# A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness than melts at a rate of 50 cm3/min. When the thickness of ice is 5 cm, then the rate at which the thickness of ice decreases, is ?

Jitender Singh IIT Delhi
8 years ago
Ans:
Let the thickness of ice to be ‘a’ at any instant ‘t’.
Volume ‘V’:
$V = \frac{4}{3}\pi (10+a)^{3}$
$\frac{\partial V}{\partial t} = 4\pi (10+a)^{2}.\frac{\partial a}{\partial t} = 50 \frac{cm^{3}}{min}$
$4\pi (10+5)^{2}.\frac{\partial a}{\partial t} = 50 \frac{cm^{3}}{min}$
$\frac{\partial a}{\partial t} = \frac{1}{18\pi } \frac{cm}{min}$
Thanks & Regards
Jitender Singh
IIT Delhi