JEE Main 2026 Apr 6 Shift 1, Physics Q6: Law of Rotation And Work-Energy Theorem

JEE Main2026Apr 6, Shift 1Physics

A solid sphere of radius $4$ cm and mass $5$ kg is rotating (rotation axis is passing through the centre of the sphere) with an angular velocity of $1200$ rpm. It is brought to rest in $10$ s by applying a constant torque. The torque applied and the number of rotations it made before it comes to rest are _____ and ______ respectively.

A
$0.128 π$ Nm, $100$
B
$0.0128 π$ Nm, $50$
C
$0.128 π$ Nm, $50$
D
$0.0128 π$ Nm, $100$

Solution

Convert: $ω_{i} = 1200$ rpm $= 1200 \cdot \frac{2 π}{60} = 40 π$ rad/s. Final $ω_{f} = 0$ in $t = 10$ s.

Angular deceleration: $α = \frac{ω _{f} - ω _{i}}{t} = - 4 π$ rad/s $^{2}$ .

Moment of inertia of solid sphere: $I = \frac{2}{5} M R^{2} = \frac{2}{5} (5) (0.04)^{2} = 3.2 \times 1 0^{- 3}$ kg m $^{2}$ .

Torque magnitude: $τ = I ∣ α ∣ = 3.2 \times 1 0^{- 3} \cdot 4 π = 0.0128 π$ Nm.

Number of revolutions: $ω_{f}^{2} = ω_{i}^{2} - 2∣ α ∣ θ \Rightarrow θ = \frac{( 40 π ) ^{2}}{2 ( 4 π )} = 200 π$ rad $= \frac{200 π}{2 π} = 100$ revolutions.

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JEE Main 2026 Apr 6 Shift 1, Physics Q6: Law of Rotation And Work-Energy Theorem

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