Q97 / 180 · NEET 2019 (All India) · Physics

A solid cylinder of mass $2 k g$ and radius $4 c m$ is rotating about its axis at the rate of $3 rpm$ . The torque required to stop after $2 π$ revolutions is

Correct answer
A
$2 \times 10^{- 6} N m$
B
$2 \times 10^{- 3} N m$
C
$12 \times 10^{- 4} N m$
D
$2 \times 10^{6} N m$

Solution

From rotational kinematic equations,

ω_{f}^{2} - ω_{i}^{2} = 2 α θ

0^{2} - {(3 \frac{2 π}{60})}^{2} = 2 α 2 π (2 π)

α = - \frac{1}{800} r a d s^{- 2}

∴

Torque required

τ = |I α|

= \frac{2 {(4 \times 10^{- 2})}^{2}}{2} \times \frac{1}{800}

= \frac{16 \times 10^{- 4}}{800}

= 2 \times 10^{- 6} N m

Hence, the correct answer is option: 'A'.

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