$A$ cone of radius $r$ and height $h$ rests on a rough horizontal surface,the coefficient of friction between the cone and the surface being $\mu$. $A$ gradually increasing horizontal force $F$ is applied to the vertex of the cone. The largest value of $\mu$ for which the cone will slide before it topples is

$A$ block is projected up an inclined plane of angle $\theta = 30^o$ with an initial velocity of $5 \, m/s$. It comes to rest in $0.5 \, s$. What is the coefficient of friction?

An inclined plane is bent in such a way that the vertical cross-section is given by $y = \frac{x^2}{4}$,where $y$ is in the vertical direction and $x$ is in the horizontal direction. If the upper surface of this curved plane is rough with a coefficient of friction $\mu = 0.5$,the maximum height in $cm$ at which a stationary block will not slip downward is............$cm$.

$A$ small block starts slipping down from a point $B$ on an inclined plane $AB$,which is making an angle $\theta$ with the horizontal. The section $BC$ is smooth and the remaining section $CA$ is rough with a coefficient of friction $\mu$. It is found that the block comes to rest as it reaches the bottom (point $A$) of the inclined plane. If $BC = 2AC$,the coefficient of friction is given by $\mu = k \tan \theta$. The value of $k$ is

There are two inclined surfaces of equal length $(L)$ and the same angle of inclination $45^{\circ}$ with the horizontal. One of them is rough and the other is perfectly smooth. $A$ given body takes $2$ times as much time to slide down the rough surface than on the smooth surface. The coefficient of kinetic friction $(\mu_k)$ between the object and the rough surface is close to:

The minimum force required to move a body up an inclined plane is three times the minimum force required to prevent it from sliding down the plane. If the coefficient of friction between the body and the inclined plane is $\frac{1}{2 \sqrt{3}}$,then the angle of the inclined plane is (in $^{\circ}$)