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

hi dear
here is your answer

Answer 2

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


→ 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 .

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


✔✔ Hence, rise in level = 90 cm ✅✅ .


THANKS



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