A wheel with moment of inertia 10 kg `m^(2)` is rotating at `(2//pi)` rps on its axis. The frictional torque acting on it, if it makes 20 rotations befor stopping is

A wheel having moment of inertia 12 kg m^(2) about its axis, rotates at 50 rpm about this axis.Then the torque that can stop the wheel in one minute is in (SI units) is (pi)/(N) .Then the value of N is ?

A wheel of moment of inertia 30 kg m^(2) is rotating at 10 rotations per minute. The work done in increasing its speed to 5 times its initial value will be

A wheel of moment of inertia 30 kg m^(2) is rotating at 10 rotations per minute. The work done in increasing its speed to 5 times its initial value will be

A wheel having moment of inertia 2 kg m^(2) about its vertical axis, rotates at the rate of 60rpm about this axis. The torque which can stop the wheel's rotation in one minute would be

A wheel having moment of inertia 4 kg m^(2) about its axis, rotates at rate of 240 rpm about it. The torque which can stop the rotation of the wheel in one minute is :-

A wheel having moment of inertia 4 kg m^(2) about its axis, rotates at rate of 240 rpm about it. The torque which can stop the rotation of the wheel in one minute is :-

A wheel having moment of inertia 4 kg m^(2) about its axis, rotates at rate of 240 rpm about it. The torque which can stop the rotation of the wheel in one minute is :-

A wheel having moment of inertia 2 kg m^2 about its axis, rotates at 50 rpm about this axis. Find the torque that can stop the wheel in one minute.