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

Hint: The rate of flow of fluid through two capillary tubes connected in series is the same. The capillary tubes connected in series are analogous to the resistors connected in series.

Formula used:
Hagen-Poiseuille equation is,
\[V = \dfrac{{\pi p{r^4}}}{{8\eta l}}\]
where V is the rate of flow of fluid whose coefficient of viscosity \[\eta \], p is the pressure difference across the end of the tube with radius r and length l.

Complete step by step solution:
The lengths of the two tubes connected in series across a pressure head P (equal to 1 m of water) is the same. Let the lengths of the two tubes connected in series are l.
\[\dfrac{{{l_1}}}{{{l_2}}} = 1\]
It is given that the radii of the two tubes are in the ratio 1:2
\[\dfrac{{{r_1}}}{{{r_2}}} = \dfrac{1}{2}\]

Using the Poiseuille flow equation, the rate of flow through respective tube is,
\[{V_1} = \dfrac{{\pi {P_1}r_1^4}}{{8\eta {l_1}}}\]
And, \[{V_2} = \dfrac{{\pi {P_2}r_2^4}}{{8\eta {l_2}}}\]
Here, \[{P_1}\] is the pressure head across the first tube and \[{P_2}\] is the pressure head across the second tube.

The total pressure head is the sum of the individual pressure head in series.
\[P = {P_1} + {P_2}\]
The flow rate is same in both the tube,
\[{V_1} = {V_2} \\ \]
\[\Rightarrow \dfrac{{\pi {P_1}r_1^4}}{{8\eta {l_1}}} = \dfrac{{\pi {P_2}r_2^4}}{{8\eta {l_2}}} \\ \]
\[\Rightarrow {P_1}r_1^4 = {P_2}r_2^4 \\ \]
\[\Rightarrow {P_2} = {P_1}{\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^4} \\ \]
\[\Rightarrow {P_2} = {P_1}{\left( {\dfrac{1}{2}} \right)^4} \\ \]
\[\Rightarrow {P_2} = \dfrac{{{P_1}}}{{16}}\]

Putting in the expression for the total pressure head, we get
\[{P_1} + \dfrac{{{P_1}}}{{16}} = P \\ \]
\[\Rightarrow \dfrac{{17{P_1}}}{{16}} = P \\ \]
\[\Rightarrow {P_1} = \dfrac{{16P}}{{17}}\]
The total pressure head is given as 1 m of water.
\[{P_1} = \dfrac{{16 \times 1m}}{{17}} \\ \]
\[\therefore {P_1} = 0.94\,m\]
Hence, the pressure head across the first capillary is approximately 0.94 m of water.

Therefore, the correct option is D.

Note: The tubes connected in parallel are analogous to the resistors connected in parallel connection and the tubes connected in series are analogous to the resistors connected in series. This analogy between the fluid flow rate and electric current flow was given by the Hagen-Poiseuille equation.