Two blocks $A$ and $B$ are released from the top of a rough inclined plane so that $A$ slides along the plane and $B$ falls down freely. Which will have higher velocity on reaching the ground?

$A$ particle is placed at rest inside a hollow hemisphere of radius $R$. The coefficient of friction between the particle and the hemisphere is $\mu = \frac{1}{\sqrt{3}}$. The maximum height up to which the particle can remain stationary is

$A$ block of mass $m$ is stationary on a rough plane of mass $M$ inclined at an angle $\theta$ to the horizontal,while the whole setup is accelerating upwards at an acceleration $a$. If the coefficient of friction between the block and the plane is $\mu$,then the force that the plane exerts on the block is

$A$ block of $10 \, kg$ mass is placed on a rough inclined surface as shown in the figure. The acceleration of the block will be

$A$ block of mass $2 \, kg$ is placed on a rough inclined plane as shown in the figure $(\mu = 0.2)$ so that it just touches the spring. The block is allowed to move downwards. The spring will be compressed to a maximum of .............. $cm$.

The time taken by an object to slide down a $45^{\circ}$ rough inclined plane is $n$ times the time it takes to slide down a perfectly smooth $45^{\circ}$ inclined plane. The coefficient of kinetic friction between the object and the inclined plane is: