An incompressible liquid is kept in a container having a weightless piston with a hole. A capillary

Q: Why do we subtract $\frac{2T}{r}$ when calculating the pressure inside the capillary?

The liquid surface inside the capillary forms a concave meniscus. The pressure just below a concave surface is less than the pressure just above it (which is atmospheric pressure $P_0$) by an amount equal to the excess pressure $\frac{2T}{r}$.

Q: Why is the pressure at point A equal to the pressure at point B?

In a continuous, static fluid, the pressure at any two points on the same horizontal level must be equal. If it weren't, there would be a horizontal pressure gradient causing the fluid to flow, which contradicts the static condition.

Q: Does the volume of the liquid entering the capillary affect the air volume?

The problem states the air is compressed from $V_0$ to $\frac{100}{101}V_0$ *with the movable piston*. This implies the final volume of the air is exactly $\frac{100}{101}V_0$, regardless of the small volume of liquid that might have entered the capillary. We assume the given volume change accounts for all factors.

Q: What if the contact angle wasn't zero?

If the contact angle $\theta$ was not zero, the upward force due to surface tension would be $2\pi r T \cos\theta$, and the excess pressure would be $\frac{2T \cos\theta}{r}$. You would simply include the $\cos\theta$ term in your calculations.

Q: Why did we convert everything to SI units before calculating?

Mixing units (like cm, mm, and Pascals) is the most common source of errors in physics problems. Converting everything to standard SI units (meters, kilograms, seconds) ensures that all derived quantities (like pressure in Pascals) are consistent.

Question

An incompressible liquid is kept in a container having a weightless piston with a hole. A capillary tube of inner radius $0.1 	ext{ mm}$ is dipped vertically into the liquid through the airtight piston hole, as shown in the figure. The air in the container is isothermally compressed from its original volume $V_0$ to $\frac{100}{101}V_0$ with the movable piston. Considering air as an ideal gas, the height (h) of the liquid column in the capillary above the liquid level in cm is_______.
[Given: Surface tension of the liquid is $0.075 	ext{ Nm}^{-1}$, atmospheric pressure is $10^5 	ext{ N m}^{-2}$, acceleration due to gravity (g) is $10 	ext{ m s}^{-2}$, density of the liquid is $10^3 	ext{ kg m}^{-3}$ and contact angle of capillary surface with the liquid is zero]

Sigma Percentile · Accepted Answer

### The Setup: A Delicate Balance

Imagine a container filled with a liquid, sealed by a movable piston. Piercing through this piston is a delicate capillary tube. Before we even touch the piston, nature is already at work. Due to the surface tension of the liquid, it creeps up the narrow capillary tube, defying gravity.

We can calculate this initial rise, let's call it $h_0$, using Jurin's Law. The formula is $h_0 = \frac{2T \cos	heta}{ho g r}$.

Plugging in the given values—surface tension $T = 0.075 	ext{ Nm}^{-1}$, density $ho = 10^3 	ext{ kg m}^{-3}$, gravity $g = 10 	ext{ m s}^{-2}$, and radius $r = 0.1 	ext{ mm}$ (which is $10^{-4} 	ext{ m}$)—we find that the liquid initially rises to a height of $0.15 	ext{ m}$, or $15 	ext{ cm}$. This is our baseline, the starting point of our physical journey.

### The Squeeze: Isothermal Compression

Now, the action begins. The piston is pushed down, compressing the trapped air. The problem states this happens isothermally, meaning the temperature remains constant. This is a classic scenario for Boyle's Law, which tells us that the product of pressure and volume is a constant ($P_0 V_0 = P V$).

The air is compressed to a new volume $V = \frac{100}{101} V_0$. By substituting this into Boyle's Law, we can easily find the new pressure $P$ inside the container. The initial volume $V_0$ cancels out beautifully, leaving us with $P = \frac{101}{100} P_0$. The pressure has increased slightly, and this extra push is going to force more liquid up the capillary tube.

### The Master Equation: Pressure Equilibrium

To find exactly how high the liquid goes, we need to look at the pressure at the liquid's surface level. Let's pick two points on this horizontal plane: Point A, which is outside the capillary tube, and Point B, which is directly inside it.

In a static fluid, points at the same horizontal level must have the exact same pressure. If they didn't, the fluid would flow until they did!

The pressure at Point A is simply the pressure of our newly compressed air, $P$. 
The pressure at Point B is a bit more complex. Starting from the top of the capillary (which is open to the atmosphere at $P_0$), we drop in pressure across the meniscus by $\frac{2T}{r}$, and then we gain hydrostatic pressure $ho g h$ as we go down the liquid column of height $h$.

So, our master equation becomes:
$$P = P_0 - \frac{2T}{r} + ho g h$$

### The Final Calculation

We already know that the pressure drop due to surface tension, $\frac{2T}{r}$, is exactly what caused the initial rise $h_0$. So, we can replace $\frac{2T}{r}$ with $ho g h_0$.

Substituting our new pressure $P = \frac{101}{100} P_0$ into the equation, we get:
$$\frac{101}{100} P_0 = P_0 - ho g h_0 + ho g h$$

Subtracting $P_0$ from both sides leaves us with:
$$\frac{P_0}{100} = ho g (h - h_0)$$

This tells us that the **extra** height the liquid rises ($h - h_0$) is directly proportional to the extra pressure we applied. Rearranging for this extra height gives:
$$h - h_0 = \frac{P_0}{100 ho g}$$

Now, it's just a matter of plugging in the numbers. With $P_0 = 10^5 	ext{ Pa}$, $ho = 10^3 	ext{ kg m}^{-3}$, and $g = 10 	ext{ m s}^{-2}$, the right side simplifies to $\frac{10^5}{10^6} 	ext{ m}$, which is $0.1 	ext{ m}$ or $10 	ext{ cm}$.

Finally, we add this $10 	ext{ cm}$ increase to our initial $15 	ext{ cm}$ rise. 
$$h = 15 	ext{ cm} + 10 	ext{ cm} = 25 	ext{ cm}$$

The liquid column stands proudly at **$25 	ext{ cm}$** above the liquid level.

Enter Numerical Value:

Visualized Solution

\text{Analyzing the Setup}

\text{Initial Capillary Rise } (h_0)

\text{Calculating } h_0

\text{Value of } h_0

\text{Isothermal Compression of Air}

\text{New Pressure of Air } (P)

\text{Pressure Balance Equation}

\text{Equating Pressures}

\text{Solving for } h - h_0

\text{Final Substitution}

\text{Final Answer}

\text{The Way Forward}

The Sigma Insight: Surface Tension and Capillary Action

The Setup

The Squeeze

The Master Equation

The Final Calculation

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