Question

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

• A. \(\frac{1}{18π} \text{cm}/\text{min}\)
• B. \(\frac{1}{36π} \text{cm}/\text{min}\)
• C. \(\frac{5}{6π} \text{cm}/\text{min}\)
• D. \(\frac{1}{54π} \text{cm}/\text{min}\)

Hint:

The volume of ice is a spherical shell; differentiate volume with respect to time.

Answer 1

The Correct Option is A

Step 1: Surface-area idea:
The rate of change of volume equals the outer surface area times the rate at which the outer radius changes: $\frac{dV}{dt} = A\frac{dR}{dt}$.

Step 2: Area:
With $R = 10 + 5 = 15$ cm, $A = 4\pi(15)^2 = 900\pi\ \text{cm}^2$.

Step 3: Rate:
$\frac{dR}{dt} = \frac{-50}{900\pi} = -\frac{1}{18\pi}$ cm/min. The iron core stays at radius 10, so the thickness falls at this rate.

Final Answer:
The thickness decreases at 1/(18 pi) cm per minute, option (A). \[ \boxed{\frac{1}{18\pi}\text{ cm/min}} \]