In the figure , mass of a ball is `9/5` times mass of the rod, Length of rod is 1m.The level of ball is same as rod level.Find out time taken by the ball to reach at upper end of rod.

In the arrangement shown in figure the mass of the ball is eta times as that of the rod. The length of the rod is L the masses of the pulleys and the threads as well as the friction, are negligible. The ball is set on the same level as the lower end of the rod and then released. How soon will the ball be opposite the upper end of the rod?

In the arrangement shown in figure the mass of the ball is eta times as that of the rod. The length of the rod is L the masses of the pulleys and the threats as well as the friction, are negligible. The ball is set on the same level as the lower and of the rod and then released. How soon will the ball be opposite the upper and of the rod?

if a mass of ball is m/2 and that of a rod is m the length of the rod is 20 cm the ball is set on the same level as the lower end of the rod and then released. If pulleys and strings are light, friction negligible how soon will the ball be opposite top the upper end of the rod?

In the pulley system shown in the figure, the mass of A is half of that of rod B. The rod length is 500 cm. The mass of pulleys and the threads may be neglected The mass A is set at the same level as the lower end of the rod and then released After releasing the mass A, it would reach the top end of the rod B in time (Assume, g = 10 m//s^2 )

In the figure, a ball of mass m is tied with two strings of equal as shown. If the rod is rotated with angular velocity omega_(1) when

In the figure, a ball of mass m is tied with two strings of equal as shown. If the rod is rotated with angular velocity omega_(1) when