Answer:
The factor is [tex]\frac{1}{9}[/tex] to obtain velocity pf liquid flowing through second pipe.
Option (D) is correct
Explanation:
Given:
Diameter of first pipe [tex]d _{1} = 1[/tex] cm
Diameter of second pipe [tex]d _{2} = 3[/tex] cm
From the equation of continuity
[tex]Av = constant[/tex]
Where [tex]A =[/tex] area of tube, [tex]v =[/tex] velocity of fluid
[tex]A_{1} v_{1} = A_{2} v_{2}[/tex]
Where [tex]A = \pi r^{2} = \pi \frac{d^{2} }{4}[/tex]
By putting above value,
[tex]\pi \frac{d_{1} ^{2} }{4} v_{1} = \pi \frac{d_{2} ^{2} }{4} v_{2}[/tex]
[tex]d_{1} ^{2} v_{1} = d_{2} ^{2} v_{2}[/tex]
So velocity at the second pipe end is,
[tex]v_{2} = \frac{d_{1} ^{2} v_{1} }{d_{2} ^{2} }[/tex]
By putting values of diameter,
[tex]v_{2} = \frac{1}{9} v_{1}[/tex]
[tex]\frac{v_{2} }{v_{1} } = \frac{1}{9}[/tex]
Therefore, the factor is [tex]\frac{1}{9}[/tex] to obtain velocity pf liquid flowing through second pipe.
