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  • A liquid flows steadily through a series combination of three capillary tubes of radii r, 2 r and 3 r, all of the same length L. If the pressure difference across the combination is <img alt="

A liquid flows steadily through a series combination of three capillary tubes of radii r, 2 r and 3 r, all of the same length L. If the pressure difference across the combination is 28 \mathrm{~cm} of mercury, the pressure difference (in cm of Hg ) across the tube of radius 2 \mathrm{r} is very nearly equal to

Option: 1

1.6


Option: 2

3.2


Option: 3

7.0


Option: 4

9.6


Answers (1)

best_answer

\mathrm{Q_1=Q_2=Q_3 \text {. Therefore, }}

\mathrm{\begin{aligned} & & \frac{\pi p_1 r^4}{8 \eta L} & =\frac{\pi p_2(2 r)^4}{8 \eta L}=\frac{\pi p_3(3 r)^4}{8 \eta L} \\\\ \Rightarrow & & p_1 & =16 p_2=81 p_3 \\\\ & \therefore & p_1 & =16 p_2 \end{aligned}}

\mathrm{\text { and } \quad p_3=\frac{16}{81} p_2}

\mathrm{\text { Given } \quad p_1+p_2+p_3=28 \mathrm{~cm} \text { of } \mathrm{Hg}}

\mathrm{\begin{aligned} & \Rightarrow \quad 16 p_2+p_2+\frac{16}{81} p_2=28 \\ & \Rightarrow \quad\left(16+1+\frac{16}{81}\right) p_2=28 \\ & \Rightarrow \quad p_2=\frac{81 \times 28}{1393}=1.6 \mathrm{~cm} \mathrm{of} \mathrm{Hg} \end{aligned}}

So the correct choice is (a). 

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Gaurav

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