Find ratio of surface tension of water to that of mercury if water rises to height of 2.72 cm in a capillary tube while mercury depresses by 1 cm in same tube. Given angle of contact for water and mercury is 0° and 140° respectively. (Take `cos 140^@ = -0.7660` )

Answer

`00.15`

Explanation

As, `T = (rh rho g)/(2 cos theta)`
When tube is dipped in water:
`T_1 prop (h, rho_1)/(cos theta_1) =(2.72xx10^2xx 10^3)/(cos theta) =27.2`
When tube is dipped in mercury:
`T_2 prop (h_2 rho_2)/(cos theta_2) =((-10^(-2))xx13.6xx10^3)/(-0.7660) =136/0.7660`
`:. T_1/T_2 =27.2/136 xx 0.7660 = 0.7660/5 = 0.1532`
`:. T_1/T_2 = 0.15`
.... (Rounding off to 2 decimal places)