Liquid rises to a height of 5.0 cm in a capillary tube and mercury falls to a depth of 2.0 cm in the same capillary tube. If the density of liquid is `1.2 g//c c`, of mercury is `13.6 g//c c` and angles of contact of liquid and mercury with capillary tube are `0^(@)` and `135^(@)` respectively. find the ratio of the surface tension for mercury and liquid.

Explanation

For liquid `h_(1) = 5.0 cm`
` theta_(1) = 0^(@), rho_(1)=1.2 g//c c`
For mercury, `h_(2) = - 2.0 cm`,
`theta_(2) = 135^(@), rho_(2) = 13.6 g//c c`
As, `(2S cos theta)/(r rho g) or S=(hr rho g)/(2 cos theta)` ltbr gt or `S prop (h rho)/(cos theta)`
`:. (S_2)/(S_1) = (h_(2) rho_(2))/(cos theta_(2)) xx (cos theta_(1))/(h_(1)rho_(1))`
`=((-2.0)xx(13.6) xx (cos 0^(@)))/((cos 135^(@)) xx 5.0 xx 1.2)`
`=(2.0 xx 13.6 xx(1))/((1//sqrt(2)) xx 5.0 xx1.2) = 6.41`.