Correct Answer - 2.55 cm ; 0.029 g wt
Here, `r = 0.6 mm = 0.06 cm, S = 75 dyn e cm^(-1),`
`rho = 1 g cm^(-3), theta = 0^@, g = 980 cm s^(-2)`
`h = (2S cos theta)/(r rho g) = (2xx75xxcos 0^@)/(0.006 xx 1xx980) = 2.551 cm`
Weight of water raised `= (pi r^2 h) rho g`
`= 3.14 xx(0.06)^2 xx 2.551 xx 1 xx 980 = 28.26 dyn e`
` =(28.26)/(980) = 0.029 g wt`