Question

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

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

Answer 1

The correct option is A 2 cm

Rise of a liquid in a capillary tube,

$h=\frac{2Scos\theta }{r\rho g}$

Here, $r=0.07cm=0.07\times {10}^{-2}m$

For water, $S=0.07\times N{m}^{-1},\rho ={10}^{3}kg{m}^{-3}$
Angle of contact $\theta =0^o$

$\therefore h=\frac{2\times \left(0.07N{m}^{-1}\right)\times 1}{(0.007\times {10}^{-2}m)\left({10}^{3}kg{m}^{-3}\right)\left(10m{s}^{-2}\right)}$

$h=2\times {10}^{-2}m=2cm$