Step 1: Check Capillary Rise Height

First, we determine if the water rises to the top of the tube. The radius $r$ of the inner section can be found from the area $A_i$.

Given $A_i = 0.2 \text{ mm}^2 = 0.2 \times 10^{-6} \text{ m}^2$.

$$ r = \sqrt{\frac{A_i}{\pi}} = \sqrt{\frac{0.2 \times 10^{-6}}{\pi}} \approx 2.5 \times 10^{-4} \text{ m} $$

The standard capillary rise height $h$ is given by:

$$ h = \frac{2\sigma}{\rho_w g r} $$ $$ h = \frac{2 \times 0.07}{10^3 \times 10 \times 2.5 \times 10^{-4}} = \frac{0.14}{2.5} = 0.056 \text{ m} = 5.6 \text{ cm} $$

The tube is half dipped. Total length $l = 6 \text{ cm}$, so length above water is $3 \text{ cm}$.

Since the theoretical rise $h (5.6 \text{ cm}) > \text{Length above water} (3 \text{ cm})$, the water will rise to the very top of the tube.

Step 2: Force Analysis

For the tube to be in equilibrium, the upward forces must balance the downward forces.

Forces acting Downwards:

• Weight of the glass tube: $W_g = mg$
• Weight of the water column raised above the general reservoir level: $W_{water\_in} = \rho_w A_i (l/2) g$
• Surface Tension force on the outer perimeter (pulling down due to wetting): $F_{ST} = s_o \sigma$

Forces acting Upwards:

• Applied Force (External): $F$
• Buoyant Force on the submerged glass volume: $F_B$

The equation of equilibrium is:

$$ F + F_B = W_g + W_{water\_in} + F_{ST} $$ $$ F = W_g + W_{water\_in} + F_{ST} – F_B $$

Step 3: Calculating Buoyant Force ($F_B$)

The buoyant force acts only on the submerged volume of the glass material. Since half the tube is dipped, half the mass of the glass is submerged.

$$ \text{Volume of Glass Submerged} = \frac{1}{2} \times \frac{m}{\rho_g} $$ $$ F_B = (\text{Vol}_{sub}) \rho_w g = \frac{m}{2 \rho_g} \rho_w g $$ $$ F_B = mg \left( \frac{\rho_w}{2 \rho_g} \right) $$

Step 4: Formulating the Final Equation

Substituting all terms into our equilibrium equation:

$$ F = mg + \rho_w A_i \frac{l}{2} g + s_o \sigma – mg \left( \frac{\rho_w}{2 \rho_g} \right) $$

Grouping the $mg$ terms:

Step 5: Substitution and Result

We substitute the values (converting all to SI units):

• $m = 0.012 \text{ kg}$
• $g = 10 \text{ m/s}^2$
• $\rho_w = 1 \text{ g/cm}^3$, $\rho_g = 2 \text{ g/cm}^3 \implies \frac{\rho_w}{\rho_g} = 0.5$
• $A_i = 0.2 \text{ mm}^2 = 0.2 \times 10^{-6} \text{ m}^2$
• $l/2 = 3 \text{ cm} = 0.03 \text{ m}$
• $s_o = 16 \text{ mm} = 0.016 \text{ m}$
• $\sigma = 0.07 \text{ N/m}$

$$ F = (0.012)(10) \left( 1 – \frac{1}{2(2)} \right) + (1000)(0.2 \times 10^{-6})(0.03)(10) + (0.016)(0.07) $$

Term 1 (Modified Weight):

$$ 0.12 \times (1 – 0.25) = 0.12 \times 0.75 = 0.09 \text{ N} $$

Term 2 (Weight of Water):

$$ 1000 \times 2 \times 10^{-7} \times 0.3 = 6 \times 10^{-5} \text{ N} \quad (\text{Negligible}) $$

Term 3 (Surface Tension):

$$ 0.016 \times 0.07 = 0.00112 \text{ N} $$

Total Force:

$$ F \approx 0.09 + 0.00112 = 0.09112 \text{ N} $$