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Question

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

When radius r=8 cm, then the increase in radius in the next 1/2 min is

A
0.025 cm
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B
0.050 cm
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C
0.075 cm
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D
0.01 cm
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Solution

The correct option is A 0.025 cm
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.
​​​​
Therefore, increase of S in 12 min=10×(12)=5 cm2/min.

If r1 is the radius of the balloon after 12 min, then 4πr21=4π(8)2+5
r2182=54π=0.397 nearly.
Required increase in the radius =r1=8.0258
=0.025 cm

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