When a long glass capillary tube of radius 0.015 cm is dipped in a liquid, the liquid rises to a hei

Q: Why does the liquid rise in the capillary tube?

The liquid rises due to capillary action, which is driven by the adhesive forces between the liquid and the glass being stronger than the cohesive forces within the liquid itself. This creates an upward force that pulls the liquid up until it's balanced by the weight of the liquid column.

Q: What happens if the contact angle is greater than $90^\circ$?

If the contact angle is greater than $90^\circ$ (like mercury on glass), the cosine of the angle becomes negative. This means the liquid will experience a capillary depression instead of a rise, and the level inside the tube will be lower than the outside.

Q: Why do we need to convert the units to meters?

Physics formulas are designed to work seamlessly with standard SI units. If you mix centimeters and meters, your final answer will be off by several orders of magnitude. Always convert to base units before calculating!

Q: What does the term 'millinewton per meter' mean?

A millinewton ($\text{mN}$) is one-thousandth of a Newton ($10^{-3} \text{ N}$). So, $1 \text{ mN/m}$ is equal to $10^{-3} \text{ N/m}$. It's a convenient unit for surface tension because the values are typically quite small.

Q: Would the liquid overflow if the tube was shorter than the calculated height?

No, the liquid will never overflow. If the tube is shorter than the theoretical capillary height, the liquid will rise to the top and stop. The meniscus will adjust its shape (its radius of curvature will increase) to balance the forces exactly at the top edge.

Question

When a long glass capillary tube of radius $0.015	ext{ cm}$ is dipped in a liquid, the liquid rises to a height of $15	ext{ cm}$ within it. If the contact angle between the liquid and glass is close to $0^\circ$, the surface tension of the liquid, (in millinewton $	ext{m}^{-1}$) to the nearest integer is
(Take, $ho_{	ext{(liquid)}} = 900	ext{ kgm}^{-3}, g = 10	ext{ ms}^{-2}$)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are observing a classic physics experiment: a long, thin glass capillary tube is gently lowered into a beaker of liquid. Almost like magic, the liquid begins to climb up the inside of the tube, defying gravity. This phenomenon is known as **capillary action**, and it is driven by the surface tension of the liquid.

The liquid will continue to rise until the upward pull of the surface tension is perfectly balanced by the downward weight of the liquid column that has been lifted. This delicate balance is beautifully captured by **Jurin's Law**.

### The Master Equation

Jurin's Law gives us the relationship between the height of the liquid column and the physical properties of the system:

$$h = \frac{2S \cos	heta}{ho g r}$$

Here, $h$ is the height of the liquid column, $S$ is the surface tension we want to find, $	heta$ is the angle of contact between the liquid and the glass, $ho$ is the density of the liquid, $g$ is the acceleration due to gravity, and $r$ is the radius of the capillary tube.

Since our goal is to find the surface tension $S$, we can easily rearrange this equation by cross-multiplying:

$$S = \frac{ho g r h}{2 \cos	heta}$$

### The Crucial Step: Unit Conversion

Before we rush into plugging in the numbers, we must pause and look at the units. Physics formulas are unforgiving if you mix different unit systems. We are given the radius $r = 0.015 	ext{ cm}$ and the height $h = 15 	ext{ cm}$. We **must** convert these to meters to align with the standard SI units of density ($	ext{kg/m}^3$) and gravity ($	ext{m/s}^2$).

$$r = 0.015 	ext{ cm} = 15 	imes 10^{-5} 	ext{ m}$$
$$h = 15 	ext{ cm} = 15 	imes 10^{-2} 	ext{ m}$$

We are also told that the contact angle $	heta$ is close to $0^\circ$. The cosine of $0^\circ$ is simply $1$.

### Final Calculation

Now, we substitute all our pristine, SI-compliant values into our rearranged equation:

$$S = \frac{900 	imes 10 	imes (15 	imes 10^{-5}) 	imes (15 	imes 10^{-2})}{2 	imes 1}$$

Let's group the numbers and the powers of ten to make the arithmetic cleaner:

$$S = \frac{(900 	imes 10) 	imes (15 	imes 15) 	imes (10^{-5} 	imes 10^{-2})}{2}$$

$$S = \frac{9000 	imes 225 	imes 10^{-7}}{2}$$

$$S = \frac{2025000 	imes 10^{-7}}{2}$$

$$S = 1012500 	imes 10^{-7} 	ext{ N/m}$$

Adjusting the decimal point to make the number more readable, we get:

$$S = 101.25 	imes 10^{-3} 	ext{ N/m}$$

The question specifically asks for the answer in millinewtons per meter ($	ext{mN/m}$). Since $1 	ext{ mN} = 10^{-3} 	ext{ N}$, our value is exactly $101.25 	ext{ mN/m}$. Rounding this to the nearest integer, we arrive at our final answer:

**$$S \approx 101 	ext{ mN/m}$$**

Enter Numerical Value:

Visualized Solution

$Capillary Rise$

$Jurin’s Law$

$Rearranging for Surface Tension$

$Substituting Values$

$Calculation$

$Final Answer$

$The Way Forward$

The Sigma Insight: Surface Tension and Capillary Action

Analyzing the Setup

The Master Equation

The Crucial Step

Final Calculation

Similar Questions

A thin liquid film formed between a U-shaped wire and a light slider supports a weight of $1.5 \times 1 0^{- 2} N$ (see figure). The length of the slider is $30 cm$ and its weight negligible. The surface tension of the liquid film is

A small spherical droplet of density $d$ is floating exactly half immersed in a liquid of density $ρ$ and surface tension $T$ . The radius of the droplet is (take note that the surface tension applies an upward force on the droplet)

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Capillary Rise

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Jurin’s Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rearranging for Surface Tension

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward