A hollow cylinder of mass M and radius R is rotating about its axis of symmetry and a solid sphere of same mass and radius is rotating about an axis passing through its centre. If torques of equal magnitude are applied to them, then the ratio of angular accelerations produced is

A hollow cylinder of mass M and radius R is rotating about its axis of

1.	A hollow cylinder of mass M and radius R is rotating about its axis of symmetry and a solid sphere of same mass and radius is rotating about an axis passing through its centre. If torques of equal magnitude are applied to them, then the ratio of angular accelerations produced is
Answer» `(2)/(5)` `(5)/(2)` `(5)/(4)` `(4)/(5)` Solution :Moment of inertia of hollow cylinder about its AXIS of symmetry `I_(1) = MR^(2)` Moment of inertia of SOLID sphere about an axis passing through its centre`I_(2) = (2)/(5) MR^(2)` Let `alpha_(1)` and `alpha_(2)` be ANGULAR acceleration produced in the cylinder and the sphere respectively on applying same torque `tau` in each CASE . then `alpha_(1) = (tau)/(I_(1)) (As rau = I alpha) and alpha_(2) = (tau)/(I_(2))` Their corresponding ratio is `(alpha_(1))/(alpha_(2)) = (I_(2))/(I_(1)) = ((2)/(5) MR^(2))/(MR^(2)) = (2)/(5)`