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 $0.07 \,N \, m^{-1}$ is $(g = 10\, m\, s^{-2})$

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

Answer 1

Answer: (A)

Rise of a liquid in a capillary tube,
$h=\frac{2S\,cos\,\theta}{r\rho g}$
Here, $r = 0.07 \,cm = 0.07 \times 10^{-2} \,m $
For water, $S=0.07\,N \, m^{-1}, \rho=10^{3}\, kg \, m^{-3}$
Angle of contact $\theta=0^{\circ} $
$\therefore h=\frac{2\times\left(0.07\,N\,m^{-1}\right)\times1}{\left(0.07\times10^{-2}\,m\right)\left(10^{3}\,kg\,m^{-3}\right)\left(10\,m\,s^{-2}\right)}$
$=2 \times10^{-2}\,m=2\, cm$