Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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:

$\frac{\pi}{2} > \theta_1 > \theta_2 > \theta_3 \ge 0$

$0 \le \theta_1 < \theta_2 < \theta_3 < \frac{\pi}{2}$

$\frac{\pi}{2} < \theta_1 < \theta_2 < \theta_3 < \pi$

$\pi > \theta_1 > \theta_2 > \theta_3 > \frac{\pi}{2}$

Step-by-Step Solution

Capillary Rise Formula: The height $h$ to which a liquid rises in a capillary tube of radius $r$ is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$ where $T$ is surface tension, $\theta$ is the angle of contact, $\rho$ is the density of the liquid, and $g$ is acceleration due to gravity .
Analyze Dependencies: We are given that $h, T, r,$ and $g$ are the same for all three liquids. Rearranging the formula for the variable terms: $\cos \theta = \frac{h r g}{2T} \cdot \rho$ Since $\frac{h r g}{2T}$ is a constant ( $K$ ), we have: $\cos \theta = K \rho$ This implies $\cos \theta \propto \rho$ .
Apply Conditions: We are given densities $\rho_1 > \rho_2 > \rho_3$ . Therefore: $\cos \theta_1 > \cos \theta_2 > \cos \theta_3$
Behavior of Cosine: For capillary rise ( $h>0$ ), the angle of contact $\theta$ must be acute, i.e., $0 \le \theta < \frac{\pi}{2}$ . In this range, the cosine function is a decreasing function (as angle increases, cosine value decreases). Therefore, the inequality for the angles reverses: $\theta_1 < \theta_2 < \theta_3$
Conclusion: The correct relationship is $0 \le \theta_1 < \theta_2 < \theta_3 < \frac{\pi}{2}$ .

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started