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A non conducting ring of mass m and radius R has a charge Q uniformly distributed over its circumference. The ring is placed on a rough horizontal surface such that the plane of the ring is parallel to the surface. A vertical magnetic field $B = B_{0} t^{2}$ tesla perpendicular to plane of ring is switched on. After 2 second from switching on the magnetic field the ring is just about to rotate about vertical axis through its centre and perpendicular to its plane. Then

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the induced electric field is quadratic in time t

the torque on the ring due to magnetic field is $B_{0} Q R^{2} t$

until 2 seconds, the friction force does not come into play

the friction coefficient between the ring and the surface is $\frac{2 B_{0} Q R}{m g}$

answer is B, D.

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Detailed Solution

$\oint \vec{E} . \vec{d l} = \frac{d ϕ}{d t} = A \frac{d B}{d t} (∵ N = 1 a n d \cos θ = 1)$

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