Capillary Rise Dependence on Radius Surface Tension and Acceleration

To find the correct statements, we analyze the capillary rise of a liquid in a tube.

1. Derivation of Capillary Rise ($h$)

Consider a vertical capillary tube of inner radius $r$ dipped in a liquid of density $\rho$ and surface tension $\sigma$. Let $\theta$ be the angle of contact between the liquid and the inner wall of the tube.

The upward force acting on the liquid column due to surface tension along the line of contact is: $F_{\text{up}} = (2\pi r) \sigma \cos\theta$

Neglecting the mass of the liquid in the meniscus, the downward gravitational force (weight) acting on the liquid column of height $h$ in an effective gravitational field $g_{\text{eff}}$ is: $W = m g_{\text{eff}} = (\rho \pi r^2 h) g_{\text{eff}}$

At equilibrium, the upward surface tension force balances the downward weight of the liquid column: $2\pi r \sigma \cos\theta = \rho \pi r^2 h g_{\text{eff}}$

Solving for the height of the liquid column $h$: $h = \frac{2\sigma \cos\theta}{\rho r g_{\text{eff}}}$

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2. Analysis of the Options

• Option (A): For a given material of the capillary tube and a given liquid, $\sigma$, $\rho$, and $\theta$ are constant. Under a constant gravitational acceleration $g$, we have: $h \propto \frac{1}{r}$ Therefore, as the radius $r$ increases, the height $h$ decreases.
  Hence, Statement (A) is true.

• Option (B): From the relation $h = \frac{2\sigma \cos\theta}{\rho r g_{\text{eff}}}$, it is evident that: $h \propto \sigma$ Thus, $h$ depends directly on the surface tension $\sigma$.
  Hence, Statement (B) is false.

• Option (C): When the system is placed in a lift accelerating upwards with a constant acceleration $a$, the effective acceleration due to gravity experienced by the liquid column becomes: $g_{\text{eff}} = g + a > g$ Substituting this into the expression for $h$: $h' = \frac{2\sigma \cos\theta}{\rho r (g + a)} < h$ Thus, the height $h$ decreases.
  Hence, Statement (C) is true.

• Option (D): The height $h$ is directly proportional to $\cos\theta$, not to the contact angle $\theta$ itself ($h \propto \cos\theta$).
  Hence, Statement (D) is false.

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Conclusion

The correct statements are A and C.