A spherical iron ball of radius 10cm is coated with a layer of ice of uniform thickness that melts a rate of $50 {cm}^{3} / \min$ When the thickness of the ice 5cm, then the rate at which the thickness (in cm/min) of the ice decreases, is:

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$\frac{5}{6 π}$

$\frac{1}{18 π}$

$\frac{1}{9 π}$

$\frac{1}{36 π}$

answer is B.

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Detailed Solution

$\frac{dv}{dt} = 50 r - 15 V = \frac{4}{3} {πr}^{3} \Rightarrow \frac{dv}{dt} = 4 {πr}^{2} \frac{dr}{dt} \Rightarrow 50 = 4 π \times 15 + 15 + \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{1}{18 π}$

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