Question

Water flows through a cylindrical pipe, whose inner radius is 1 cm, at the rate of 70 cm/s in an empty tank. How much water will flow in the tank in half an hour?
(
use pie as 22/7

Answer 1

Internal radius of the pipe , r = 1 cm .

→ Length of water flowing in 1 sec , h = 80 cm .


▶ Then, Volume of water flowing in 1 second

= πr²h .

= ( π × 1 × 1 × 80 ) cm³ .

= 80π cm³ .


▶ Volume of water flowing in 30 minutes [ Half an hour ]

= ( 80π × 60 × 30 ) cm³ .

= 144,000π cm³ .


→ Radius of cylindrical tank, R = 40 cm .

Let the rise in level of water be H cm .


▶ Volume of water in the tank

= πR²H .

= ( π × 40 × 40 × H ) cm³ .

= 1600πH cm³ .


▶ Volume of water in the tank = Volume of water flown through a pipe .




✔✔ Hence, rise in level = 90 cm ✅✅ .


THANKS

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Answer 2

Answer:

Step-by-step explanation:

$\pi = 22/7 , r = 1 cm , h = 70 cm \\ v = \pi r2h\\v = 22/7 *1*2*70\\v = 440$