Question

The rise of the water level in a capillary tube of radius $0.07\text{cm}$ when dipped vertically in a beaker containing water (assuming it to be perfectly wetting) will be.
Given : Surface tension of water is $0.07{\text{Nm}}^{-1}$ and $g=10{\text{ms}}^{-2}$

• A. $2\text{cm}$
• B. $4\text{cm}$
• C. $1.5\text{cm}$
• D. $3\text{cm}$

Answer 1

The correct option is A $2\text{cm}$

Rise of a liquid in a capillary tube,
$h=\frac{2T\cos \theta }{r\rho g}$
Here, $r=0.07\text{cm}=0.07\times {10}^{-2}\text{m}$
For water, $T=0.07{\text{Nm}}^{-1},\rho ={10}^3{\text{kg m}}^{-3}$
Angle of contact $\theta =0^{\circ }$, for perfectly wetting characteristics.
$\Rightarrow h=\frac{(2\times 0.07\times 1)}{(0.07\times {10}^{-2})\times \left({10}^3\right)\times \left(10\right)}$
$\therefore h=2\times {10}^{-2}\text{m}=2\text{cm}$