A solid cylinder and a solid sphere, having same mass $$M$$ and radius $$R$$, roll down the same inclined plane from top without slipping. They start from rest. The ratio of velocity of the solid cylinder to that of the solid sphere, with which they reach the ground, will be :

A spherical shell of 1 kg mass and radius R is rolling with angular speed $$\omega$$ on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin O is $${a \over 3}$$ R^{2}$$\omega$$. The value of a will be :

A ball is spun with angular acceleration $$\alpha$$ = 6t^{2} $$-$$ 2t where t is in second and $$\alpha$$ is in rads^{$$-$$2}. At t = 0, the ball has angular velocity of 10 rads^{$$-$$1} and angular position of 4 rad. The most appropriate expression for the angular position of the ball is :

A $$\sqrt {34} $$ m long ladder weighing 10 kg leans on a frictionless wall. Its feet rest on the floor 3 m away from the wall as shown in the figure. If E_{f} and F_{w} are the reaction forces of the floor and the wall, then ratio of $${F_w}/{F_f}$$ will be :

(Use g = 10 m/s^{2}.)