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A ring type flywheel of mass 200 kg and diameter 2m is rotating at the rate of $300 / π$ revolution per minute then,
List – I List – II
(a) Moment of inertia of the flywheel is ---- $kg . m^{2}$ (p) 10
(b) The kinetic energy of rotation of the flywheel is ----kJ (q) 200
(c) The flywheel if subjected to a rotating torque of 200 N. m will come to rest in time t =…sec (r) 5
(d) Initial angular velocity is --- $rad / s$ (s) 15
(t) 50

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$(a) \to (r), (b) \to (p), (c) \to (s), (d) \to (q)$

$(a) \to (q), (b) \to (s), (c) \to (p), (d) \to (r)$

$(a) \to (p), (b) \to (q), (c) \to (s), (d) \to (r)$

$(a) \to (q), (b) \to (p), (c) \to (p), (d) \to (p)$

answer is D.

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Detailed Solution

Moment of inertia is

$I = M R^{2} = (200) {(1)}^{2} = 200 {kg.m}^{2}$

Kinetic energy is

$\begin{matrix} K = \frac{1}{2} I ω^{2} \\ = \frac{1}{2} (200) {(\frac{300}{π} (\frac{2 π}{60}))}^{2} \\ = 10000 J = 10 kJ \end{matrix}$

$τ = I α, α = \frac{τ}{I} = \frac{- 200}{200} = - 1 {rad/s}^{2}$

$t = \frac{ω - ω_{0}}{α} = \frac{0 - (\frac{300}{π} (\frac{2 π}{60}))}{- 1} = 10 \sec$

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