Question

Three liquids of densities 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-

• A. $\frac{\pi}{2} > \theta_1 > \theta_2 > \theta_3 \geq 0 $
• B. $0 \leq \theta_1 < \theta_2 < \theta_3 < \frac{\pi}{2}$
• C. $\frac{\pi}{2} < \theta_1 < \theta_2 < \theta_3 < \pi$
• D. $\pi > \theta_1 > \theta_2 > \theta_3 > \frac{\pi}{2}$

Answer 1

The Correct Option is B

The given formula is:

\(h = \frac{2T \cos\theta_{c}}{r \rho g}\)

Where:

• h is the height of the fluid column,
• T is the surface tension,
• r is the radius of the tube,
• ρ is the density of the fluid,
• g is the acceleration due to gravity,
• θ_c is the contact angle between the fluid and the solid surface.

 

The equation shows that the height h is directly proportional to the cosine of the contact angle θ_c and the density of the fluid ρ.

The relationship can be expressed as:

\(\cos \theta_{c} \propto \rho\)

Which means, the cosine of the contact angle is directly proportional to the density of the fluid.

Comparing Densities and Contact Angles:

If ρ₁ > ρ₂ > ρ₃, then the corresponding contact angles will follow the order:

\(\cos \theta_{c_1} > \cos \theta_{c_2} > \cos \theta_{c_3}\)

Thus, the angles will satisfy the relation:

\(0 \le \theta_{c_1} < \theta_{c_2} < \theta_{c_3} < \frac{\pi}{2}\)

In other words, as the density increases, the contact angle decreases.