A spherical iron ball of 10 cm radius 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 ice is 5 cm , then the rate (in $\mathrm{cm} / \mathrm{min}$ ) at which of the thickness of ice decreases, is:

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Application of Derivatives

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

50 cm^{3} / min

. When the thickness of ice is 5 cm , then the rate (in

cm / min

) at which of the thickness of ice decreases, is:

Answer: (C)

With ice thickness

x

, the outer radius is

10 + x

and the volume of ice plus ball is

V = \frac{4}{3} π (10 + x)^{3}

, so

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

. The ice melts at

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

^{3}

/min, and at

x = 5

the radius is

15

, so

- 50 = 4 π (15)^{2} \frac{d x}{d t}

. This gives

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

, so the thickness shrinks at

\frac{1}{18 π}

cm/min.

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Single correct option