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 $2.72\,cm$ 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 $13.6$ , $cos 140 ^\circ = - 0 . 7660$

Answer 1

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