A fly wheel of moment of inertia `4xx10^(-3)kgm^(2)` is makign 10 revolution per second. The torque required to stop it in 5 second is:

A wheel of moment of inertia 5xx10^(-3) kg m^(2) is making 20 revolutions per second. If it is stopped in 20 s, then its angular retardation would be

A wheel has moment of inertia 5 xx 10^(-3) kg m^(2) and is making 20 "rev" s^(-1) . The torque needed to stop it in 10 s is….. xx 10^(-2) N-m

A wheel has moment of inertia 5 xx 10^(-3) kg m^(2) and is making 20 "rev" s^(-1) . The torque needed to stop it in 10 s is….. xx 10^(-2) N-m

A wheel of moment of inertia 2 xx 10^3 kg m^2 is rotating at uniform angular speed of 4 rad/s,.then . the torque required to stop it in one second is,

A wheel has a moment of inertia of 5xx10^(-3) kg m"^(2) and is making "20 rev/s". What is the magnitude of the torque required to stop it in 10 s?

A fly wheel of moment of inertia I is rotating at n revolutions per sec. The work needed to double the frequency would be -