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 :-
- A 2.5 \(cm ^2\)/sec
- B 2.66 \(cm ^2\)/sec
- C 2.61 \(cm ^2\)/sec
- D 2.59 \(cm ^2\)/sec
Answer & Solution
Correct Answer
(B) 2.66 \(cm ^2\)/sec
Step-by-step Solution
Detailed explanation
\(V = \frac{4}{3}\pi r^3 \Rightarrow \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\) \(4 = 4\pi (3)^2 \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{4}{36\pi} = \frac{1}{9\pi}\) \(A = 4\pi r^2 \Rightarrow \frac{dA}{dt} = 8\pi r \frac{dr}{dt}\)…
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