If the volume of a spherical ball is increasing at the rate of $4 \pi \, cc/sec$,then find the rate of increase of its radius (in $cm/sec$),when the volume is $288 \pi \, cc$.

The displacement of a particle at time $t$ is $s = t^{3} - 4t^{2} - 5t$. The velocity of the particle at $t = 2 \text{ sec}$ is:

The diameter of one of the bases of a truncated cone is $100 \, mm$. If the diameter of this base is increased by $21 \%$ such that it still remains a truncated cone with the height and the other base unchanged,the volume also increases by $21 \%$. The radius of the other base (in $mm$) is

The volume of a sphere is increasing at the rate of $1200 \text{ cm}^3/\text{s}$. The rate of increase in its surface area when the radius is $10 \text{ cm}$ is: (in $\text{ cm}^2/\text{s}$)

If $pv = 81$,then $\frac{dp}{dv}$ at $v = 9$ is equal to

$A$ point moves along the arc of the parabola $y = 2x^2$. Its abscissa increases uniformly at the rate of $2 \text{ units/sec}$. At the instant the point is passing through $(1, 2)$,its distance from the origin is increasing at the rate of