A capillary tube of radius `r` is immersed in water and water rises to a height of `h` mass of water in the capillary tube is `5xx10^(-3)` kg the same capillary tube is now immersed in a liquid whose surface tension in `sqrt(2)` times the surface tension of water. The angle of contact between the capillary tube and this liquid is `45^(@)` the mass of liquid which rises into the capillary tube now is (in kg)

Explanation

Height of water rise in a capillary tube
`h=(2T cos theta)/(rdg),h_(1)=(2T cos theta_(1))/(rdg),h_(2)=(2T_(2)cos theta_(2))/(rdg)`
`"Given, "h_(1)=h, T_(1)=T, theta_(1)=0," "therefore h=(2T)/(rdg)`
`"Given "T_(2)=sqrt2T, theta=45^(@), cos 45^(@)=(1)/(sqrt2)`
`therefore h_(2)=(2sqrt2Txx(1)/(sqrt2))/(rdg)`
From Eqs (i) and (ii), we observe `h_(2)=h`
Hence, same mass of liquid rises into the capillary as before `5xx10^(-3)kg`.