Question

Water flows out through a circular pipe, whose internal diameter is 2 cm, at the rate of 0.7m per

second into a cylindrical tank, the radius of whose base is 40 cm. By how much the height of

water will rise in half hour?​

Answer 1

Step-by-step explanation:

diameter is given so divide by 2 so that u get radius of the circle then convert the m into cm or vice versa and apply the formula and get your answer

Answer 2

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