Question

water is flowing through a cylindrical pipe of internal diameter 2 cm into a cylindrical tank of base 40 cm at the rate of 0.4 m per second . determine the rise in the level of the water in the tank in half an hour

Answer 1

/>Given, internal diameter of cylindrical pipe = 2 cm

So, radius of cylindrical pipe (r) = 2/ 2 = 1 cm

Area of cross section of pipe= r2= (1)2= cm2

Speed of water= 0.4 m/s =0.4 x 100 cm/s = 40 cm/s

Thus,

Volume of water flown out in half an hour = x 40 x 30 x 60 = 72000 cm2

And,

Radius of cylindrical tank (R) = 40 cm

Let level of water rise to the height of h cm.

Thus,

Volume of cylindrical tank = R2h = (40)2h = 1600h cm2.

Now,

Volume of cylindrical tank = Volume of water flown out in half an hour

1600h = 7200

h = 45cm

Hence, level of water rise to the height of 45 cm.

Hope it helps

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Answer 2

Step-by-step explanation:

Given diameter of the circular pipe = 2 cm

So, the radius of the circular pipe = 2/2 = 1 cm

Height of the circular pipe = 0.7 m = 0.7 * 100 = 70 cm

Now, volume of the water flows in 1 second = πr2 h

                                                                = 3.142 * 12 * 70

                                                                = 3.142 * 70

Volume of the water flows in 1/2 hours =  3.142 * 70 * 30 * 60

Now, volume of the water flows = Volume of the cylinder

=> 3.142 * 70 * 30 * 60 = πr2 h

=> 3.142 * 70 * 30 * 60 = 3.142 * (40)2 h

=> 70 * 30 * 60 = 40 * 40 * h

=> h = (70 * 30 * 60)/(40 * 40)

=> h = (70 * 3 * 6)/(4 * 4)

=> h = 1260/16

=> h = 78.85 cm

So, the level of water rise in the tank in half an hour is 78.75 cm

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