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

Answer

D

Explanation

Rise of a liquid in a capillary tube is given by
`h = (2 S cos theta)/(r rhog)`
For water , `S = 0.07 Nm ^(-1) , rho = 10 ^(3) kgm^(-3)`
Angle of contact `theta = 0^(@)`
`therefore h=(2 xx (0.07 Nm ^(-1)) xx 1 )/(( 0.07 xx 10 ^(-2) m ) (10^(3) kgm ^(-3)) (10 ms ^(-2)))`
`= 2 xx 10 ^(-2) m = 2 cm`