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

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Application of Derivatives

A spherical iron ball of radius 10 cm is coated with a layer of ice of uniform thickness that melts at a rate of

50 cm^{3} / min

. When the thickness of the ice is 5 cm, then the rate at which the thickness (in cm/min) of ice decreases is

Answer: (D)

The ice forms a shell of thickness

h

on the

10 cm

ball, with volume

V = \frac{4}{3} π [(10 + h)^{3} - 1 0^{3}]

. Differentiating,

\frac{d V}{d t} = 4 π (10 + h)^{2} \frac{d h}{d t}

. With melting at

\frac{d V}{d t} = - 50 cm^{3} min^{- 1}

and

h = 5 cm

- 50 = 4 π (15)^{2} \frac{d h}{d t} = 900 π \frac{d h}{d t}

, so

\frac{d h}{d t} = - \frac{1}{18 π}

; the thickness is shrinking at

\frac{1}{18 π} cm min^{- 1}

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option