Three liquids have same surface tension and densities $\varrho_{1}, Q_{2}$, and…
Three liquids have same surface tension and densities $\varrho_{1}, Q_{2}$, and $\varrho_{3}\left(\varrho_{1}>\varrho_{2}>\varrho_{3}\right)$.
In three identical capillaries, rise of liquid is same. The corresponding angles of contact $\theta_{1}, \theta_{2}$ and $\theta_{3}$ are related as
$\theta_{1}>\theta_{2}>\theta_{3}$
$\theta_{1} < \theta_{2}>\theta_{3}$
$\theta_{1}>\theta_{2} < \theta_{3}$
$\theta_{1} < \theta_{2} < \theta_{3}$
Solution
Rise of a liquid in a capillary tube is given by,
$\mathrm{h}=\frac{2 \mathrm{T} \cos \theta}{\mathrm{r} \rho \mathrm{g}}$
or $\cos \theta=h r \rho g / 2 T$
where $\theta=$ angle of contact,
$\mathrm{r}=$ radius of capillary tube
$\mathrm{T}=$ surface tension
$\rho=$ density of liquid
now given that $h$, $T$ and $r$ are constants for all three liquids,
and $\rho_{1}>\rho_{2}>\rho_{3}$,
therefore $\cos \theta_{1}>\cos \theta_{2}>\cos \theta_{3}$
or $\theta_{1} < \theta_{2} < \theta_{3}$
now as the liquid is rising in all three capillaries therefore angles of contact will be acute,
$0 \leq \theta_{1} < \theta_{2} < \theta_{3} < \frac{\pi}{2}$