water is flowing at the rate of 2.52 km/h through a cylindrical pipe into a cylindrical tank, the radius of whose base is 40 cm, If the increase in the level of water in the tank, in half an hour is 3.15 m, find the internal diameter of th pipe.

Explanation

Let the internal radius of the pipe be r.
Length of water flowing rhrough the pipe in 1 hour,
h = 2.52 km = 2520 m.
Volume of water flowing through the pipe in 1 hour
`=pir^(2)h=(pixxr^(2)xx2450)m^(3)`.
Volume of water flowing through the pipe in half an hour
`=(1/2xxpir^(2)xx2520)m^(3)=(1260pir^(2))m^(2)`.
Radius of the cylindracal tank, `R=40cm=40/100m=2/5m`.
Increase in level of water, H = 3,15 m.
Volume of water filled in the tank
`=piR^(2)H=(pixx2/5xx2/5xx315/100)m^(3)=((63pi)/(125))m^(3)`.
Now, volume of water flown in half an hour = volume of water filled in the tank
`rArr" "1260pir^(2)=(63pi)/(125)rArrr^(2)=(63)/(125xx1260)=1/(2500)`
`rArr" "r=sqrt(1/2500)=1/50m=(1/50xx100)cm=2cm.`
Hence, the internal diameter of the pipe `= (2xx2) cm = 4 cm`.