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: Understanding the Concept:
The ball has radius 10 cm and ice thickness \(t\). Outer radius is \(R = 10 + t\). The volume of ice is the volume of a spherical shell: \(V = \frac43\pi(R^3 - 10^3)\).

Step 2: Differentiate:
\[ \frac{dV}{dt} = 4\pi R^2\frac{dR}{dt} \]
Because the ice melts at a rate of \(50\ \text{cm}^3/\text{min}\), \(\frac{dV}{dt} = -50\).

Step 3: Evaluate:
When thickness is 5 cm, \(R = 15\) cm, so \(R^2 = 225\).
\[ \frac{dR}{dt} = \frac{-50}{4\pi \times 225} = -\frac{50}{900\pi} = -\frac{1}{18\pi} \]
The thickness \(t = R - 10\) changes at the same rate as \(R\), so it decreases at \(\frac{1}{18\pi}\) cm/min.

Step 4: Why the other options are wrong.
\(\frac1{36\pi}\) results from using \(\frac{4}{3}\pi R^2\) or R = 15 with the wrong formula, \(\frac{1}{54\pi}\) from a different slip, and \(\frac{5}{6\pi}\) from using \(R = 5\).

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