$A$ body is released from a height of $30 \ m$ vertically downwards. The speed of the body at which potential energy is twice that of kinetic energy is (Acceleration due to gravity $= 10 \ m \ s^{-2}$)

$A$ particle of mass $m$ is moving along a circle of radius $R$ such that its tangential acceleration $a_t$ varies with distance covered $x$ as $a_t = \alpha x^2$, where $\alpha$ is a constant. The kinetic energy $K$ of the particle varies with the distance as $K = \beta x^c$, where $\beta$ and $c$ are constants. The values of $\beta$ and $c$ are:

The displacement-time graph of a particle moving in a straight line is as shown in the figure. Select the correct alternative.

$A$ particle of mass $20\,g$ is released with an initial velocity $5\,m/s$ along the curve from the point $A,$ as shown in the figure. The point $A$ is at height $h = 10\,m$ from point $B.$ The particle slides along the frictionless surface. When the particle reaches point $B,$ its angular momentum about $O$ will be ......... $kg \cdot m^2/s$. [Take $g = 10\,m/s^2$]

$A$ body of mass $0.25 \ kg$ travels along a straight line from $x=0$ to $x=2 \ m$ with a speed $v=k x^{3/2}$ where $k=2$ $SI$ units. The work done by the net force during this displacement is: (in $J$)

$A$ particle is released from height $S$ above the surface of the earth. At a certain height,its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively: