MHT CET · Maths · Application of Derivatives
The surface area of a spherical ball is increasing at the rate of \(4 \pi \mathrm{~cm}^2 /\) second . The rate at which the radius is increasing when the surface area is \(16 \pi \mathrm{~cm}^2\) is
- A \(0.5 \mathrm{~cm} /\) second
- B \((-\infty, 0)\)
- C \(0.125 \mathrm{~cm} /\) second
- D \(1 \mathrm{~cm} /\) second
Answer & Solution
Correct Answer
(A) \(0.5 \mathrm{~cm} /\) second
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