The radius of a spherical balloon is increasing at the rate of 0.5 cm/s. The rate of increase of its surface area when the radius is 10 cm is: A. 40π cm²/s B. 20π cm²/s C. 10π cm²/s D. 80π cm²/s Answer: A) 40π cm²/s Explanation: Surface area S = 4πr². dS/dt = 8πr (dr/dt) = 8π × 10 × 0.5 = 40π cm²/s.

imo class 12 application of derivatives

The radius of a spherical balloon is increasing at the rate of 0.5 cm/s. The rate of increase of its surface area when the radius is 10 cm is:

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The radius of a spherical balloon is increasing at the rate of 0.5 cm/s. The rate of increase of its surface area when the radius is 10 cm is:

Answer: A) 40π cm²/s

Explanation: Surface area S = 4πr². dS/dt = 8πr (dr/dt) = 8π × 10 × 0.5 = 40π cm²/s.