Question

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. $0.94 \, m$

Answer 1

Answer: (D)

$0.94 \, m$

Answer 2

Here, $l_{1}=l_{2}=1 m$ and $\frac{r_{1}}{r_{2}}=\frac{1}{2}$ As $V=\frac{\pi P_{1}r_{1}^{4}}{8 \eta l}=\frac{\pi P_{2}r_{2}^{4}}{8 \eta l}$ or $\frac{P_{1}}{P_{2}}=\left(\frac{r_{2}}{r_{1}}\right)^{4}=16$ $\therefore P_{1}=16 P_{2}$ Since, both tubes are connected in series, hence pressure difference across the combination is $P=P_{1}+P_{2}$ $\Rightarrow 1=P_{1}+\frac{P_{1}}{16}$ or $P_{1}=\frac{16}{17}=0.94\,m$