Question

A glass tube of 2.5 mm diameter is immersed vertically in a fluid.
Assume contact angle is zero. The surface tension of the fluid is 0.1 N/m, density of the fluid is \( 1000\ kg/m^3 \) and acceleration due to gravity is \( 10\ m/s^2 \).
The approximate capillary rise is ______ mm.

• A. 2
• B. 8
• C. 16
• D. 32

Hint:

Use \( h = \dfrac{4\sigma\cos\theta}{\rho g d} \) with \( d \) in metres.

Answer 1

The Correct Option is C

Step 1: Set up the force balance directly:
Rather than quoting the capillary rise formula from memory, we can derive it by balancing forces on the liquid column that rises in the tube.
The upward force comes from surface tension acting along the wetted circumference, and the downward force is the weight of the liquid column that has been lifted.

Step 2: Write the two forces in terms of diameter $d$:
The circumference of the tube is $\pi d$, so the vertical component of the surface tension force (with contact angle zero, so the full tension acts vertically) is:
\[ F_{up} = \sigma \times \pi d \]
The weight of the raised column of height $h$ and cross-section area $\frac{\pi d^2}{4}$ is:
\[ F_{down} = \rho g \times \frac{\pi d^2}{4} \times h \]

Step 3: Equate the forces and solve for $h$:
\[ \sigma \pi d = \rho g \frac{\pi d^2}{4} h \]
Cancelling $\pi d$ from both sides:
\[ \sigma = \rho g \frac{d}{4} h \implies h = \frac{4\sigma}{\rho g d} \]
This is the same relation used before, but now built from scratch using a force balance instead of recalling it.

Step 4: Plug in the numbers:
$\sigma = 0.1$, $d = 0.0025$ m, $\rho = 1000$, $g = 10$.
\[ h = \frac{4 \times 0.1}{1000 \times 10 \times 0.0025} = \frac{0.4}{25} = 0.016\ m = 16\ mm \]

Final Answer:
The force balance approach confirms the same 16 mm answer independently. \[ \boxed{h = 16\ mm} \]