A fly wheel of M.I $6 \times 10^{- 2}$ kgm² is rotating with an angular velocity of 20 rad s^-1.The torque required to bring it to rest 4s is

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1.6 Nm

0.6 Nm

0.8 Nm

0.3 Nm

answer is D.

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Detailed Solution

$\begin{array}{l} τ = Iα; α = \frac{ω - ω_{0}}{t}; τ = 6 \times 10^{- 2} \times \frac{ω}{t} \\ = 6 \times 10^{- 2} \times \frac{20}{4} = 0 .3 N - m \end{array}$

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