Question

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

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

Answer 1

Answer: (A)

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