Question

A glass rod of radius $r_1$ is inserted symmetrically into a vertical capillary tube of radius $r_2$ ($r_1 < r_2$) such that their lower ends are at the same level. The arrangement is dipped in water. The height to which water will rise into the tube will be ($\rho$ = density of water, $T$ = surface tension in water, $g$ = acceleration due to gravity)

• A. $\frac{2T}{(r_2 - r_1)\rho g}$
• B. $\frac{T}{(r_2^2 - r_1^2)\rho g}$
• C. $\frac{T}{(r_2 - r_1)\rho g}$
• D. $\frac{2T}{(r_2^2 - r_1^2)\rho g}$

Hint:

For any non-standard capillary cross-section, remember that the height of rise is always $h = \frac{T \cdot P_{\text{wetted}}}{A \cdot \rho \cdot g}$, where $P_{\text{wetted}}$ is the perimeter in contact with the liquid and $A$ is the cross-sectional area.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
When a solid cylinder is placed inside a concentric hollow tube, an annular spacing is formed. When dipped in a wetting liquid like water, surface tension forces lift a column of liquid up into this narrow annular gap until balanced by gravity.

Step 2: Key Formula or Approach:
At equilibrium height $h$, the upward surface tension force equals the downward weight of the water column:
$$F_{\text{up}} = W_{\text{down}}$$ For water and clean glass, the contact angle $\theta \approx 0^\circ$, so $\cos\theta \approx 1$.
The liquid is in contact with two vertical solid walls: the inner circumference of the outer tube and the outer circumference of the inner rod.
$$F_{\text{up}} = T \cdot (2\pi r_2 + 2\pi r_1) = 2\pi T (r_2 + r_1)$$ The weight of the liquid column is given by:
$$W_{\text{down}} = \text{Volume} \times \rho \times g = \pi(r_2^2 - r_1^2)h\rho g$$

Step 3: Detailed Explanation:
Equating the forces to find the equilibrium state:
$$2\pi T (r_2 + r_1) = \pi(r_2^2 - r_1^2)h\rho g$$ We can factor the difference of squares on the right-hand side: $(r_2^2 - r_1^2) = (r_2 - r_1)(r_2 + r_1)$. $$2\pi T (r_2 + r_1) = \pi(r_2 - r_1)(r_2 + r_1)h\rho g$$ Canceling out the common term $\pi(r_2 + r_1)$ from both sides:
$$2T = (r_2 - r_1)h\rho g$$ Isolating the column height $h$:
$$h = \frac{2T}{(r_2 - r_1)\rho g}$$

Step 4: Final Answer:
The height to which water rises in this annular configuration is $\frac{2T}{(r_2 - r_1)\rho g}$, which is option (A).