Question

If the ratio of surface tension of water to that of mercury is $\frac{x}{100}$ then find $x$ . if water rises to height of $2.72cm$ in a capillary tube while mercury depresses by $1cm$ in same tube. Given angle of contact for water and mercury is $0^{\circ }$ and $140^{\circ }$ respectively. (Take specific gravity of mercury is $13.6$ , $cos140^{\circ }=-0.7660$

Answer 1

Answer: 15

As, $T=\frac{rh\rho g}{2cos\theta }$
When tube is dipped in water:
$\frac{h_1{\rho }_1}{cos{\theta }_1}=\frac{2.72\times 10^{-2}\times 10^3}{1}=27.2$
When tube is dipped in mercury:
$\frac{h_2{\left(\rho \right)}_2}{cos{\left(\theta \right)}_2}=\frac{\left(-{\left(10\right)}^{-2}\right)\times 13.6\times {\left(10\right)}^3}{-0.7660}=\frac{136}{0.7660}$
$\therefore \frac{T_1}{T_2}=\frac{27.2}{136}\times 0.7660=\frac{0.7660}{5}=0.1532$
$\therefore \frac{T_1}{T_2}=0.15$
.... (Rounding off to $2$ decimal places)