Rankers Physics

Surface Tension and Viscosity: Practice Problem & Solution

Three liquids of densities $\rho_1$, $\rho_2$ and $\rho_3$ (with $\rho_1 > \rho_2 > \rho_3$), having the same value of surface tension $T$, rise to the same height in three identical capillaries. The angles of contact $\theta_1$, $\theta_2$ and $\theta_3$ obey: (2016 - II)
$\frac{\pi}{2} < \theta_1 < \theta_2 < \theta_3 < \pi$
$\pi > \theta_1 > \theta_2 > \theta_3 > \frac{\pi}{2}$
$\frac{\pi}{2} > \theta_1 > \theta_2 > \theta_3 \ge 0$
$0 \le \theta_1 < \theta_2 < \theta_3 < \frac{\pi}{2}$

Solution Explained:

To solve this problem, we apply the core principles of Surface Tension and Viscosity. Understanding the underlying formula is key to arriving at the correct answer below:

Capillary rise $h = \frac{2T \cos\theta}{r \rho g}$. Since $h, T, r, g$ are constant, $\cos\theta \propto \rho$. Since $\rho_1 > \rho_2 > \rho_3$, $\cos\theta_1 > \cos\theta_2 > \cos\theta_3$. For acute angles, this means $$0 \le \theta_1 < \theta_2 < \theta_3 < \frac{\pi}{2}$$.

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