JEE Main 2026 — Apr 4, Shift 2 — Physics solution

JEE Main2026Apr 4, Shift 2Physics

A uniform disc of radius R and mass M is free to oscillate about the axis A as shown in the figure. For small oscillations the time period is ________. ( $g$ is acceleration due to gravity)

A
$2 π \frac{5 R}{4 g}$
B
$2 π \frac{2 R}{3 g}$
C
$2 π \frac{3 R}{2 g}$
D
$2 π \frac{3 R}{g}$

Solution

This is a physical pendulum pivoted at the rim. Distance from pivot to centre of mass is $R$ .

Moment of inertia about A (parallel-axis theorem):

$I_{A} = \frac{M R ^{2}}{2} \cdot \frac{1}{2} + M R^{2}$ . Correcting: disc moment about its diameter is $\frac{M R ^{2}}{4}$ (axis A lies along the rim, tangent to the disc in its plane).

$I_{A} = \frac{M R ^{2}}{4} + M R^{2} = \frac{5 M R ^{2}}{4}$ .

Restoring torque for angular displacement $θ$ : $τ = - M g R sin θ \approx - M g R θ$ .

$α = \frac{τ}{I _{A}} = - \frac{M g R}{\frac{5 M R ^{2}}{4}} θ = - \frac{4 g}{5 R} θ$ .

$ω = \frac{4 g}{5 R}$ , so $T = 2 π \frac{5 R}{4 g}$ .

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