In what time would a cistern be filled by 3 pipes ,whose diameter are 1cm ,4/3cm and 2 cm,running together ,when the largest alone will fill in 61 min,the amount of water flowing in by each pipe being proportional to the square of its diameter ?
Solution(By Examveda Team)
In one minute the pipe of 2cm diameter fills 1/61 of the cistern.
In one minute the pipe of 1cm diameter fills 1/61* 1/4 of the cistern
In one minute the pipe of 11/3 cm diameter fills 1/61* 4/9 of the cistern
In one minute (1/61+1/(61*4)+4/(61*9))=1/36 of the cistern is filled.
The whole cistern is filled in 36 minutes.
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Volume of three pipes work togther = volume of of largest pipe work alone
We know that, Volume = capacity X time
Capcity of all three pipe = 1^2 + (4/3)^2 + 2^2 =61/9
Now using volume formula,
61/9 * time taken = volume of alone third pipe= 61 * 4
So, time taken = 9*4 =36min
For all three pipes together:
2^2 = 4
1^2 = 1
4/3^2 = 16/9
4 + 1 + 16/9 = 61/9 is proportional to the amount of water flowing in.
We know that for the largest pipe: 2^2 = 4 is proportional to the amount of water flowing in and this takes 61 minutes.
So we need to calculate:
61 x (4/(61/9)) = 61 x (36/61) = 36 minutes.
(4k+16/9k+1*k)T = C..........(1)
k = some pipe const same for all
C = capacity of cistern
T = Required time
Also 4k*61min = C........(2)
Substituting (2)in (1), we get
T=36 min.