Water rises to a height of 10 cm. in a certain capillary tube. If in the same tube, level of Hg is depressed by 3.42 cm., compare the surface tension of water and mercury. Sp. Gr. Of Hg is 13.6 the angle of contact for water is zero and that for Hg is `135^@.`

Answer

0.152

Explanation

Let `S_1, S_2` be the surface tension of water and mercury respectively. Let h be the height of water raised due to S.T. and `h_2` the depression of mercury. Then `h_1 = 10 cm , h_2 = - 3.42 cm.` Angle of contant for water and capillary tube, `theta_1 = 0^@` and for mercury and capillary tube, `theta_2 = 135^@.`
`h_1 = (2S_1 cos theta_1)/(r rho_1g) and h_2 = (2S_2 cos theta_2)/(r rho_2g)`
`:.(h_1)/(h_2) = (S_1 cos theta_1rho_2)/(S_2 cos theta_2 rho_1) or (S_1)/(S_2) = (h_1 rho_1 cos theta_2)/(h_2 rho cos theta_1)`
`or (S_1)/(S_2) = (10 xx 1xxcos 135^@)/((-3.42)xx 13.6 xx cos 0^@)`
`=(10xx(-0.7071))/((-3.42)xx 13.6) = 0.152`