Question

Two capillary tubes of same diameter are immersed vertically in two liquids of surface tension 60 dyne cm^-1 and 50 dyne cm^-1 respectively. The rise of liquid in the capillaries is 'h₁' and 'h₂' respectively. If the densities of liquid are 0.8 g/cm³ and 0.6 g/cm³ respectively, then ${\displaystyle \frac{{\text{h}}_1}{{\text{h}}_2}}$ is ______.

(Neglect the angle of contact)

विकल्प

• ${\displaystyle \frac{10}{3}}$

• ${\displaystyle \frac{10}{9}}$

• ${\displaystyle \frac{9}{10}}$

• ${\displaystyle \frac{3}{10}}$

Answer 1

Two capillary tubes of same diameter are immersed vertically in two liquids of surface tension 60 dyne cm^-1 and 50 dyne cm^-1 respectively. The rise of liquid in the capillaries is 'h₁' and 'h₂' respectively. If the densities of liquid are 0.8 g/cm³ and 0.6 g/cm³ respectively, then ${\displaystyle \frac{{\text{h}}_1}{{\text{h}}_2}}$ is ${\displaystyle \underline{\frac{9}{10}}}$.

Explanation:

The rise of liquid is given by

${\displaystyle \text{h}=\frac{2T\cos \theta }{r\rho g}}$

${\displaystyle \therefore \frac{{\text{h}}_1}{{\text{h}}_2}=\frac{T_1}{T_2}\cdot \frac{{\rho }_2}{{\rho }_1}=\frac{60}{50}\cdot \frac{0.6}{0.5}=\frac{9}{10}}$