NEET 2026 — Physics solution

NEET2026Physics

A thin wire of length ‘L’ and linear mass density ‘m’ is bent into a circular ring (in x-y plane) with centre ‘C’ as shown in figure. The moment of inertia of the ring about an axis yy’ will be:

A
$\frac{3 m L ^{3}}{8 π}$
B
$\frac{3 m L ^{2}}{8 π ^{2}}$
C
$\frac{3 m L ^{3}}{8 π ^{2}}$
D
$\frac{3 m L ^{2}}{8 π}$

Solution

Total mass $M = m L$ and $L = 2 π R$ , so $R = \frac{L}{2 π}$ .

Moment of inertia about a tangent (axis yy’ in the plane of the ring, tangent to it) is

$I_{y y^{'}} = \frac{M R ^{2}}{2} + M R^{2} = \frac{3}{2} M R^{2}$

$= \frac{3}{2} (m L) (\frac{L}{2 π})^{2} = \frac{3 m L ^{3}}{8 π ^{2}}$ .

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