Two capillaries of same length and radii in the ratio `1 : 2` are connected in series. A liquid flows through them in streamlined condition. If the pressure across the two extreme ends of the combination is 1 m of water, the pressure difference across first capillary is

Options

A. 9.4 m
B. 4.9 m
C. 0.49 m
D. 0.94 m

Answer

D

Explanation

Here , `l_(1)=l_(2)=1m` and `(r_(1))/(r_(2))=(1)/(2)`
As `V=(piP_(1)r_(1)^(4))/(8etal)=(piP_(2)r_(2)^(4))/(8etal)` or `(P_(1))/(P_(2))=((r_(2))/(r_(1)))^(4)=16`
`therefore P_(1)=16P_(2)`
Since, both tubes are connected in series , hence pressure difference across the combination is
`P=P_(1)+P_(2) implies 1=P_(1)+(P_(1))/(16)`
or `P_(1)=(16)/(17)=0.94`m