One end of a uniform glass capillary tube of radius $r=0.025 \mathrm{~cm}$ is immersed vertically in water…
One end of a uniform glass capillary tube of radius $r=0.025 \mathrm{~cm}$ is immersed vertically in water to a depth $h=1 \mathrm{~cm}$. The excess pressure in $\mathrm{N} / \mathrm{m}^2$ required to blow an air bubble out of the tube
(Surface tension of water $=7 \times 10^{-2} \mathrm{~N} / \mathrm{m}$ Density of water $=10^3 \mathrm{~kg} / \mathrm{m}^3$. Acceleration due to gravity $=10 \mathrm{~m} / \mathrm{s}^2$ )
$0.0048 \times 10^5$
$0.0066 \times 10^5$
$1.0048 \times 10^5$
$1.0066 \times 10^5$
Solution
Excess pressure
$
p=h d g+h^{\prime} d g \Rightarrow p=d g\left(h+h^{\prime}\right)
$
where $h$ is capillary rise $=\frac{2 T}{r d g}$
$
\begin{aligned}
& =\frac{2 \times 7 \times 10^{-2}}{25 \times 10^{-5} \times 10^3 \times 10} \\
& =0.056 \mathrm{~m} \\
\therefore \quad p & =10^3 \times 10[0.056+0.01] \\
& =0.066 \times 10^4 \\
& =0.0066 \times 10^5 \mathrm{Nm}^{-2}
\end{aligned}
$