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A circular disc of moment of inertia $3.5 k g m^{2}$ is rotating with angular speed 60 rad/s about an axis passing Through it’s centre and $⊥^{r e}$ to it’s plane. The torque required to stop the disc in 5 Seconds is

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84 N-m

40 N-m

10.5 N-m

42 N-m

answer is D.

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Detailed Solution

$I = 3.5 K g m^{2}$
W= 60 rad/s
T=5sec
$τ = I (\frac{W}{t})$

$τ = 3.5 \times \frac{60}{5} = 42 N - m$

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