Question

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

• A. $\frac{1}{36\pi }$
• B. $\frac{5}{6\pi }$
• C. $\frac{1}{18\pi }$
• D. $\frac{1}{9\pi }$

Answer 1

The correct option is C $\frac{1}{18\pi }$


Let the thickness of ice as $h\text{cm}$, then the volume of spherical ball is
$\frac{dV}{dt}=4\pi (10+h{)}^2\frac{dh}{dt}$
$-50=4\pi (10+5{)}^2\frac{dh}{dt}$
$\frac{dh}{dt}=\frac{-1}{18\pi }\frac{\text{cm}}{\text{min}}(-\text{signify the decrease in thickness})$