© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Write down the formula for capillary rise (or depression)
The height (or depth) to which a liquid rises (or is depressed) in a capillary tube is given by
$h = \dfrac{2S \cos \theta}{r \,\rho\, g},$
where:
$S$ is the surface tension of the liquid,
$\theta$ is the contact angle with the tube material,
$r$ is the radius of the capillary tube,
$\rho$ is the density of the liquid,
$g$ is the acceleration due to gravity.
Step 2: Set up the equality for the same magnitude of rise and depression
According to the problem statement, mercury is depressed by an amount $h$ in a tube of radius $r_1$, and water rises by the same amount $h$ in a tube of radius $r_2$. Therefore, we can write:
$$
\dfrac{2S_{\text{Hg}} \cos \theta_{\text{Hg}}}{r_1 \rho_{\text{Hg}} \, g}
=
\dfrac{2S_{\text{water}} \cos \theta_{\text{water}}}{r_2 \rho_{\text{water}} \, g}.
$$
Step 3: Rearrange to find $r_1 / r_2$
Canceling the common factor of $2g$ on both sides, we get:
$$
\dfrac{S_{\text{Hg}} \cos \theta_{\text{Hg}}}{r_1 \rho_{\text{Hg}}}
=
\dfrac{S_{\text{water}} \cos \theta_{\text{water}}}{r_2 \rho_{\text{water}}}
\quad\Longrightarrow\quad
\dfrac{r_1}{r_2}
=
\dfrac{S_{\text{Hg}} \cos \theta_{\text{Hg}} \,\rho_{\text{water}}}
{S_{\text{water}} \cos \theta_{\text{water}} \,\rho_{\text{Hg}}}.
$$
Step 4: Substitute given values
From the question:
The ratio of surface tensions $S_{\text{Hg}} : S_{\text{water}}$ is $7.5:1$.
The ratio of densities $\rho_{\text{Hg}} : \rho_{\text{water}}$ is $13.6:1$.
The contact angles are $\theta_{\text{Hg}} = 135^\circ$ and $\theta_{\text{water}} = 0^\circ$. Hence,
$\cos 135^\circ$ has magnitude $|\cos 135^\circ| = \dfrac{\sqrt{2}}{2}$ (but is negative for depression),
and $\cos 0^\circ = 1$.
Using magnitudes in the ratio, we get:
$$
\dfrac{r_1}{r_2}
\approx
\dfrac{\bigl(7.5 \times \dfrac{\sqrt{2}}{2}\bigr)\times 1}
{1 \times 1 \times 13.6}
\approx
\dfrac{7.5 \times 0.707}{13.6}
\approx
0.39 \approx \dfrac{2}{5}.
$$
Step 5: Conclude the final ratio
Hence, the ratio of the radii is close to
$$
\dfrac{r_1}{r_2} = \dfrac{2}{5}.
$$
