Question

What is the diameter of a capillary tube in which mercury is depressed by $1.21cm$ and surface tension for mercury is $540\times 10^{-3}N{m}^{-1}$ the angle of contact is $120^{\circ }$ and density of mercury is $13.6\times 10^{2}kg{m}^{-3}?$

• A. $0.06mm$
• B. $10mm$
• C. $2mm$
• D. $6mm$

Answer 1

Answer: (A)

$h=\frac{2S\cos \theta }{\mathrm{rpg}}$
$\Rightarrow r=\frac{2S\cos {\theta }^0}{h\rho g}$
$D=\frac{4S\cos \theta }{h\rho g}$
$\therefore D=\frac{4\times 540\times 10^{-3}\times \cos 120^{\circ }}{-1.21\times 10^{-2}\times 13.6\times 10^3\times 9.8}$
$D=\frac{4\times 540\times 10^{-3}\times (-0.5)}{-1.21\times 10^{-2}\times 13.6\times 10^3\times 9.8}$
$D=\frac{540\times 10^{-4}}{806.34}=0.06mm$