If the surface area of a spherical bubble is increasing at the rate of $4 \text{ cm}^2/\text{sec}$,then the rate of change in its volume (in $\text{cm}^3/\text{sec}$) when its radius is $8 \text{ cm}$ is:

The distance covered by a particle in $t$ seconds is given by $x = 3 + 8t - 4t^2$. After $1$ second, the velocity will be:

By dropping a stone in a quiet lake,a wave in the form of a circle is generated. The radius of the circular wave increases at the rate of $2.1 \text{ cm/sec}$. Then the rate of increase of the enclosed circular region,when the radius of the circular wave is $10 \text{ cm}$,is (Given $\pi = \frac{22}{7}$)

$A$ particle moves in a straight line in such a way that its velocity at any point is given by $v^2 = 2 - 3x$,where $x$ is measured from a fixed point. The acceleration is

$A$ point is moving on the curve $y=x^3-3x^2+2x-1$ and the $y$-coordinate of the point is increasing at the rate of $6 \text{ units/sec}$. When the point is at $(2, -1)$,the rate of change of the $x$-coordinate of the point is:

$A$ spherical iron ball of $10 \ cm$ radius is coated with a layer of ice of uniform thickness that melts at a rate of $50 \ cm^3/min$. When the thickness of ice is $5 \ cm$,then the rate (in $cm/min$) at which the thickness of ice decreases is: