Question

A spherical iron ball of $10cm$ 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 (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 ${}^{\prime }{h}^{\prime }.$
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 $\frac{dv}{dt}=50c{m}^3/min$ and $h=5cm$
$\Rightarrow 50=\frac{4\pi }{3}\left(3{\left(10+5\right)}^2\right)\frac{dh}{dt}$
$\Rightarrow \frac{dh}{dt}=\frac{50}{4\pi \times 15^2}=\frac{1}{18\pi }cm/min$