Question

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

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

Answer 1

The correct option is B $\frac{1}{18\pi }cm/min$

Let the thickness of ice be $r$ and let $V$ be the volume of ice on the ball.
$V=\frac{4}{3}\pi (10+r{)}^3-\frac{4}{3}\pi .(10{)}^3$
Now, differentiating w.r.t. $t$, we get
$\frac{dV}{dt}=\frac{4}{3}\pi \times 3(10+r{)}^2\times \frac{dr}{dt}-0$
$\Rightarrow 50=\frac{4}{3}\pi \times 3(10+r{)}^2\times \frac{dr}{dt}$
When $r=5$
$\Rightarrow \frac{dr}{dt}=\frac{50}{4\pi \times 15\times 15}=\frac{1}{18\pi }cm/min$