Water rises to a height of `10 cm` in a capillary tube and mercury falls to a depth of `3.42 cm` in the same capillary tube. If the density of mercury is `13.6 g//c.c.` and the angles of contact for mercury and for water are `135^@` and `0^@`, respectively, the ratio of surface tension for water and mercury is

Options

A. `1:3`
B. `1:4`
C. `1:5.5`
D. `1:6.5`

Answer

D

Explanation

`h=(2T cos theta)/(rrhog)`
`10=(2T_(1)cos 0^@)/(rrho_(1)g)` …(i)
`-3.42=(2T_(2) cos 135^(@))/(r rho_(2)g)`…(ii)
from these two equations, we get
`(T_(1))/(T_(2))=(10.(cos135^(@)))/((-3.42)(cos0^(@))rho_(2))`
`=((10)(-1//sqrt(2))(1))/((-3.42)(1)(13.6))`
`= 1:6.5`