$A$ small disc is placed on the top of a smooth hemisphere of radius $R$. What is the smallest horizontal velocity $v$ that should be given to the disc for it to leave the hemisphere immediately at the top and not slide down it? [There is no friction]

$A$ motorcycle driver doubles their velocity when taking a turn. The force exerted outwardly will be

$A$ stone of mass $2 \,kg$ is tied at one end of a string of length $2 \,m$ and is whirled in a horizontal circle. If the string can withstand a maximum tension of $64 \,N$, then the permissible maximum number of rotations per minute is

$A$ particle moves in a horizontal circle on the smooth inner surface of a hemispherical bowl of radius $R$. The plane of motion is at a depth $d$ below the centre of the hemisphere. The speed of the particle is:

$A$ string of length $L$ fixed at one end carries a body of mass $m$ at the other end. The mass is revolved in a circle in the horizontal plane about a vertical axis passing through the fixed end of the string. The string makes an angle $\theta$ with the vertical. The angular frequency of the body is $\omega$. The tension in the string is

$A$ particle describes a horizontal circle of radius $r$ in a conical funnel with a smooth inner surface with a speed of $v = 0.5 \text{ m/s}$. The height $h$ of the plane of the circle from the vertex of the funnel is (acceleration due to gravity, $g = 10 \text{ m/s}^2$): (in $\text{ cm}$)