Water rises to a height of 20 mm in a capillary tube having cross sectiona area. If area of cross section of the tube is made `A/4` ,then water will rise to a height of

Options

A. 4 cm
B. 2 cm
C. 3 cm
D. 6 cm

Explanation

To solve the problem step by step, let's break it down:

### Step 1: Understand the Problem
We have a capillary tube where water rises to a height of 20 mm when the cross-sectional area is \( A \). We need to find out how high the water will rise if the cross-sectional area is reduced to \( \frac{A}{4} \).

### Step 2: Use the Capillary Rise Formula
The height of the liquid column in a capillary tube is given by the formula:
\[ H = \frac{2T \cos \theta}{R \rho g} \]
Where:
- \( H \) is the height of the liquid column,
- \( T \) is the surface tension of the liquid,
- \( \theta \) is the contact angle,
- \( R \) is the radius of the capillary tube,
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity.

### Step 3: Relate the Radii of the Capillary Tubes
Let’s denote:
- For the first capillary tube (area \( A \)), the radius is \( R_1 \).
- For the second capillary tube (area \( \frac{A}{4} \)), the radius is \( R_2 \).

Since the area of a circle is given by \( A = \pi R^2 \), we can write:
\[ A = \pi R_1^2 \quad \text{and} \quad \frac{A}{4} = \pi R_2^2 \]
From the second equation, we have:
\[ R_2^2 = \frac{A}{4\pi} \]
Substituting \( A = \pi R_1^2 \) into this gives:
\[ R_2^2 = \frac{\pi R_1^2}{4\pi} = \frac{R_1^2}{4} \]
Thus,
\[ R_2 = \frac{R_1}{2} \]

### Step 4: Calculate the Height for the Second Capillary
Using the capillary rise formula for the second capillary tube:
\[ H_2 = \frac{2T \cos \theta}{R_2 \rho g} \]
Substituting \( R_2 = \frac{R_1}{2} \):
\[ H_2 = \frac{2T \cos \theta}{\left(\frac{R_1}{2}\right) \rho g} = \frac{2T \cos \theta \cdot 2}{R_1 \rho g} = \frac{4T \cos \theta}{R_1 \rho g} \]

### Step 5: Relate \( H_2 \) to \( H_1 \)
From the first capillary tube, we know:
\[ H_1 = \frac{2T \cos \theta}{R_1 \rho g} \]
Given \( H_1 = 20 \, \text{mm} \):
\[ H_1 = \frac{2T \cos \theta}{R_1 \rho g} = 20 \, \text{mm} \]
Thus,
\[ \frac{2T \cos \theta}{R_1 \rho g} = 20 \]

### Step 6: Substitute \( H_1 \) into \( H_2 \)
Now we can express \( H_2 \) in terms of \( H_1 \):
\[ H_2 = 4 \cdot H_1 = 4 \cdot 20 \, \text{mm} = 80 \, \text{mm} \]

### Step 7: Convert to Centimeters
To convert \( H_2 \) from mm to cm:
\[ H_2 = \frac{80 \, \text{mm}}{10} = 8 \, \text{cm} \]

### Final Answer
The height to which water will rise in the capillary tube with area \( \frac{A}{4} \) is **8 cm**.

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