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Mercury has an angle of contact equal to $140^{\circ}$ with sodalime glass. A narrow tube of radius $1 \mathrm{~mm}$ made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the mercury surface outside? Surface tension of mercury at the temperature of the experiment is $0.465 \mathrm{Nm}^{-1}$. Density of mercury $=13.6 \times 10^3 \mathrm{kgm}^{-3}$.
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Given, $\theta=140^{\circ}, \mathrm{r}=1 \times 10^{-3} \mathrm{~m}, \mathrm{~s}=0.465 \mathrm{Nm}^{-1}$;
$$
\rho=13.6 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}, \cos 140^{\circ}=-\cos 40^{\circ}=-0.766
$$
Now, $\mathrm{h}=\frac{2 \mathrm{~s} \cos \theta}{\mathrm{r \rho g}}=\frac{2 \times 0.465 \times \cos 140^{\circ}}{10^{-3} \times 13.6 \times 10^3 \times 9.8}$
$$
=-5.34 \mathrm{~mm}
$$
Negative value indicates that mercury level is depressed in the tube.
$$
\rho=13.6 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}, \cos 140^{\circ}=-\cos 40^{\circ}=-0.766
$$
Now, $\mathrm{h}=\frac{2 \mathrm{~s} \cos \theta}{\mathrm{r \rho g}}=\frac{2 \times 0.465 \times \cos 140^{\circ}}{10^{-3} \times 13.6 \times 10^3 \times 9.8}$
$$
=-5.34 \mathrm{~mm}
$$
Negative value indicates that mercury level is depressed in the tube.
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