CUET · MATHS · PYQ PAPER 2025
The rate of change of volume of a sphere with respect to its surface area, when radius is 4 cm, is equal to
- A \(64 \pi cm^3 / cm ^2\)
- B \(32 \pi cm^3 / cm ^2\)
- C \(\frac{1}{2} cm^3 / cm ^2\)
- D \(2 cm^3 / cm ^2\)
Answer & Solution
Correct Answer
(D) \(2 cm^3 / cm ^2\)
Step-by-step Solution
Detailed explanation
\( V = \frac{4}{3} \pi r^3 \) \( A = 4 \pi r^2 \) \( \frac{dV}{dr} = 4 \pi r^2 \) \( \frac{dA}{dr} = 8 \pi r \) \( \frac{dV}{dA} = \frac{dV/dr}{dA/dr} = \frac{4 \pi r^2}{8 \pi r} = \frac{r}{2} \) \( \frac{dV}{dA} \Big|_{r=4} = \frac{4}{2} = 2 \) \( 2 cm^3 / cm^2 \)
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