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

 To solve this problem, we need to analyze the capillary rise phenomenon and how it relates to the angles of contact for three liquids with the same surface tension rising to the same height in identical capillaries.

The capillary rise \(h\) in a tube is given by the formula:

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

where \(T\) is the surface tension, \(\theta\) is the contact angle, \(r\) is the radius of the capillary, \(\rho\) is the density of the liquid, and \(g\) is the acceleration due to gravity.

Given that the height \(h\) of the liquid column is the same for all three liquids, it implies that the value of \(\cos \theta\) must be the same for all three since other factors remain constant.

Since \(\cos \theta\) correlates inversely with the angle \(\theta\):

• If \(0 \leq \theta \lt \frac{\pi}{2}\), then \(\cos \theta\) is positive.
• As \(\theta\) increases within this range, \(\cos \theta\) decreases.

Therefore, for the height to be the same when the surface tension \(T\) is the same for all three liquids, the ordering of the angles will be such that:

\(0 \leq \theta_1 < \theta_2 < \theta_3 < \frac{\pi}{2}\)

This ordering implies that higher angles have lower values of \(\cos \theta\), thus allowing for equal heights given the constraint. Therefore, the correct answer is the second option: \(0 \leq \theta_1 < \theta_2 < \theta_3 < \frac{\pi}{2}\).