Question

A spherical iron ball $10$ cm in radius is coated with a layer of ice of uniform thickness
than melts at a rate of $50c{m}^3/min$. When the thickness of ice is $5$ cm, then the rate at
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. $\frac{5}{6\pi }cm/min$

Answer 1

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

Let inner radius of ice layer = $R_i$ and outer radius = $R_o$.

Therefore thickness = $R_o-R_i$.

During melting, only the outer radius decreases whereas the inner radius is unchanged. So the thickness changes only due to the change in outer radius.

$\therefore \frac{d}{dt}(R_o-R_i)=\frac{d{R}_o}{dt}$.

Volume of ice = $V=\frac{4\pi }{3}(R_o^3-R_i^3)\Rightarrow \frac{dV}{dt}=4\pi R_o^2\frac{d{R}_o}{dt}=50㎤/min$

When thickness of layer is $5cm$, $R_o=15cm$.

So, $\frac{d{R}_o}{dt}=\frac{1}{4\pi R_o^2}\frac{dV}{dt}=\frac{50}{900\pi }=\frac{1}{18\pi }cm/min$.

So, rate of decrease of thickness is $\frac{1}{18\pi }cm/min$.