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

(B) $\frac{1}{18 \pi} \mathrm{~cm} / \mathrm{min}$

(A) $\frac{1}{36 \pi} \mathrm{~cm} / \mathrm{min}$

(C) $\frac{1}{54 \pi} \mathrm{~cm} / \mathrm{min}$

(D) $\frac{5}{6 \pi} \mathrm{~cm} / \mathrm{min}$

JEE Main · 2005 · 08 May 2005 · Offline · Application of Derivatives

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 cm^{3} / min

. When the thickness of ice is 5 cm , then the rate at which the thickness of ice decreases, is

Answer: (B)

The ice forms a shell outside the fixed iron ball, but since the inner radius of

10

cm is constant, the rate of change of the ice volume is

\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}

, where

r

is the outer radius. When the ice is

5

cm thick,

r = 15

cm, and the ice melts at

\frac{d V}{d t} = - 50

^{3}

/min. Thus

- 50 = 4 π (15)^{2} \frac{d r}{d t}

, giving

\frac{d r}{d t} = - \frac{50}{900 π} = - \frac{1}{18 π}

cm/min. Since the thickness equals

r - 10

, it decreases at the rate

\frac{1}{18 π}

cm/min.

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option