$A$ simple pendulum is released from point $A$ as shown in the figure. If $m$ and $l$ represent the mass of the bob and the length of the pendulum respectively,the gain in kinetic energy at point $B$ is:

$A$ body falls freely under gravity. Its speed is $v$ when it has lost an amount $U$ of the gravitational potential energy. Then its mass is

$A$ bullet of mass $10 \; g$ leaves a rifle at an initial velocity of $1000 \; m/s$ and strikes the earth at the same level with a velocity of $500 \; m/s$. The work done in joule $(J)$ in overcoming the resistance of air will be:

$A$ 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:

$A$ ball at rest is dropped from a height of $12 \,m$. It loses $25 \%$ of its kinetic energy on striking the ground and bounces back to a height '$h$'. Then the value of '$h$' is (in $\,m$)

$A$ body falls freely from a height $h$ in a gravitational field. Match the following:

Column-$I$	Column-$II$
$(1)$ Potential Energy $=$ Kinetic Energy	$(a)$ At height $\frac{2h}{3}$
$(2)$ Potential Energy $= 2 \times$ Kinetic Energy	$(b)$ Constant at every point
	$(c)$ At height $\frac{h}{2}$