Understanding Capillary Action and Mass Rise

Capillary action describes how liquids flow in narrow spaces without assistance from, or even in opposition to, external forces like gravity. In the context of a capillary tube dipped in water, the liquid rises due to surface tension acting upwards along the tube's inner wall.

The height ($h$) a liquid rises in a capillary tube is given by the formula:

$h = \frac{2 \gamma \cos \theta}{\rho g r}$

Here, $\gamma$ is the surface tension, $\theta$ is the contact angle, $\rho$ is the liquid density, $g$ is acceleration due to gravity, and $r$ is the tube's radius. Importantly, this shows that the height $h$ is inversely proportional to the radius $r$ ($h \propto \frac{1}{r}$).

Calculating Mass of Water Risen

The mass ($M$) of the water column that rises in the tube depends on its volume ($V$) and density ($\rho$). The volume of this cylindrical column is $V = \pi r^2 h$. Therefore, the mass is:

$M = \rho \times V = \rho \times (\pi r^2 h)$

Let's substitute the expression for $h$ into the mass formula:

$M = \rho \times \pi r^2 \times \left( \frac{2 \gamma \cos \theta}{\rho g r} \right)$

After simplification, the mass $M$ becomes:

$M = \frac{2 \pi r \gamma \cos \theta}{g}$

This formula reveals that the mass ($M$) of the risen liquid is directly proportional to the radius ($r$) of the capillary tube ($M \propto r$), assuming the liquid ($\gamma$, $\rho$, $\theta$) and conditions ($g$) remain constant.

Mass Risen in the Two Capillary Tubes

We are given that $M$ is the mass of water rising in a capillary tube of radius $r$. Let's call this $M_1$. So:

$M_1 = M \quad (\text{for radius } r)$

Now, consider the second capillary tube with radius $2r$. Let the mass of water rising in this tube be $M_2$. Since mass is directly proportional to the radius ($M \propto r$):

$M_2 \propto 2r$

This means $M_2$ will be twice the mass that rises in a tube of radius $r$. Therefore:

$M_2 = 2 \times M = 2M \quad (\text{for radius } 2r)$

Determining the Total Mass

The question asks for the total mass of water when both tubes (one with radius $r$ and the other with radius $2r$) are placed simultaneously in water.

Total Mass = Mass in the first tube + Mass in the second tube

Total Mass $= M_1 + M_2$

Substituting the values we found:

Total Mass $= M + 2M$

Total Mass $= 3M$

So, the total mass of water that rises when both capillary tubes are placed in water is $3M$.