The rise in the water level in a capillary tube of radius 0.07 cm when dipped veryically in a beaker containing water of surface tension `0.07 N m^(-1)` is (g = `10 m s^(-2)`)

Options

A. 2cm
B. 4cm
C. 1.5 cm
D. 3 cm

Answer

A

Explanation

Rise of water in a capillary tube is `h= (2S cos theta)/(r rho g)`
Here, `r= 0.07 cm = 0.07 xx 10^(-2)m`
`S= 0.07 Nm^(-1), rho = 10^(3) kg m^(-3), theta = 0^(@)`
`:. H= (2 xx (0.07 N m^(-1)) xx 1)/((0.07 xx 10^(-2)m) (10^(3) kg m^(-3)) (10 ms^(-2)))`
`=2 xx 10^(-2) m = 2cm`