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?


plss be fast​

Answer 1

AnSwEr :

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

Now, volume of the water flows in 1 second = πr2 h= 3.142 × 12 × 70

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

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 = 1260/16=> h = 78.85 cm

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

Hope you understand ♪♦♪♦

Answer 2

Answer:

78.75cm

Step-by-step explanation:

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 = πr2h = π x 1 x 1 x 70 = 70π

Volume of water flowing in 60 sec = 70π x 60  

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

Volume of water in Tank = πr2H = π x 40 x 40 x H  

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

π x 40 x 40 x H = 70π x 60 x 30  H = 78.75cm.