$A$ ladder $5 \ m$ long rests against a vertical wall. If its top slides downwards at the rate of $10 \ cm/sec$,then the foot of the ladder is sliding at the rate of . . . . . . $m/sec$,when it is $4 \ m$ away from the wall.

$A$ particle moves in a straight line such that $s = \sqrt{t}$,then its acceleration is proportional to

The rate of change of the volume of a sphere with respect to its surface area,when the radius is $5 \ m$ is

$A$ tank in the shape of a rectangular parallelepiped has a volume of $27 \ m^3$. This tank is filled with water such that the rate of change of the water level is thrice the rate of change of the water quantity falling into the tank. The height of the tank (in meters) is:

$A$ water tank has the shape of an inverted right circular cone whose semi-vertical angle is $\tan^{-1}\left(\frac{1}{2}\right)$. Water is poured into it at a constant rate of $5 \text{ m}^3/\text{min}$. The rate at which the level of water is rising in $\text{m/min}$ at the instant when the depth of water in the tank is $10 \text{ m}$ is:

$A$ particle is moving in a straight line. Its displacement at time $t$ is given by $s = -4t^2 + 2t$. Then its velocity and acceleration at time $t = \frac{1}{2} \text{ s}$ are: