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

Given, `l_(1)=l_(2)=1m " and" (r_(1))/(r_(2))=(1)/(2)`
`V=(pi p_(1)r_(1)^(4))/(8 eta l)=(pi p_(2)r_(2)^(4))/(8 eta l) implies (p_(1))/(p_(2))=((r_(2))/(r_(1)))^(4)=16 implies p_(1)=16p_(2)`
Since, both tubes are connected in series, hence pressure difference across combination,
`p=p_(1)+p_(2) implies p_(1)=(p_(1))/(16) implies p_(1)=(16)/(17)=0.94 m`