A ring of moment of inertia $10^{- 2} k g m^{2}$ about an axis passing through its centre and perpendicular to its plane is placed with its centre at origin. The current is passed through ring so that magnetic moment is $\vec{M} = (4 \hat{i} - 3 \hat{j}) A - m^{2}$ . When magnetic field $\vec{B} = (3 \hat{i} + 4 \hat{j})$ is switched on at $t = 0$ ,

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Angular acceleration of the ring at $t = 0$ is 5000 $r a d / s^{2}$

Maximum angular velocity of the ring is 200 rad/sec

Maximum angular velocity of the ring is 100 rad/sec

Angular acceleration of the ring at $t = 0$ is 3000 ${rad/s}^{2}$

answer is A, C.

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Detailed Solution

$\vec{τ} = \vec{M} \times \vec{B} = (4 \hat{i} - 3 \hat{j}) \times (3 \hat{i} + 4 \hat{j}) = 25 \hat{k} (N - m)$

Moment of inertia about an axis passing through centre of mass and parallel to $\vec{τ}$ ,

$I = \frac{10^{- 2}}{2} ({kg.m}^{2}) α = \frac{τ}{I} = \frac{25}{0.5 \times 10^{- 2}} = 5000 r a d / \sec^{2} \frac{1}{2} I ω^{2} = U_{i} - U_{f} = - M B \cos 90^{0} - (- M B \cos 0^{0}) ω = \sqrt{\frac{2 M B}{I}} = {(\frac{2 (5) (5)}{(0.5) 10^{- 2}})}^{1 / 2} = 100 rad/sec$