$A$ ladder of $5 \ m$ long rests against a vertical wall with the lower end on the horizontal ground. The lower end of the ladder is pulled along the ground away from the wall at the rate of $3 \ m/sec$. The height of the upper end (in meters) while it is descending at the rate of $4 \ m/sec$,is

$A$ water tank has the shape of a right circular cone with its axis vertical and vertex downwards. Its semi-vertical angle is $\tan^{-1} \frac{3}{4}$. Water is poured into it at a constant rate of $6 \text{ m}^3/\text{hr}$. The rate (in $\text{m}^2/\text{hr}$) at which the wet curved surface area of the tank is increasing when the depth of water in the tank is $4 \text{ m}$ is:

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

The rate of change of the area of a circle with respect to its radius at $r = 3 \text{ cm}$ is . . . . . . $\text{cm}^2/\text{cm}$. (in $\pi$)

The displacement of a particle at time $t$ is given by $x = At^2 + Bt + C$,where $A, B$ and $C$ are constants. If $v$ is the velocity,then $4Ax - v^2 = ....$

If the radius of a circular blot of oil is increasing at the rate of $2 \text{ cm/min}$,then the rate of change of its area when its radius is $3 \text{ cm}$ is: