Water is flowing at the rate of 10 km/h through a pipe of diameter 14 cm into a rectangular tank which is 70 m long and 22 m wide. Find the time in which the level of water in the tank will rise by 50 cm. (use Π=22/7)
Answers
Let the level of water in the pond rises by 21 cm in t hours.
Speed of water = 15 km/hr
Diameter of the pipe = 14/100 m
Radius of the pipe (r) = 7/100 m
Volume of water flowing out of the pipe in 1 hour
= π r 2 h
= (22/7) x (7/100) x (7/100) x 15000
= 231 m3
Volume of water flowing out of the pipe in t hours = 231 t m3.
Volume of water in the cuboidal pond
= 50 x 44 x (21/100)
= 462 m3
Volume of water flowing out of the pipe in t hours = Volume of water in the cuboidal pond
So, 231 t = 462
t = 2
Thus, the required time is 2 hours.
Answer:
Step-by-step explanation:
Let the time taken be t hrs.
Since water is flowing at rate of 10 km/hr. So ,we assume height of pipe is 10 km.
Volume of water in pipe in t time =Volume of water in tank
πr^2h*t=l*b*h
t=lbh/πr^2h
(Convert all the units in meter)
t=(70*22*(50/100))/((22/7)*(7/100)*(7/100)*(10*1000))
t=5hrs