If M is the mass of water that rises in a capillary tube of radius r, then mass of water which will

Q: Why doesn't the water keep rising forever?

Because the upward surface tension force is eventually balanced by the downward gravitational pull on the mass of the water column. Once these forces are equal, the water stops rising.

Q: Doesn't a wider tube mean the water rises to the same height, so mass should be 4 times?

No, the height of capillary rise is inversely proportional to the radius ($h \propto 1/r$). So in a tube twice as wide, the water only rises half as high. The area is $4$ times, but height is $1/2$, making the volume (and mass) exactly $2$ times.

Q: How did we get the upward force as $2\pi r T \cos\theta$?

Surface tension $T$ acts along the boundary of the meniscus, which is a circle of circumference $2\pi r$. The vertical component of this force is $T \cos\theta$, so the total upward force is the product of these two.

Q: What if the liquid was mercury instead of water?

Mercury has an obtuse angle of contact, so $\cos\theta$ is negative. The capillary force acts downwards, causing a capillary depression instead of a rise.

Q: Does the atmospheric pressure play a role here?

Atmospheric pressure acts on both the open surface of the beaker and the top of the meniscus, so it cancels out and doesn't affect the height of the capillary rise.

Question

If $M$ is the mass of water that rises in a capillary tube of radius $r$, then mass of water which will rise in a capillary tube of radius $2r$ is

Options:

(A) $2M$

(B) $4M$

(C) $\frac{M}{2}$

(D) $M$

Sigma Percentile · Accepted Answer

### The Magic of Capillary Action

Imagine dipping a narrow glass tube into a beaker of water. Almost magically, the water defies gravity and climbs up the tube! This fascinating phenomenon is known as **capillary action**. It happens because the adhesive forces between the water molecules and the glass walls are stronger than the cohesive forces between the water molecules themselves.

In our problem, we are given a capillary tube of radius $r$. When dipped in water, a certain amount of water rises in it, and the total mass of this raised water column is $M$. The question asks: what would be the mass of the water if we used a wider tube with double the radius, $2r$?

### The Physics of the Climb

To solve this, we need to understand why the water stops rising. It reaches a state of dynamic equilibrium where two opposing forces perfectly balance each other out.

First, there is the **upward pull** caused by surface tension. This force acts along the entire circumference of the meniscus (the curved surface of the water). If the surface tension is $T$ and the angle of contact is $	heta$, the vertical component of this force is given by:

$$F_{	ext{up}} = (2\pi r) T \cos	heta$$

Second, there is the **downward pull** of gravity acting on the mass of the water column itself. This is simply its weight:

$$F_{	ext{down}} = Mg$$

### The Master Equation

Since the water column is in equilibrium, these two forces must be equal. Equating them gives us our master equation:

$$Mg = 2\pi r T \cos	heta$$

Now, let's look closely at this equation. For a specific liquid (like water) and a specific material (like glass), the surface tension $T$, the angle of contact $	heta$, and the acceleration due to gravity $g$ are all **constants**.

If we rearrange the equation to solve for mass, we get:

$$M = \left( \frac{2\pi T \cos	heta}{g} ight) r$$

This reveals a beautiful and simple relationship: **the mass of the raised water is directly proportional to the radius of the tube** ($M \propto r$).

### The Final Calculation

Armed with this proportionality, the rest is straightforward. If we take a new tube with double the radius ($r' = 2r$), the new mass $M'$ will also double!

$$M' \propto 2r \implies M' = 2M$$

### The Twist: What Happens to the Height?

There is a fascinating catch here that often confuses students. Even though the wider tube holds *more* mass, the water actually doesn't rise as high!

Let's prove this. We know that mass is density times volume ($M = ho \pi r^2 h$). If we substitute this into our proportionality $M \propto r$, we get:

$$ho \pi r^2 h \propto r$$

Dividing both sides by $r^2$, we find that the height $h$ is **inversely proportional** to the radius ($h \propto \frac{1}{r}$).

So, in a tube that is twice as wide, the water only rises to **half** the original height. However, because the tube is much wider (its cross-sectional area is four times larger), the total volume—and therefore the total mass—is still double!

Animated Solution for Physics - Properties of Solids and Liquids: If $M$ is the mass of water that rises in a capillary tube of radius $r$ , then mass of water which will rise in a capillary tube of radius $2 r$ is

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Visualized Solution

$C a p i l l a r y R i se$

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The Sigma Insight: Surface Tension and Capillary Action

The Magic of Capillary Action

The Physics of the Climb

The Master Equation

The Final Calculation

The Twist

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