Question

Water rises to a height of 10 cm in a capillary tube and mercury falls to a depth of 3.5 cm in the same capillary tube. If the density of mercury is 13.6 g/cc and its angle of contact is $135{}^{\circ }$ and density of water is 1 g/cc and its angle of contact is $0{}^{\circ }$ , then the ratio of surface tensions of the two liquids is: $\left(\text{cos }135{}^{\circ }=0.7\right)$

• A. 1 : 14
• B. 5 : 34
• C. 1 : 5
• D. 5 : 25

Answer 1

Answer: (B)

The rise or fall of liquid in the capillary tube is given by $h\rho g=2T\cos \theta \Rightarrow h=\frac{2T\cos \theta }{h\rho g}$ Hence, surface tension $T=\frac{hr\rho g}{2\cos \theta }$ where r is radius of capillary Given: For watery $h_1=10cm,$ $h_2=3.5cm$ (for mercury) Density of watery $d_1=1g/cc,$ Density of mercury $d_2=13.6g/cc$ Angle of contact ${\theta }_1=0$ (for water), angle of contact ${\theta }_2={135}^o$ °(for mercury) In first case $T_1=\frac{10\times r\times 1\times g}{2\cos \theta }=\frac{10rg}{2\cos \theta }=\frac{10rg}{2}=5rg$ ?(i) $T_2=\frac{3.5\times r\times 13.6\times g}{2\cos {135}^o}=\sqrt{2}\times 3.5\times 6.8rg$ ?(ii) Dividing Eq. (ii) by Eq. (i), we get $\frac{T_1}{T_2}=\frac{5rg}{\sqrt{2}\times 3.5\times 6.8rg}=\frac{5rg}{33.65rg}=\frac{5}{34}$