If the surface area of a spherical balloon of radius $6 \text{ cm}$ is increasing at the rate $2 \text{ cm}^2/\text{sec}$,then the rate of increase in its volume in $\text{cm}^3/\text{sec}$ is

$A$ swimming pool is to be drained for cleaning. If $L$ represents the number of litres of water in the pool $t$ seconds after the pool has been plugged off to drain and $L=200(10-t)^{2}$,how fast is the water running out at the end of $5$ seconds? What is the average rate at which the water flows out during the first $5$ seconds?

Moving along the $x$-axis are two points with $x = 10 + 6t$ and $x = 3 + t^2$. The speed with which they are receding from each other at the time of encounter is ........... $cm/sec$. ($x$ is in $cm$ and $t$ is in seconds)

$A$ spherical balloon is being inflated at the rate of $35 \, cm^3/min$. The rate of increase in the surface area (in $cm^2/min$) of the balloon when its diameter is $14 \, cm$,is

$A$ particle is moving along a line according to the law $S = t^3 - 3t^2 + 4t - 2$,where $S$ is measured in meters and $t$ is measured in seconds. Then the velocity (in $m/s$) of the particle when its acceleration is zero is:

If the law of motion in a straight line is $s = \frac{1}{2}vt$,then acceleration is