Water in cylindrical tank of diameter 4m and height 10m is released through a cylindrical pipe a diameter of 10cm at the rate of 2.5km/hr how much time will it take to empty the half of the tank? Assume that the tank is full of water to begin with
Answers
Mark me a brainlist answer
.....
Now we are going to convert every measurements into m
Diameter of the cylindrical tank = 4 m
radius of the cylindrical tank = 2 m
height of the tank = 10 m
Diameter of the cylindrical pipe = 10 cm
radius of the cylindrical pipe = 10/2 = 5 cm
= 5/100 m
speed of water = 2.5 km/hr
1000 m = 1 km
= 2.5 x 1000
= 2500 m/hr
Volume of water discharged from
the cylindrical pipe = (1/2) Volume of cylindrical tank
Area of cross section x time x speed = (1/2) Π r² h
Π r²x time x speed = (1/2) Π r² h
(5/100)²x time x 2500 = (1/2) (2)² (10)
(5/100) x (5/100) x time x 2500 = (1/2) x (2) x 2 x (10)
Time = 2 x 10 x (100/5) x (100/5) x (1/2500)
Time = 80/25
= 3.2 hour
1 hour = 60 minute
= 192 minute
= 180 + 12
3 hour and 12 minute
Therefore time taken to empty the half of the tank is 3 hour and 12 minute
Now we are going to convert every measurements into m
Diameter of the cylindrical tank = 4 m
radius of the cylindrical tank = 2 m
height of the tank = 10 m
Diameter of the cylindrical pipe = 10 cm
radius of the cylindrical pipe = 10/2 = 5 cm
= 5/100 m
speed of water = 2.5 km/hr
1000 m = 1 km
= 2.5 x 1000
= 2500 m/hr
Volume of water discharged from
the cylindrical pipe = (1/2) Volume of cylindrical tank
Area of cross section x time x speed = (1/2) Π r² h
Π r²x time x speed = (1/2) Π r² h
(5/100)²x time x 2500 = (1/2) (2)² (10)
(5/100) x (5/100) x time x 2500 = (1/2) x (2) x 2 x (10)
Time = 2 x 10 x (100/5) x (100/5) x (1/2500)
Time = 80/25
= 3.2 hour
1 hour = 60 minute
= 192 minute
= 180 + 12
3 hour and 12 minute
Therefore time taken to empty the half of the tank is 3 hour and 12 minute