Question

The surface area of a solid hemisphere is increasing at the rate of \( 8 \, \text{cm}^2/\text{sec} \) (retaining its shape). Then the rate of change of its volume (in \( \text{cm}^3/\text{sec} \)), when the radius is \( 5 \,\text{cm} \), is

• A. \( \frac{50}{3} \)
• B. \( \frac{20}{3} \)
• C. \( \frac{40}{3} \)
• D. \( \frac{25}{3} \)
• E. \( \frac{80}{3} \)

Hint:

In related rates, first find \( \frac{dr}{dt} \), then substitute into required derivative.

Answer 1

The Correct Option is C

Concept: \[ \text{Surface area} = 3\pi r^2, \quad \text{Volume} = \frac{2}{3}\pi r^3 \]

Step 1: Differentiate surface area.
\[ \frac{dS}{dt} = 6\pi r \frac{dr}{dt} \] \[ 8 = 6\pi \cdot 5 \cdot \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{4}{15\pi} \]

Step 2: Differentiate volume.
\[ \frac{dV}{dt} = 2\pi r^2 \frac{dr}{dt} \] \[ = 2\pi \cdot 25 \cdot \frac{4}{15\pi} = \frac{200}{15} = \frac{40}{3} \]