A) 1 : 14
B) 5 : 34
C) 1 : 5
D) 5 : 25
Correct Answer: B
Solution :
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 \[\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{5rg}{\sqrt{2}\times 3.5\times 6.8rg}=\frac{5rg}{33.65rg}=\frac{5}{34}\]
You need to login to perform this action.
You will be redirected in
3 sec
