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. 1:05
• D. 5:25

Answer 1

Answer: (B)

5:34

Answer 2

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 ${{h}_{1}}=10\,cm,$ ${{h}_{2}}=3.5\,cm$ (for mercury) Density of watery ${{d}_{1}}=1g/cc,$ Density of mercury ${{d}_{2}}=13.6\,g/cc$ 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) ${{T}_{2}}=\frac{3.5\times r\times 13.6\times g}{2\cos {{135}^{o}}}=\sqrt{2}\times 3.5\times 6.8\,rg$ ?(ii) Dividing E (ii) by E (i), we get $\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{5rg}{\sqrt{2}\times 3.5\times 6.8rg}=\frac{5rg}{33.65rg}=\frac{5}{34}$