Q132 / 150 · MHT CET 2023 (10 May Shift 1) · Physics

Three liquids have same surface tension and densities $\rho_1, \rho_2$, and $\rho_3\left(\rho_1>\rho_2>\rho_3\right)$. In three identical capillaries rise of liquid is same. The corresponding angles of contact $\theta_1, \theta_2$ and $\theta_3$ are related as

  • A
    $\theta_1>\theta_2>\theta_3$
  • B
    $\theta_1>\theta_3>\theta_2$
  • Correct answer
    C
    $\theta_1 < \theta_2 < \theta_3$
  • D
    $\theta_1=\theta_2=\theta_3$

Solution

Rise in capillary tube, $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{rpg}}$ Given that, $\mathrm{h}, \mathrm{T}, \mathrm{r}$ and $\mathrm{g}$ are constant. $\begin{array}{ll} \therefore \quad & \frac{\cos \theta}{\rho}=\text { constant } \\ & \text { i.e., } \frac{\cos \theta_1}{\rho_1}=\frac{\cos \theta_2}{\rho_2}=\frac{\cos \theta_3}{\rho_3} \\ & \text { as } \rho_1>\rho_2>\rho_3 \\ & \cos \theta_1>\cos \theta_2>\cos \theta_3 \\ \therefore \quad & \theta_1 < \theta_2 < \theta_3 \end{array}$

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

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