In a capillary tube , water rises to a height of 4 cm If its cross section area were one forth , the water would have to rises a height of

Options

A. 2 cm
B. 6 cm
C. 4 cm
D. 8 cm

Explanation

To solve the problem, we need to understand the relationship between the height of water (H) in a capillary tube and its cross-sectional area (A). The height of the liquid in a capillary tube is given by the formula:

\[ H = \frac{2T \cos \theta}{r \rho g} \]

Where:
- \( T \) is the surface tension of the liquid,
- \( \theta \) is the angle of contact,
- \( r \) is the radius of the capillary tube,
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity.

Given that the height of water in the original capillary tube is \( H_1 = 4 \) cm, we want to find the new height \( H_2 \) when the cross-sectional area is reduced to one-fourth of the original area.

### Step 1: Understand the relationship between area and radius
The cross-sectional area \( A \) of a capillary tube is related to its radius \( r \) by the formula:

\[ A = \pi r^2 \]

If the area is reduced to one-fourth, we can express this as:

\[ A_2 = \frac{1}{4} A_1 \]

### Step 2: Relate the radius to the area
Since \( A \) is proportional to \( r^2 \), we can write:

\[ A_2 = \pi r_2^2 \]
\[ A_1 = \pi r_1^2 \]

From the relationship \( A_2 = \frac{1}{4} A_1 \), we have:

\[ \pi r_2^2 = \frac{1}{4} \pi r_1^2 \]

This simplifies to:

\[ r_2^2 = \frac{1}{4} r_1^2 \]
\[ r_2 = \frac{1}{2} r_1 \]

### Step 3: Use the relationship between height and radius
From the formula for height, we know that height \( H \) is inversely proportional to the radius \( r \):

\[ H \propto \frac{1}{r} \]

Thus, we can express the relationship between the heights and radii of the two scenarios:

\[ \frac{H_1}{H_2} = \frac{r_2}{r_1} \]

### Step 4: Substitute the values
We know \( H_1 = 4 \) cm and \( r_2 = \frac{1}{2} r_1 \):

\[ \frac{4}{H_2} = \frac{\frac{1}{2} r_1}{r_1} \]
\[ \frac{4}{H_2} = \frac{1}{2} \]

### Step 5: Solve for \( H_2 \)
Cross-multiplying gives:

\[ 4 = \frac{1}{2} H_2 \]

Multiplying both sides by 2:

\[ H_2 = 8 \text{ cm} \]

### Final Answer
The height to which water would rise in the capillary tube with one-fourth the cross-sectional area is **8 cm**.

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