A capillary tube made of glass of radius 0.15 mm is dipped vertically in a beaker filled with methyl

Q: Why is the angle of contact $30^\circ$ and not $60^\circ$?

The $60^\circ$ is the total angle between the two tangents from opposite sides. By symmetry, each tangent makes half of that angle, which is $30^\circ$, with the vertical wall. The angle of contact is defined as the angle between the tangent to the liquid surface and the solid surface inside the liquid, which corresponds to this $30^\circ$.

Q: What is the difference between small $r$ and capital $R$?

Small $r$ is the physical radius of the glass capillary tube itself. Capital $R$ is the radius of curvature of the spherical meniscus formed by the liquid. They are related by $r = R \cos \theta$.

Q: Why do we use $\cos \theta$ in the ascent formula?

The upward force pulling the liquid column is the vertical component of the surface tension force. Since the surface tension acts along the tangent at an angle $\theta$ to the vertical, its vertical component is $T \cos \theta$.

Q: What would happen if the angle of contact was greater than $90^\circ$?

If $\theta > 90^\circ$ (like for mercury and glass), $\cos \theta$ becomes negative. This means the liquid would experience a capillary *descent* or depression instead of an ascent, and the meniscus would be convex upwards.

Q: Why did we get $1.0005$ in the denominator?

The denominator is $r \rho g$. Substituting the values in SI units: $(0.15 \times 10^{-3} \text{ m}) \times (667 \text{ kg/m}^3) \times (10 \text{ m/s}^2) = 1.0005$. It's a carefully designed problem where the denominator simplifies beautifully to approximately $1$.

Question

A capillary tube made of glass of radius $0.15 	ext{ mm}$ is dipped vertically in a beaker filled with methylene iodide, which rises to height $h$ in the tube. It is observed that the two tangents drawn from liquid-glass interfaces (from opposite sides of the capillary) make an angle of $60^\circ$ with one another. Then, $h$ is close to (Given, surface tension $= 0.05 	ext{ Nm}^{-1}$, density $= 667 	ext{ kg m}^{-3}$ and $g = 10 	ext{ ms}^{-2}$)

Options:

(A) 0.049 m

(B) 0.087 m

(C) 0.137 m

(D) 0.172 m

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a delicate glass capillary tube dipped vertically into a beaker filled with methylene iodide. As soon as the tube touches the liquid, the fluid begins to climb up the narrow passage, defying gravity. This phenomenon, known as capillary action, occurs due to the interplay between the cohesive forces within the liquid and the adhesive forces between the liquid and the glass.

The liquid rises to a certain height $h$ and forms a curved surface at the top, called a meniscus. Because the liquid wets the glass, this meniscus is concave upwards.

### Decoding the Geometry

The problem gives us a fascinating geometric clue: the tangents drawn from the points of contact of the meniscus (on opposite sides of the capillary) make an angle of $60^\circ$ with each other.

Let's visualize this. If we draw these two tangents extending downwards into the liquid, they intersect at a $60^\circ$ angle. By the sheer symmetry of the cylindrical tube, each tangent must make exactly half of this angle with the vertical wall.

Therefore, the angle each tangent makes with the vertical is $\frac{60^\circ}{2} = 30^\circ$. By definition, the angle of contact $	heta$ is the angle between the tangent to the liquid surface and the solid surface, measured inside the liquid. Thus, we have brilliantly deduced that our angle of contact is **$	heta = 30^\circ$**.

### The Master Equation

With the angle of contact in hand, we can bring in the heavy artillery: the **Capillary Ascent Formula**. The height $h$ to which a liquid rises in a capillary tube is governed by the balance between the upward surface tension force and the downward weight of the liquid column.

The formula is given by:
$$h = \frac{2T \cos 	heta}{r ho g}$$

Here, $T$ is the surface tension, $r$ is the radius of the capillary tube, $ho$ is the density of the liquid, and $g$ is the acceleration due to gravity. Notice how the term $\cos 	heta$ appears. This is because only the vertical component of the surface tension force ($T \cos 	heta$) contributes to lifting the liquid.

### Final Calculation

Now, it's just a matter of plugging in the numbers carefully. We must ensure all units are in the standard SI system to avoid any catastrophic silly mistakes.

Given values:
- Surface tension, $T = 0.05 	ext{ Nm}^{-1}$
- Angle of contact, $	heta = 30^\circ$
- Radius of capillary, $r = 0.15 	ext{ mm} = 0.15 	imes 10^{-3} 	ext{ m}$
- Density, $ho = 667 	ext{ kg m}^{-3}$
- Gravity, $g = 10 	ext{ ms}^{-2}$

Substituting these into our master equation:
$$h = \frac{2 	imes 0.05 	imes \cos 30^\circ}{0.15 	imes 10^{-3} 	imes 667 	imes 10}$$

Let's simplify the numerator and denominator separately. The numerator becomes:
$$2 	imes 0.05 	imes \frac{\sqrt{3}}{2} = 0.05 	imes 1.732 = 0.0866$$

The denominator evaluates to:
$$0.15 	imes 10^{-3} 	imes 667 	imes 10 = 0.15 	imes 6.67 = 1.0005$$

Dividing the two gives us our final height:
$$h = \frac{0.0866}{1.0005} \approx 0.0866 	ext{ m}$$

Rounding off to three decimal places, we get **$h \approx 0.087 	ext{ m}$**. This perfectly matches option (b). A beautiful and elegant application of geometry and fluid mechanics!

Select Answer:

Visualized Solution

Visualizing the Capillary Setup

Decoding the Angle of Contact

Relating Radii $r$ and $R$

The Capillary Ascent Formula

Substituting the Values

Executing the Calculation

Final Conclusion

The Way Forward

The Sigma Insight: Surface Tension and Capillary Action

Analyzing the Setup

Decoding the Geometry

The Master Equation

Final Calculation

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