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

$V = \frac{4}{3} π (10 + x)^{3} \Rightarrow \frac{d V}{d t} = 4 π (10 + x)^{2} \frac{d x}{d t}$ …(i)

It is given that at $x = 5, \frac{d V}{d t} = 50 {cm}^{3} / \min$

$∴ 50 = 4 π (10 + 5)^{2} \frac{d x}{d t} \Rightarrow \frac{d x}{d t} = \frac{1}{18 π} cm / \min$

A sphere of $10 c m$ radius has a uniform thickness of ice around it. If ice is melting at the rate of $50 c m 3 / m i n$ when thickness is $5 c m$ ,then rate of change of thickness is