B) $\frac{2 M L ^{2}}{5 π ^{2}}$
(C) $\frac{M L ^{2}}{2 π ^{2}}$
(D) $\frac{M L ^{2}}{4 π ^{2}}$

Answer: (A)

Solution

A wire of length

L

bent into a semicircle has

π R = L

, so its radius is

R = \frac{L}{π}

. Every element of the wire lies at the same distance

R

from the centre, so about an axis through the centre perpendicular to the plane of the wire

I = \int R^{2} d m = M R^{2} = M (\frac{L}{π})^{2}

The moment of inertia is

\frac{M L ^{2}}{π ^{2}}

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Single correct option

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