If the surface tension of water is 0.06 Nm, then the capillary rise in a tube of diameter 1mm is `(theta=0^(@)`)

Options

A. 1.22cm
B. 2.44cm
C. 3.12cm
D. 3.86 cm

Answer

B

Explanation

Here, `S= 0.06 N m^(-1), theta = 0^(@), r =(1)/(2) mm = 0.5 xx 10^(-3)m, rho = 10^(3) kg m^(-3), g = 9.8 ms^(-2)`
The rise of water in a capillary tub is
`h= (2S cos theta)/(r rho g) = (2 xx 0.06 xx cos 0^(@))/(0.5 xx 10^(-3) xx 10^(3) xx 9.8)`
`=2.44 xx 10^(-2) m= 2.44cm`