Answer:
[tex]\dfrac{Q_B}{Q_A}=\sqrt{5}[/tex]
Explanation:
Lets take
Length of pipe B = L
Length of pipe A = 5 L
Discharge in pipe A = Q₁
Discharge in pipe B = Q₂
We know that head loss in the pipe given as
[tex]h_f=\dfrac{FLQ^2}{12.1d^5}[/tex]
F=Friction factor, Q=Discharge,L=length
d=Diameter of pipe
here all only Q and L is varying and all other quantity is constant
So we can say that
LQ²= Constant
L₁Q₁²=L₂Q²₂
By putting the values
5LQ₁²=LQ²₂
[tex]\dfrac{Q_2}{Q_1}=\sqrt{5}[/tex]
Therefore
[tex]\dfrac{Q_B}{Q_A}=\sqrt{5}[/tex]