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 cross first capillary is

The correct answer is 0.94 m.

Step-by-step derivation:

1. According to Poiseuille's Law, the rate of flow of a liquid ($Q$) through a capillary tube of length $L$ and radius $r$ under a pressure difference $P$ is given by the formula:
$Q=\frac{\pi P{r}^4}{8\eta L}$
where $\eta$ is the coefficient of viscosity of the liquid.

2. We can define the liquid flow resistance ($R$) of the capillary tube as the ratio of the pressure difference to the rate of flow:
$R=\frac{P}{Q}=\frac{8\eta L}{\pi r^4}$

3. Since the two capillaries have the same length ($L_1=L_2=L$) and the same liquid flows through both, the resistance is inversely proportional to the fourth power of the radius:
$R\propto \frac{1}{r^4}$

4. The ratio of the radii of the two capillaries is given as $r_1:r_2=1:2$. Therefore, the ratio of their resistances is:
$\frac{R_1}{R_2}={\left(\frac{r_2}{r_1}\right)}^4={\left(\frac{2}{1}\right)}^4=16$
This gives $R_1=16R_2$.

5. In a series connection, the rate of flow ($Q$) of the liquid is the same through both capillaries. Thus:
$Q=\frac{P_1}{R_1}=\frac{P_2}{R_2}$
where $P_1$ is the pressure difference across the first capillary and $P_2$ is the pressure difference across the second capillary. This relation implies:
$\frac{P_1}{P_2}=\frac{R_1}{R_2}=16\Rightarrow P_2=\frac{P_1}{16}$

6. The total pressure difference across the combination of the two capillaries is 1 m of water. Therefore:
$P_{\text{total}}=P_1+P_2=1\text{m}$

7. Substitute $P_2$ in terms of $P_1$ into the equation:
$P_1+\frac{P_1}{16}$
$\frac{17}{16}$
$P_1=\frac{16}{17}$

Thus, the pressure difference across the first capillary is approximately 0.94 m of water.