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Water flows through a cylindrical pipe, whose inner radius is $1 cm,$ at the rate of $80 cm / s$ in an empty cylindrical tank, the radius of whose base is $40 cm$ .What is the rise of water level in tank in half an hour?

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80cm

70cm

90cm

100cm

answer is C.

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Detailed Solution

Given that a cylindrical pipe, whose inner radius is 1cm, the rate of water flowing is

80 cm / s

and the radius of the base of the cylindrical tank is 40cm.
We have the following formula,
Volume of cylinder =

π r 2 h

The volume of water in the tank will be equal to the volume of water flowing through a pipe.
Then,
r = 1 cm
Length of water flowing in 1 sec= 80cm
Volume of water flowing in 1 second will be,

= π × 1 × 1 × 80 = 80 π c m 3

The volume of water flowing in 30 minutes will be,

= 80 π × 60 × 30 c m 3

= 144, 000 π c m 3 .

Radius of cylindrical tank (R) = 40 cm
We are considering the rise in the level of water as H cm.
Then the volume of water in the tank,

= π R 2 H

= π × 40 × 40 × H c m 3

= 1600 π H c m 3

Volume of water in the tank is equal to the volume of water flowing through a pipe.

⇒ 1600 π H = 144, 000 π H = 90 c m

Thus the rise of the water level in the tank in half an hour is 90cm.
Therefore the correct option is 3.

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