MHT CET · Physics · Mechanical Properties of Fluids
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 \(\theta_1>\theta_3>\theta_2\)
- C \(\theta_1 < \theta_2 < \theta_3\)
- D \(\theta_1=\theta_2=\theta_3\)
Answer & Solution
Correct Answer
(C) \(\theta_1 < \theta_2 < \theta_3\)
Step-by-step Solution
Detailed explanation
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.
\(\begin{array}{ll}
\therefore \quad & \frac{\cos \theta}{\rho}=\text { constant } \\
& \text { i.e., } \frac{\cos \theta_1}{\rho_1}=\frac{\cos \theta_2}{\rho_2}=\frac{\cos \theta_3}{\rho_3} \\
& \text { as } \rho_1>\rho_2>\rho_3 \\
& \cos \theta_1>\cos \theta_2>\cos \theta_3 \\
\therefore \quad & \theta_1 < \theta_2 < \theta_3
\end{array}\)
\(\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{rpg}}\)
Given that, \(\mathrm{h}, \mathrm{T}, \mathrm{r}\) and \(\mathrm{g}\) are constant.
\(\begin{array}{ll}
\therefore \quad & \frac{\cos \theta}{\rho}=\text { constant } \\
& \text { i.e., } \frac{\cos \theta_1}{\rho_1}=\frac{\cos \theta_2}{\rho_2}=\frac{\cos \theta_3}{\rho_3} \\
& \text { as } \rho_1>\rho_2>\rho_3 \\
& \cos \theta_1>\cos \theta_2>\cos \theta_3 \\
\therefore \quad & \theta_1 < \theta_2 < \theta_3
\end{array}\)
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