JEE Advanced Model Paper 3 with solutions for online practice

A ring of moment of inertia 10−2kgm2 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

→

M

= (4

^

i

−3

^

j

) A - m2. When magnetic field

→

B

= (3

^

i

+4

^

j

) is switched on at t=0. Then the angular acceleration of ring at t = 0 is 5 × 10n ad/ s2 then the value of n is