Question

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 $50{\text{cm}}^3/\text{min}$. When the thickness of ice is $5\text{cm}$, then the rate (in $\text{cm/min}$) at which the thickness of ice decreases, is :

• A. $\frac{5}{6\pi }$
• B. $\frac{1}{54\pi }$
• C. $\frac{1}{36\pi }$
• D. $\frac{1}{18\pi }$

Answer 1

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

Let thickness of ice be $x\text{cm}$.
Therefore, net radius of sphere $=(10+x)\text{cm}$

Volume of sphere $V=\frac{4}{3}\pi (10+x{)}^3$

$\Rightarrow \frac{dV}{dt}=4\pi (10+x{)}^2\frac{dx}{dt}$
At $x=5,\frac{dV}{dt}=50{\text{cm}}^3/\text{min}$
$\Rightarrow 50=4\pi \times 225\times \frac{dx}{dt}$
$\Rightarrow \frac{dx}{dt}=\frac{1}{18\pi }\text{cm/min}$