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Question

A spherical balloon is being inflated so that its volume increases uniformly at the rate of 40 cm3/min.

At radius r=8 cm, its surface area increases at the rate

A
8 cm2/min
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B
10 cm2/min
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C
20 cm2/min
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D
None of these
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Solution

The correct option is B 10 cm2/min
Let V be the volume and r the radius of the balloon at any time.
Then, V=(43)πr3
dVdt=4πr2drdt
40=4πr2drdt (Given)
drdt=10πr2 ...(1)

Now, let S be the surface area of the balloon when its radius is r. Then
S=4πr2
dSdt=8πrdrdt

From (1) and (2),
dSdt=8πr10πr2=80r

When r=8, the rate of increase of surface area is, S=808=10 cm2/min.

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