Question

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

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

Answer 1

Answer: (A)

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, $S=0.07N{m}^{-1},\rho =10^{3}kg{m}^{-3}$
Angle of contact $\theta =0^{\circ }$
$\therefore h=\frac{2\times \left(0.07N{m}^{-1}\right)\times 1}{\left(0.07\times 10^{-2}m\right)\left(10^{3}kg{m}^{-3}\right)\left(10m{s}^{-2}\right)}$
$=2\times 10^{-2}m=2cm$