Question:
water is flowing through a circular pipe whose internal diameter is 4cm at the rate of 1.4m
Answer:
The complete question is:
Water is flowing through a circular pipe whose internal diameter is 4 cm at the rate of 1.4 m/sec. into a cylindrical tank,
the radius of whose base is 80 cm. By how much will the water level rise in 2 hours?
Solution:
Given, internal diameter of the circular cube = 4 cm
So, the radius of the circular cube r = 4/2 = 2 cm
and height of the circular cube h = 1.4 m = 1.4 * 100 cm
Now, volume of the water flowing from cylinder in 2 hours = volume of the cylinder * 2 hours
= πr2 h * 2* 60 * 60
= (22/7)* 22 * 1.4 * 100 * 2* 60 * 60
= 22 * 4 * 0.2 * 2 * 100 * 60 * 60
= 88 * 20 * 7200 cm3
Now, it is going down in cylinder having radius = 80 cm
So, volume of the cylinder = πr2 h
=> 88 * 20 * 7200 = πr2 h
=> 88 * 20 * 7200 = π(80)2 h
=> 88 * 20 * 7200 = π * 80 * 80 * h
=> 88 * 20 * 7200 = (22/7) * 80 * 80 * h
=> 7 * 88 * 20 * 7200 = 22 * 80 * 80 * h
=> h = (7 * 88 * 20 * 7200)/(22 * 80 * 80)
=> h = (7 * 4 * 20 * 7200)/(80 * 80)
=> h = (7 * 4 * 20 * 72)/(8 * 8)
=> h = (7 * 4 * 20 * 9)/8
=> h = (7 * 20 * 9)/2
=> h = 7 * 10 * 9
=> h = 630 cm
=> h = 630/100 m
=> h = 6.3 m
So, the water level rise in 2 hours is 6.3 m
Not what you are looking for? Go ahead and submit the question, we will get back to you.
