A disc of mass m and radius R is released from rest.If it takes `t_1` sec to slide down the smooth plane and `t_2` sec is the time taken to roll down the rough plane, such that `t_2/t_1 = sqrt3/x`. Find x

A circular disc reaches from top to bottom of an inclined plane of length 'L'. When it slips down the plane, it takes time 't_(1)' . When it rolls down the plane, it takes time t_(2) . The value of (t_(2))/(t_(1))" is "sqrt((3)/(x)) . The value of x will be____.

A ring takes time t_(1) in slipping down an inclined plane of length L, whereas it takes time t_(2) in rolling down the same plane. The ratio of t_(1) and t_(2) is -

A body sliding down on a smooth inclined plane slides down 1//4th distane in 2s. I will slide down the complete plane in

A block released from rest from the top of a smooth inclined plane of angle theta_1 reaches the bottom in time t_1 . The same block released from rest from the top of another smooth inclined plane of angle theta_2 reaches the bottom in time t_2 If the two inclined planes have the same height, the relation between t_1 and t_2 is

A body is released from the top of a tower of height h. It takes t sec to reach the ground. Where will be the ball after time (t)/(2) sec?