A sphere of $10 cm$ radius has a uniform thickness of ice around it. If  ice is melting at the rate of $50 cm3/min$ when thickness is $5 cm$,then rate of change of thickness is 

Let the thickness of ice be $x$ cm and $V$ be the total 

volume of the sphere and ice. Then

$V=\frac{4}{3}\pi (10+x{)}^3\Rightarrow \frac{dV}{dt}=4\pi (10+x{)}^2\frac{dx}{dt}$                                       …(i)

It is given that at $x=5,\frac{dV}{dt}=50{\mathrm{cm}}^3/\min$

$\therefore 50=4\pi (10+5{)}^2\frac{dx}{dt}\Rightarrow \frac{dx}{dt}=\frac{1}{18\pi }\mathrm{cm}/\min$