water is flowing at the rate of 0.70 m/s through a circular pipe, whose internal diameter is 2 cm into a cylindrical tank the radius of whose base is 40 cm. Determine the increase in water level in half hour. (if answer is 0.7875)
Answers
Answered by
0
Answer:
Step-by-step explanation:
internal diameter = 2 cm
radius = 2/2 cm = 1cm
water flows at the rate of 0.7 m/s.
In 1 second it covers distance = 0.7 m = 70 cm
volume of water flows in 1 second = πr²d = 3.14 x 12 x 70 = 219.8 m³
Thus volume of water flows in 1 and half hour = 5400 x 219.8 = 1186920 m³
radius of cylindrical tank (r) = 40 cm
Let the increase level of water = h cm
Volume of cylindrical tank = π(40)²h
Si water flows out at the rate of 0.7 m/s into a cylindrical tank.
So,volume of water flows in 1and half hour = Volume of cylindrical tank
1186920 m³= π(40)²h
h = 1186920 / ((3.14 x (40)²)
h = 236.25 m
Hence, water rise by 236.25
Arya79:
wrong i had given answer
thats the right answer given by my teacher but i am not able
to solve it
Correct solution is in one second volume of water in pipe = π r² h
= π×(1)² × 70 = 70π
in 30 min,volume = 30×60×70π = 126000π
this volume = volume of tank
⇒π×(40)² ×H = 126000 π
⇒H =126000/1600 = 78.75 cm
= π×(1)² × 70 = 70π
in 30 min,volume = 30×60×70π = 126000π
this volume = volume of tank
⇒π×(40)² ×H = 126000 π
⇒H =126000/1600 = 78.75 cm
Similar questions