Question

A spherical iron ball of $10\, cm$ 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 (in cm/min.) at which of 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

Answer: (D)

Let thickness of ice be $'h'.$
Vol. of ice $=v=\frac{4\pi}{3}\left(\left(10+h\right)^{3}-10^{3}\right)$
$\frac{dv}{dt}=\frac{4\pi}{3}\left(3\left(10+h\right)^{2}\right). \frac{dh}{dt}$
Given $\frac{dv}{dt}=50cm^{3}/ min$ and $h=5cm$
$\Rightarrow 50=\frac{4\pi}{3}\left(3\left(10+5\right)^{2}\right) \frac{dh}{dt}$
$\Rightarrow \frac{dh}{dt}=\frac{50}{4\pi\times15^{2}}=\frac{1}{18\pi} cm/ min$