Question
Question: Water flows through a cylindrical pipe, whose inner radius is 1 cm, at the rate of\[80{cm}/{\sec }\;...
Water flows through a cylindrical pipe, whose inner radius is 1 cm, at the rate of80cm/sec in an empty cylindrical tank, the radius of whose base is 40 cm. What is the rise of water level in a tank in half an hour?
Solution
As the water flows through a cylindrical pipe, so, we will make use of the formula of volume of the cylinder. The volume of the water in a cylindrical tank equals the volume of the water that flows from the circular pipe in half an hour. So, we will equate this condition to find the value of the rise in water level in the tank in half an hour.
Formula used:
V=πr2h
Complete answer:
From the given information, we have the data as follows.
Water flows through a cylindrical pipe, whose inner radius is 1 cm, at the rate of80cm/sec in an empty cylindrical tank, the radius of whose base is 40 cm.
The formula that we are using to solve this question is given as follows.
The volume of a cylinder.
V=πr2h
Where r is the radius of the cylinder and h is the height of the cylinder.
The volume of the water in a cylindrical tank equals the volume of the water that flows from the circular pipe in half an hour.
The radius of the cylindrical tank is, r1=40cm
The radius of the circular pipe is, r2=1cm
The height of the cylindrical tank is, h1=?
The speed of water = 80cm/sec
The height of the circular pipe is, h2=80×60×30=144000cm
πr12h1=πr22h2
Substitute the values in the above equation.