Question

When a capillary tube is dipped in a liquid, liquid rises $0.06\text{m}$ in it. Surface tension of liquid is $0.03\text{N/m}$, density of liquid is ${10}^3{\text{kg/m}}^3$. If contact angle is ${60}^{\circ }$, then radius of capillary is (Assume $g=10{\text{m/s}}^2$)

• A. $1\text{mm}$
• B. $0.5\text{mm}$
• C. $0.05\text{mm}$
• D. $0.01\text{mm}$

Answer 1

The correct option is C $0.05\text{mm}$

Given:
$h=0.06\text{m},\theta ={60}^{\circ }$
$T=0.03\text{N/m},\rho ={10}^3{\text{kg/m}}^3$
$g=10{\text{m/s}}^2$
We know that,
$h=\frac{2T\cos \theta }{r\rho g}$
$\Rightarrow r=\frac{2T\cos \theta }{h\rho g}=\frac{2\times 0.03\times \cos {60}^{\circ }}{0.06\times {10}^3\times 10}\Rightarrow r=0.05\text{mm}$