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 $135°$ and $0°$, 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, $\left(\frac{r_1}{r_2}\right)$, is then close to

• A. $3/5$
• B. $2/3$
• C. $2/5$
• D. $4/5$

Answer 1

The correct option is C $2/5$

Given:
$\frac{T_{Hg}}{T_{\text{water}}}=7.5$
$\frac{{\rho }_{Hg}}{{\rho }_W}=13.6$
${\theta }_{Hg}=135°,{\theta }_W=0°$

$\frac{\cos {\theta }_{Hg}}{\cos {\theta }_W}=\frac{\cos 135°}{\cos 0°}=\frac{1}{\sqrt{2}}$

As we know that

$\frac{2T\cos \theta }{r\rho g}=h$

$\frac{(r_1{)}_{Hg}}{(r_2{)}_W}=\left(\frac{T_{Hg}}{T_W}\right)\left(\frac{{\rho }_W}{{\rho }_{Hg}}\right)\left(\frac{\cos {\theta }_{Hg}}{\cos {\theta }_W}\right)=7.5\times \frac{1}{13.6}\times \frac{1}{\sqrt{2}}=0.4=\frac{2}{5}$

$\frac{\left(r_1\right)}{\left(r_2\right)}=\frac{2}{5}$

Hence option $\left(B\right)$ is correct.