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 tu8be 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=(2Tcostheta)/(rdg),h_(1)=(2T_(1)costheta_(1))/(rdg),h_(2)=(2T_(2)costheta_(2))/(rdg)`
Given, `h_(1)=h,T_(1)=T,theta_(1)=0,becauseh=(2T)/(rdg)`
Given `T_(2)=sqrt(2)T,theta=3435^(@),cos45^(@)=(1)/(sqrt(2))`
`becauseh_(2)=(2sqrt(2)Txx(1)/(sqrt(2)))/(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`