$A$ block of mass $m$ is placed on a surface with a vertical cross-section given by $y = \frac{x^3}{6}$. If the coefficient of friction is $0.5$,the maximum height above the ground at which the block can be placed without slipping is:

The minimum force required to start pushing a body up a rough (frictional coefficient $\mu$) inclined plane is $F_{1}$,while the minimum force needed to prevent it from sliding down is $F_{2}$. If the inclined plane makes an angle $\theta$ with the horizontal such that $\tan \theta = 2\mu$,then the ratio $\frac{F_{1}}{F_{2}}$ is:

When an object is shot from the bottom of a long smooth inclined plane kept at an angle $60^{\circ}$ with the horizontal,it can travel a distance $x_{1}$ along the plane. But when the inclination is decreased to $30^{\circ}$ and the same object is shot with the same velocity,it can travel $x_{2}$ distance. Then $x_{1}: x_{2}$ will be

$A$ block is at rest on an inclined plane making an angle $\alpha$ with the horizontal. As the angle $\alpha$ of the incline is increased,the block starts slipping when the angle of inclination becomes $\theta$. The coefficient of static friction between the block and the surface of the inclined plane is:

$A$ chain of length $L$ rests on a rough table. If $\mu$ is the coefficient of friction,the maximum fraction of the chain that can hang over the table will be:

$A$ homogeneous rectangular brick lies on a rough inclined plane of very small inclination. Which half part of the brick exerts a greater contact force on the plane?