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CUET · MATHS · PYQ PAPER 2025

The volume of spherical balloon is increasing at the rate of 4 \(cm ^3\) / sec. The rate of increase of its surface area, when the radius is 3 cm will be :-

  1. A 2.5 \(cm ^2\)/sec
  2. B 2.66 \(cm ^2\)/sec
  3. C 2.61 \(cm ^2\)/sec
  4. D 2.59 \(cm ^2\)/sec
Verified Solution

Answer & Solution

Correct Answer

(B) 2.66 \(cm ^2\)/sec

Step-by-step Solution

Detailed explanation

\(V = \frac{4}{3}\pi r^3 \Rightarrow \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\) \(4 = 4\pi (3)^2 \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{4}{36\pi} = \frac{1}{9\pi}\) \(A = 4\pi r^2 \Rightarrow \frac{dA}{dt} = 8\pi r \frac{dr}{dt}\)…