Question

Water flows through a cylindrical pipe, whose inner radius is 1 cm, at the rate of 80 cm / s in an empty cylindrical tank, the radius of whose base is 40 cm. What is the rise of water level in tank in half an hour?

Answer 1

$\huge \bf \green{Hey \: there !! }$


→ 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 .

$\textsf \implies 1600 \cancel\pi H = 144000 \cancel\pi . \\ \\ \implies H = \frac{144000}{1600} = \frac{ \cancel{1440} \: \: ^{90} }{ \cancel{16}} . \\ \\ \huge \boxed{ \boxed{ \pink{ \therefore H = 90 \sf cm.}}}$


✔✔ Hence, rise in level = 90 cm ✅✅ .


THANKS



#BeBrainly.

Answer 2

→ Internal radius of the pipe , r = 1 cm .
→ Length of water flowing in 1 sec , h = 80 cm .

Volume of water flowing in 1 second
= πr²h .
= ( π × 1 × 1 × 80 ) cm³ .
= 80π cm³ .

Volume of water flowing in 30 minutes .
= ( 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 .
=> 1600πH = 144000π .
=> H = 144000/1600 .
•°• H = 90 cm.