$A$ sphere increases its volume at the rate of $\pi \text{ cm}^3/\text{s}$. The rate at which its surface area increases when the radius is $1 \text{ cm}$ is

If the volume of a sphere is increasing at the rate of $12 \text{ cm}^3/\text{sec}$,then the rate (in $\text{cm}^2/\text{sec}$) at which its surface area is increasing,when the diameter of the sphere is $12 \text{ cm}$,is:

$A$ square plate is contracting at the uniform rate $3 \,cm^2 / sec$. The rate at which the perimeter is decreasing, when the side of the square is $15 \,cm$, is

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

Let $x$ be the length of each of the equal sides of an isosceles triangle and $\theta$ be the angle between these sides. If $x$ is increasing at the rate $\frac{1}{12} \text{ m/hour}$ and $\theta$ is increasing at the rate $\frac{\pi}{180} \text{ rad/hour}$,then the rate at which the area of the triangle is increasing when $x=12 \text{ m}$ and $\theta=\frac{\pi}{4}$ is:

The radius of a cylinder is increasing at a rate of $3 \text{ m/s}$ and its height is decreasing at a rate of $4 \text{ m/s}$. Find the rate of change of its volume when the radius is $4 \text{ m}$ and the height is $6 \text{ m}$.