$A$ homogeneous cubical brick lies motionless on a rough inclined surface. The half of the brick which applies greater pressure on the plane is:

$A$ block of mass $1 \ kg$ is pushed up a surface inclined to the horizontal at an angle of $60^{\circ}$ by a force of $10 \ N$ parallel to the inclined surface as shown in the figure. When the block is pushed up by $10 \ m$ along the inclined surface,the work done against the frictional force is: $\left[g=10 \ m/s^2, \mu_s = 0.1\right]$

$A$ block of mass $m$ is placed on a surface having a vertical cross-section given by $y = x^2 / 4$. If the coefficient of friction is $0.5$,the maximum height above the ground at which the block can be placed without slipping is:

$A$ block of mass $m$ slides down an inclined plane at an angle of $30^{\circ}$ with an acceleration of $\frac{g}{4}$. The value of the coefficient of kinetic friction will be:

Consider a block kept on an inclined plane (inclined at $45^{\circ}$) as shown in the figure. If the force required to just push it up the incline is $2$ times the force required to just prevent it from sliding down,the coefficient of friction between the block and inclined plane $(\mu)$ is equal to

$A$ cylindrical vessel filled with water is released on an inclined surface of angle $\theta$ as shown in the figure. The friction coefficient of the surface with the vessel is $\mu (< \tan \theta)$. Then the contact angle made by the surface of water with the incline will be