(ii) A liquid of density `rho` and surface tension S rises to a height h in a capillary tube of diameter D. What is the weight of the liquid in the capillary tube? Angel of contact is `0^@`.

Explanation

Rise of liquid in the capillary tube is given as,
`h=(2Scostheta)/(rho r g) `
Here, `r=(D)/(2). Theta=0^@., cos0^@=1`
So, `h=(2Sxx1)/(rhoxxD/2xxg)=(4S)/(rho Dg)`
Now, weight of liquid in the capillary tube,
W = mass `xx` = m (volume `xx` density )`xx`g
`=(pir^2xxh)xxrhoxxg`
`=pi(D^2)/(4)xx((4S)/(rhoD g))xxrhoxxg=piDS`
This is the required result.