Question

A spherical iron ball $10$ cm in 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 at which the thickness of ice decreases, is

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

Answer 1

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

Let the thickness of the ice layer be $x$.

Then the total volume of the ice =Volume with ice - volume of the sphere
$V=\frac{4\pi }{3}\left[\right(10+x{)}^3-{10}^3]$

Differentiating with respect to time,we get
$\frac{dV}{dt}=4\pi \left[\right(10+x{)}^2].\frac{dx}{dt}$

Now at $x=5cm$, we get

$50c{m}^3/min=4\pi [15\times 15].\frac{dx}{dt}$
$\frac{1}{18\pi }cm/min=\frac{dx}{dt}$