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A capillary tube of radius $0.1 \mathrm{~mm}$ is dipped in water. The water rises to a height of $2 \mathrm{~cm}$ in the tube. If the surface tension of water is $0.072 \mathrm{Nm}^{-1}$, the contact angle between water and wall of the tube is
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Verified Answer
The correct answer is:
$\theta=\cos ^{-1}\left(\frac{1}{7.2}\right)$
Radius of capillary tube, $\mathrm{r}=0.1 \mathrm{~mm}=0.1 \times 10^{-3} \mathrm{~m}$ Height of water rises, $\mathrm{h}=2 \mathrm{~cm}=02 \mathrm{~m}$
Surface tension, $T=0.072 \mathrm{~N} / \mathrm{m}$
Rise in water is given by
$\begin{aligned}
& \mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho g r} \\
& \cos \theta=\frac{\mathrm{h} \rho \mathrm{gr}}{2 \mathrm{~T}} \\
& =\frac{2 \times 10^{-2} \times 10^3 \times 10 \times 0.1 \times 10^{-3}}{2 \times 0.072} \\
& \theta=\cos ^{-1}\left(\frac{1}{7.2}\right)
\end{aligned}$
Surface tension, $T=0.072 \mathrm{~N} / \mathrm{m}$
Rise in water is given by
$\begin{aligned}
& \mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho g r} \\
& \cos \theta=\frac{\mathrm{h} \rho \mathrm{gr}}{2 \mathrm{~T}} \\
& =\frac{2 \times 10^{-2} \times 10^3 \times 10 \times 0.1 \times 10^{-3}}{2 \times 0.072} \\
& \theta=\cos ^{-1}\left(\frac{1}{7.2}\right)
\end{aligned}$
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