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(D) Given: Lengths are equal,$l_1 = l_2 = l$. Radii ratio is $\frac{r_1}{r_2} = \frac{1}{2}$.For a liquid flowing in a streamlined condition through a

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$0.94$

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(D) Given: Lengths are equal,$l_1 = l_2 = l$. Radii ratio is $\frac{r_1}{r_2} = \frac{1}{2}$.For a liquid flowing in a streamlined condition through a capillary,the volume flow rate $V$ is given by Poiseuille's equation: $V = \frac{\pi P r^4}{8 \eta l}$.Since the capillaries are connected in series,the volume flow rate $V$ through both must be the same.Therefore,$\frac{\pi P_1 r_1^4}{8 \eta l} = \frac{\pi P_2 r_2^4}{8 \eta l}$.This simplifies to $P_1 r_1^4 = P_2 r_2^4$,or $\frac{P_1}{P_2} = \left( \frac{r_2}{r_1} \right)^4$.Given $\frac{r_1}{r_2} = \frac{1}{2}$,we have $\frac{r_2}{r_1} = 2$. Thus,$\frac{P_1}{P_2} = (2)^4 = 16$,which means $P_1 = 16 P_2$.The total pressure difference across the combination is $P = P_1 + P_2 = 1 \ m$.Substituting $P_2 = \frac{P_1}{16}$ into the equation: $P_1 + \frac{P_1}{16} = 1$.$\frac{17 P_1}{16} = 1 \implies P_1 = \frac{16}{17} \approx 0.94 \ m$.

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

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