$A$ body takes $4 \, s$ to reach the bottom of an inclined plane from the top. How much time (in $s$) will it take to cover one-fourth of the distance?

Two smooth inclined planes $A$ and $B$ each of height $20 \ m$ have angles of inclination $30^{\circ}$ and $60^{\circ}$ respectively. If $t_1$ and $t_2$ are respectively the times taken by two blocks to reach the bottom of the planes $A$ and $B$ from the top,then $t_1 - t_2 = $ (Acceleration due to gravity $= 10 \ m \ s^{-2}$)

Describe Galileo's experiment of an inclined plane regarding motion.

Two blocks of equal masses are tied to the ends of a light string. The string passes over a massless pulley fixed on a frictionless surface as shown in the figure. The acceleration of the centre of mass of the blocks is ($g$ = acceleration due to gravity).

$A$ football of radius $R$ is kept on a hole of radius $r$ (where the diameter of the hole is $2r$ and $r < R$) made on a plank kept horizontally. One end of the plank is now lifted so that it gets tilted,making an angle $\theta$ with the horizontal as shown in the figure. The maximum value of $\theta$ such that the football does not start rolling down the plank satisfies (figure is schematic and not drawn to scale) -

Two blocks each of mass $M$ are resting on a frictionless inclined plane as shown in the figure. Then: