Question

The ratio of surface tensions of mercury and water is given to be 7.5 while the ratio of their densities is 13.6. Their contact angles, with glass, are close to \[{135^ \circ }\] and \[{0^ \circ }\], respectively. It is observed that mercury gets depressed by an amount h in a capillary tube of radius \[{r_1}\], while water rises by the same amount h in a capillary tube of radius ${r_2}$. The ratio, $\left( {\dfrac{{{r_2}}}{{{r_1}}}} \right)$, is then close to:
A) $\dfrac{2}{3}$
B) $\dfrac{3}{5}$
C) $\dfrac{2}{5}$
D) $\dfrac{4}{5}$

Answer 1

Hint: The surface of liquids such as water and mercury appear as if they are stretched. This is due to the phenomenon of surface tension. Due to the surface tension, the liquid has the tendency to maintain the most minimum surface area.

Formula used:
Height of liquid column raised by surface tension is given by:
$h = \dfrac{{2S\cos \theta }}{{r\rho g}}$....................(1)
Where,
h is the height of the liquid column,
S is surface tension,
$\theta $ is the contact angle,
r is the radius of the capillary tube,

Complete step by step answer:
Given,
Ratio of surface tension for mercury and water is 7.5, i.e. $\dfrac{{{S_2}}}{{{S_1}}} = 7.5 = \dfrac{{15}}{2}$.
Contact angles of mercury and water are ${\theta _2} = {135^ \circ }$ and ${\theta _1} = {0^ \circ }$ respectively.
Ratio of their density is 13.6.
Thus, ratio is given by
$\dfrac{{{\rho _2}}}{{{\rho _1}}} = 13.6$
Therefore,
Height raised in the water column is the same as the height dip of the mercury column. Hence, $\dfrac{{{h_1}}}{{{h_2}}} = - 1$.
To find:
Value of the ratio $\dfrac{{{r_2}}}{{{r_1}}}$.
Substituting the expression for height of the liquid column in the ratio of their heights and rearranging, we obtain –
$\dfrac{{{h}_{1}}}{{{h}_{2}}}=\dfrac{\dfrac{2{{S}_{1}}\cos {{\theta }_{1}}}{{{r}_{1}}{{\rho }_{1}}g}}{\dfrac{2{{S}_{2}}\cos {{\theta }_{2}}}{{{r}_{2}}{{\rho }_{2}}g}}$
$\Rightarrow \dfrac{{{h}_{1}}}{{{h}_{2}}}=\dfrac{{{S}_{1}}}{{{S}_{2}}}\times \dfrac{\cos {{\theta }_{1}}}{\cos {{\theta }_{2}}}\times \dfrac{{{r}_{2}}}{{{r}_{1}}}\times \dfrac{{{\rho }_{2}}}{{{\rho }_{1}}}$
$\Rightarrow \dfrac{{{r}_{2}}}{{{r}_{1}}}=\dfrac{{{h}_{1}}}{{{h}_{2}}}\times \dfrac{{{S}_{2}}}{{{S}_{1}}}\times \dfrac{\cos {{\theta }_{2}}}{\cos {{\theta }_{1}}}\times \dfrac{{{\rho }_{1}}}{{{\rho }_{2}}}$
Substituting the values of the above ratios, we get –
$\dfrac{r_2}{r_1} = ( - 1) \times 7.5 \times \dfrac{{\cos ({{135}^\circ })}}{{\cos ({0^\circ })}} \times \dfrac{1}{{13.6}} = \dfrac{{7.5}}{{\sqrt 2 \times 13.6}} = 0.39$
$\therefore \dfrac{{{r_2}}}{{{r_1}}} = 0.39 \approx \dfrac{2}{5}$
The ratio, $\left( {\dfrac{{{r_2}}}{{{r_1}}}} \right)$, is approximately equal to $\dfrac{2}{5}$.

Hence, the correct option is Option C.

Note: The contact angle for water and mercury are \[{0^ \circ }\] and \[{135^ \circ }\] respectively. So, while finding the cosine term present in the height expression, a negative sign arises for mercury. Physically this means that instead of increasing, the height of mercury decreases inside the capillary tube.