Water raises 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(gm)/(c.c)` and its angle of contact is `135^(@)` and density of water is `1(gm)/(c.c)` and its angle of contact is `0^(@C)` then the ratio of surface tensions of two liquids is `(cos135^(@)=0.7)`

Options

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

Answer

B

Explanation

The rise or fall of liquid in the capillary tube is given by
`h=(2Tcostheta)/(r rhog)`
Hence, surface tension `T=(hr rhog)/(2costheta)`
where, r is radius of capillary
Given, For water `h_(1)=3.5` cm
`h_(2)=3.5cm` (for mercury)
Density of mercury `d_(1)=1g//c c`
angle of contact `theta_(1)=0` (for water)
angle of contact
`T_(1)=(10xxrxx1xxg)/(2costheta)=(10g)/(2costheta)`
`=(10rg)/(2)=5rg` . . . (i)
In second case,
`T_(2)=(3.5xxrxx13.6xxg)/(2cos135^@)`
`=sqrt(2)xx3.5xx6.8rg` . . . (ii)
On dividing eq. (ii) by Eq. (i), we get
`(T_(1))/(T_(2))=(5rg)/(sqrt(2)xx3.5xx6.8rg)`
`=(5rg)/(33.65rg)~~(5)/(34)`