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Q.

The surface area of a sphere increases at the rate of 1 sq. ft per second, then the rate of increase of volume when the radius is 3 ft is

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a

2/3 cubic ft/sec

b

3/2 cubic ft/sec

c

3 cubic ft/sec

d

2 cubic ft/sec

answer is D.

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Detailed Solution

V=3    dsdt=1 S=4πr2    dsdt=8πrdrdt                    1=24πdrdt  drdt=124π V=43πr3  drdt=4πr2drdt                               =4π×9×124π =32

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