Q102 / 150 · MHT CET 2023 (11 May Shift 1) · Physics

Two capillary tubes of the same diameter are kept vertically in two different liquids whose densities are in the ratio $4: 3$. The rise of liquid in two capillaries is ' $h_1$ ' and ' $h h_2$ ' respectively. If the surface tensions of liquids are in the ratio $6: 5$, the ratio of heights $\left(\frac{h_1}{h_2}\right)$ is (Assume that their angles of contact are same)

  • A
    $0.4$
  • B
    $0.5$
  • C
    $0.8$
  • Correct answer
    D
    $0.9$

Solution

Given: Density ratio: $\frac{\rho_1}{\rho_2}=\frac{4}{3}$ and surface tension ratio: $\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{6}{5}$ $\therefore \quad$ Rise of liquid in a capillary: $$ \mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}} $$ For constant $\theta, \mathrm{r}$ and $\mathrm{g}$, $$ \begin{aligned} \mathrm{h} & \propto \frac{\mathrm{T}}{\rho} \\ \therefore \quad & \frac{\mathrm{h}_1}{\mathrm{~h}_2}=\frac{\mathrm{T}_1 \rho_2}{\mathrm{~T}_2 \rho_1}=\frac{6 \times 3}{5 \times 4}=0.9 \end{aligned} $$

Hence, the correct answer is option: 'D'.

Relevant Questions