Step 1: Understanding the Concept:
Capillary rise is the phenomenon where a liquid ascends in a narrow tube against gravity.
The height of the liquid column is inversely proportional to the radius (and thus the diameter) of the capillary tube, according to Jurin's Law.
Step 2: Key Formula or Approach:
The formula for capillary rise is \( h = \frac{2T\cos\theta}{\rho g r} \).
Since the liquid and material of the tube are the same, surface tension \( T \), angle of contact \( \theta \), and density \( \rho \) are constant.
Therefore, \( h \propto \frac{1}{r} \), which implies \( h \times r = \text{constant} \).
Since diameter \( d = 2r \), it also follows that \( h \propto \frac{1}{d} \), or \( h_1 d_1 = h_2 d_2 \).
Step 3: Detailed Explanation:
Let the height and diameter for capillary \( P \) be \( h_P \) and \( d_P \).
We are given \( h_P = 2.4\text{ cm} \).
Let the height and diameter for capillary \( Q \) be \( h_Q \) and \( d_Q \).
We are given that the diameter of \( Q \) is \( 80% \) of the diameter of \( P \).
So, \( d_Q = 0.8 \times d_P \).
Using the inverse relationship \( h_P \times d_P = h_Q \times d_Q \):
\[ h_Q = h_P \times \left( \frac{d_P}{d_Q} \right) \]
Substitute the known values:
\[ h_Q = 2.4 \times \left( \frac{d_P}{0.8 \times d_P} \right) \]
\[ h_Q = 2.4 \times \left( \frac{1}{0.8} \right) \]
\[ h_Q = 2.4 \times \left( \frac{10}{8} \right) \]
\[ h_Q = 2.4 \times 1.25 \]
\[ h_Q = 3.0\text{ cm} \]
Step 4: Final Answer:
The rise of liquid in capillary \( Q \) is \( 3\text{ cm} \).