If the volume of spherical ball is increasing at the rate of 4 π c m 3 / sec then the rate of change of its surface area when the volume is 288 π c m 3 is

Question

If the volume of spherical ball is increasing at the rate of 4 π c m 3 / sec then the rate of change of its surface area when the volume is 288 π c m 3 is, 4 3 π c m 2 / s e c, 2 3 π c m 2 / s e c, 4 π c m 2 / s e c, 2 π c m 2 / s e c

MARKS App · Accepted Answer

V = 4 3 π r 3 ⇒ d V d t = 4 π r 2 d r d t When V - 288 π 288 π = 4 3 π r 3 ⇒ r = 6 Now d V d t = 4 π ∴ 4 π r 2 d r d t = 4 π = d r d t = 1 r 2 A = surface area = 4 π r 2 ∴ d A d t = 8 π r d r d t = 8 π r × 1 r 2 = 8 π r = 8 π 6 = 4 π 3

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