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Question

Suppose that water is emptied from a spherical tank of radius 10 cm. If the depth of the water in the tank is 4 cm and is decreasing at the rate of 2 cm/sec, them the radius of the top surface of water is decreasing at the rate of


A

1

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B

23

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C

32

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D

2

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Solution

The correct option is C

32


dhdt=2; r=10 cm

We have to find dxdt when h = 4
where x is the radius of the top surface.
From the figure r2=x2+(10h)2
2xdxdt=2(10h)dhdt
dxdt=(10h)x(2) dxdt=2(104)x=12x (i)when h=4, then x2=10262=64 or x=8 dxdt=128=32
Hence the correct option is c


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