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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, then pressure difference across first capillary is

Option: 1

9.4 m


Option: 2

4.9 m


Option: 3

0.49 m


Option: 4

0.94 m


Answers (1)

best_answer

\text { Given, } l_1=l_2=1 \mathrm{~m} \text { and } \frac{r_1}{r_2}=\frac{1}{2}

v=\frac{\pi \rho_1 r_1 4}{8 \eta l}=\frac{\pi \rho_2 r_2 4}{8 \eta l}

\Rightarrow \quad \frac{\rho_1}{\rho_2}=\left(\frac{r_2}{r_1}\right)^4=16

p_1=16 p_2

since, both tubes are connected inceries, hence pressure difference across combination.

\rho=\rho_1+p_2 \Rightarrow 1=\rho_1+\frac{p_1}{16} \Rightarrow p_1=\frac{16}{17}=0.94 \mathrm{~m}

Posted by

Deependra Verma

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