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A glass capillary tube of internal radius $\mathrm{r}=0.25$ $\mathrm{mm}$ is immersed in water. The top end of the tube projected by $2 \mathrm{~cm}$ above the surface of the water. At what angle does the liquid meet the tube? Surface tension of water $=0.7 \mathrm{~N} / \mathrm{m}$
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Verified Answer
The correct answer is:
$\quad \theta=70^{\circ}$
Water wets glass and so the angle of contact is zero. For full rise, neglecting the small mass in the meniscus
$$
\begin{aligned}
2 \pi \mathrm{T} &=\pi \mathrm{r}^{2} \mathrm{~h} \rho \mathrm{g} \Rightarrow \mathrm{h}=\frac{2 \mathrm{~T}}{\mathrm{r} \rho \mathrm{g}}\left[\because \text { water wets glass, } \theta=0^{\circ}\right] \\
&=\frac{2 \times 007}{0.25 \times 10^{-3} \times 1000 \times 9.8}
\end{aligned}
$$
As the tube is only $2 \mathrm{~cm}$ above the water and so, water will rise by $2 \mathrm{~cm}$ and meet the tube at an angle such that,
$$
\begin{aligned}
2 \pi r & \operatorname{T} \cos \theta=\pi r^{2} h^{\prime} \rho g \\
\Rightarrow & 2 T \cos \theta=h^{\prime} r \rho g \\
\Rightarrow \cos \theta &=\frac{h^{\prime} r \rho g}{2 T} \\
&=\frac{2 \times 10^{-2} \times 0.25 \times 10^{-3} \times 1000 \times 9.8}{2 \times 0.07}
\end{aligned}
$$
The liquid will meet the tube at an angle, $\theta$ $\cong 70^{\circ}$
$$
\begin{aligned}
2 \pi \mathrm{T} &=\pi \mathrm{r}^{2} \mathrm{~h} \rho \mathrm{g} \Rightarrow \mathrm{h}=\frac{2 \mathrm{~T}}{\mathrm{r} \rho \mathrm{g}}\left[\because \text { water wets glass, } \theta=0^{\circ}\right] \\
&=\frac{2 \times 007}{0.25 \times 10^{-3} \times 1000 \times 9.8}
\end{aligned}
$$
As the tube is only $2 \mathrm{~cm}$ above the water and so, water will rise by $2 \mathrm{~cm}$ and meet the tube at an angle such that,
$$
\begin{aligned}
2 \pi r & \operatorname{T} \cos \theta=\pi r^{2} h^{\prime} \rho g \\
\Rightarrow & 2 T \cos \theta=h^{\prime} r \rho g \\
\Rightarrow \cos \theta &=\frac{h^{\prime} r \rho g}{2 T} \\
&=\frac{2 \times 10^{-2} \times 0.25 \times 10^{-3} \times 1000 \times 9.8}{2 \times 0.07}
\end{aligned}
$$
The liquid will meet the tube at an angle, $\theta$ $\cong 70^{\circ}$
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