One end fo a capillary tube of radius r is immersed verticaly in water and the mass of water rises in the capillary tube is 5 g. if one end of another capillary tube of radius 2 r is immersed vertically in water, what will be the mass of water that will rise in it ?

Answer

10 g

Explanation

Let h be the height through which the liquid rises in the capillary tube of radius r. Then `h = (2S cos theta)/(r rho g)` and mass of water in the first tube is `m = pi r^2 h rho = pi r^2 xx ((2S cos theta)/(r rho g))xxrho`
` = (pi r 2 S cos theta)/(g)`
` i.e., m prop r`
` :. (m')/(m) = (r')/(r ) = (2r)/(r ) = 2 or m' = 2m = 2xx 5 = 10 g`