Step 1: Understanding the Concept:
When a capillary tube is dipped in water, water rises in it due to surface tension.
The height to which the water rises is inversely proportional to the radius of the capillary tube (Jurin's Law).
The mass of the liquid raised depends on both the volume of the cylindrical column and the density of the liquid.
Step 2: Key Formula or Approach:
The height $h$ of capillary rise is given by Jurin's Law:
\[ h = \frac{2T \cos \theta}{\rho g r} \]
where $T$ is surface tension, $\theta$ is contact angle, $\rho$ is density, and $r$ is the radius.
This implies $h \propto \frac{1}{r}$.
The mass $m$ of the liquid in the capillary is:
\[ m = \text{Volume} \times \text{Density} = (\pi r^2 h) \times \rho \]
Step 3: Detailed Explanation:
Substitute the proportionality $h \propto \frac{1}{r}$ into the mass equation.
\[ m = \pi r^2 \left(\frac{k}{r}\right) \rho \]
where $k$ is a constant equal to $\frac{2T \cos \theta}{\rho g}$.
Simplifying the expression for mass:
\[ m = (\pi k \rho) \cdot r \]
This shows that the mass of the liquid that rises in the capillary tube is directly proportional to its radius ($m \propto r$).
Let the initial mass be $m_1 = m$ for radius $r_1 = r$.
Let the new mass be $m_2$ for the new radius $r_2 = \frac{r}{5}$.
Using the direct proportionality $m \propto r$:
\[ \frac{m_2}{m_1} = \frac{r_2}{r_1} \]
\[ \frac{m_2}{m} = \frac{\frac{r}{5}}{r} = \frac{1}{5} \]
\[ m_2 = \frac{m}{5} \]
The new mass of water that will rise is $\frac{m}{5}$.
Step 4: Final Answer:
The mass of water that will rise is $\frac{m}{5}$.