The rate of change of the volume of the sphere with respect to its surface area $S$ is . . . . . . .

The rate of change of the surface area of a sphere of radius $r$ when the radius is increasing at the rate of $2 \text{ cm/sec}$ is proportional to

$A$ spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is $1 \ cm$, the ice-cream melts at the rate of $81 \ cm^3/min$ and the thickness of the ice-cream layer decreases at the rate of $\frac{1}{4\pi} \ cm/min$. The surface area (in $cm^2$) of the chocolate ball (without the ice-cream layer) is: (in $\pi$)

Displacement $s$ of a particle at time $t$ is expressed as $s=2 t^3-9 t$. Find the acceleration at the time when the velocity vanishes.

If the distance $s$ metres traversed by a particle in $t$ seconds is given by $s = t^3 - 3t^2$,then the velocity of the particle when the acceleration is zero,in $m/s$ is:

If the surface area of a sphere of radius $r$ is increasing uniformly at the rate $8 \, cm^2/s$,then the rate of change of its volume is