Question

Ice is coated uniformly around a sphere of radius 15 cm. If ice is melting at the rate of \( 80 { cm}^3/{min} \) when the thickness is 5 cm, then the rate of change of thickness of ice is

• A. \( \frac{1}{10\pi} \) cm/min
• B. \( \frac{1}{50\pi} \) cm/min
• C. \( \frac{1}{80\pi} \) cm/min
• D. \( \frac{1}{40\pi} \) cm/min
• E. \( \frac{1}{20\pi} \) cm/min

Hint:

For a sphere with changing radius, use the volume differentiation formula \( \frac{dV}{dt} = 4\pi R^2 \frac{dR}{dt} \).

Answer 1

Answer: (E)

Let \( R \) be the total radius of the ice-coated sphere and \( r \) be the radius of the original sphere. The volume of the sphere is given by: \[ V = \frac{4}{3} \pi R^3 \] Differentiating both sides with respect to time \( t \): \[ \frac{dV}{dt} = 4\pi R^2 \frac{dR}{dt} \] Given: \[ \frac{dV}{dt} = -80 { cm}^3/{min}, \quad r = 15 { cm}, \quad {thickness} = 5 { cm} \] Thus, total radius \( R \) is: \[ R = 15 + 5 = 20 { cm} \] Substituting the values: \[ -80 = 4\pi (20)^2 \frac{dR}{dt} \] \[ -80 = 1600\pi \frac{dR}{dt} \] \[ \frac{dR}{dt} = -\frac{80}{1600\pi} = -\frac{1}{20\pi} \] Thus, the rate of change of thickness is: \[ \frac{1}{20\pi} { cm/min} \] Thus, the correct answer is (E).