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}}=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 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} $