Question

Water is flowing through a cylindrical pipe of internal diameter 2cm, into a cylindrical tank of base radius 40 cm at the rate of 0.7m/sec. By how much will the water rise in the tank in half an hour?​

Answer 1

Answer: 395.64 liters

Steps:

Volume of Water flowing through the pipe is given by the formula:

Cross Sectional Area of the Pipe × Rate of Flow

According to the question,

Diameter of the pipe is 2 cm. Hence the cross sectional area of the pipe can be calculated as:

→ Area of Circle = πr²

→ Area of Circle = 3.14 × 1 × 1 = 3.14 cm²

Hence Volume of water flowing per second is given as:

→ Volume = 3.14 cm² × 0.7 m/s ( 70 cm/s )

→ Volume = 219.8 cm³ / s

Hence 219.8 cm³ of water is flowing per second. Hence for 30 minutes, the amount of water flowing is given as:

→ Volume after 30 minutes = 30 × 60 × 219.8

→ Volume after 30 minutes = 395640 cm³

1 cm³ = 0.001 liter

Hence volume of water in liters is given as: 395.64 liters.

Hence after 30 minutes, the volume of water present in the cylindrical tank is 395.64 liters.

Answer 2

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Water is flowing through a cylindrical pipe of internal diameter 2cm, into a cylindrical tank of base radius 40 cm at the rate of 0.7m/sec. By how much will the water rise in the tank in half an hour?

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For pipe, r = 1cm

Length of water flowing in 1 sec, h = 0.7m = 7cm

Cylindrical Tank, R = 40 cm, rise in water level = H

Volume of water flowing in 1 sec = r2h = T x 1x 1x 70

= 70T

Volume of water flowing in 60 sec = 7011 X 60

Volume of water flowing in 30 minutes = 70T x 60 x 30

Volume of water in Tank = Tr2H = T x 40 x 40 x H

Volume of water in Tank = Volume of water flowing in 30

minutes

TT X 40 x 40 x H = 7OTT X 60 x 30

H = 78.75cm

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