As shown in the figure,a body of mass $m$ moving vertically downward with a speed of $3\, m/s$ hits a smooth fixed inclined plane and rebounds with a velocity $v_f$ in the horizontal direction. If the angle of the inclined plane is $30^{\circ}$,the velocity $v_f$ will be:

$A$ body is sliding down a rough inclined plane. The coefficient of friction between the body and the plane is $0.5$. The ratio of the net force required for the body to slide down and the normal reaction on the body is $1:2$. Then the angle of the inclined plane is (in $^{\circ}$)

If the coefficient of friction between the block of mass $m$ and the wedge is $\mu $,then the maximum horizontal acceleration of the wedge for which the block will remain at rest with respect to the wedge is

$A$ uniform chain of total length $L$ is at rest,partially on an incline of angle $\theta = 30^{\circ}$ and partially hanging vertically. The coefficient of friction between the chain and the incline is $\mu = \frac{1}{2\sqrt{3}}$. Find the ratio $\frac{L_{\max}}{L_{\min}}$,where $L_{\max}$ is the maximum length of the chain on the incline and $L_{\min}$ is the minimum length of the chain on the incline such that the chain remains at rest.

$A$ block of mass $8\, kg$ is at rest on a rough inclined plane as shown in the figure below. The magnitude of the net force exerted by the surface on the block will be ........ $N$ $(g = 10\, m/s^2)$.

What percentage of the total length of a chain can hang off the edge of a table without sliding,given that the coefficient of friction between the chain and the table is $0.25$?