Water rises up in a glass capillary upto a heigh 10 cm, while mercury falls down by 3.5 cm in the same capillary. If the angles of contact for water-glass and mercury-glass are taken as `0^(@)` and `135^(@)` respectively, compare the surface tensions of water and mercury. [Given density of water = `10^(3)` kg `m^(-3)`, density of mercury = 13.6 `xx 10^(3)` kg `m^(-3)`]

Explanation

For water, `h_(1) = 10 cm`
= 0.1 cm
Density `rho_(1) = 10^(3) kg m^(-3)`
Let the radius of the capillary be = r
Applying the formula `h_(1)rho_(1)g=(2S_(1)cos theta_(1))/(r)`
`rArr S_(1) = (h_(1)rho_(1)gr)/(2 cos theta_(1))`
`=(0.1 xx 10^(3)gr)/(2 cos 0^(@))`
`S_(1) = 100 (gr)/(2)" "...(i)`
For mercury, `h_(2) = -3.5 cm`
= - 0.035 m
Density `rho_(2) = 13.6 xx 10^(3) kg m^(-3)`
then `S_(2) = (h_(2)rho_(2)gr)/(2 cos theta_(2))`
`=(-0.035 xx 13.6 xx 10^(3)gr)/(2 cos(135^(@)))`
`rArr S_(2)=(-0.476)/(-0.71)(gr)/(2)xx10^(3)`
`=0.67 (gr)/(2)xx10^(3)`
`S_(2) = 670 (gr)/(2)" "...(ii)`
Comparing equations (i) and (ii), we get
`(S_(1))/(S_(2))=(100)/(670)=(1)/(6.7)`
This is the required ratio.