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: 24`
C. `1: 5`
D. `5: 27`

Answer

B

Explanation

As `h= (2S cos theta)/(r rho g):. S= (hr rho g)/(2 cos theta)`
For the same capillarity, `S prop (h rho)/(cos theta)`
`:. (S_(w))/(S_(Hg)) = (h_(w) rho_(w))/(h_(Hg) rho_(Hg)) xx (cos 135^(@))/(cos 0^(@)) = (10 xx 1 xx (-0.7))/((-3.5) xx 13.6 xx 1) = (5)/(34)`