Question
A spherical iron ball $$10\,cm$$ in radius is coated with a layer of ice of uniform thickness that melts at a rate of $$50\,c{m^3}/\min .$$ 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 :
$$\frac{1}{{18\pi }}cm/\min .$$
Solution :
Given that
$$\eqalign{
& \frac{{dv}}{{dt}} = 50\,c{m^3}/\min \Rightarrow \frac{d}{{dt}}\left( {\frac{4}{3}\pi {r^3}} \right) = 50 \cr
& \Rightarrow 4\pi {r^2}\frac{{dr}}{{dt}} = 50 \cr
& \Rightarrow \frac{{dr}}{{dt}} = \frac{{50}}{{4\pi {{\left( {15} \right)}^2}}} = \frac{1}{{18\pi }}cm/\min \,\left( {{\text{here}}\,r = 10 + 5} \right) \cr} $$