Water rises to a height `h` in a capillary tube of cross-sectional area A. the height to which water will rise in a capillary tube of cross-sectional area `4A` will be

Options

A. `h`
B. `h//2`
C. `h//4`
D. `4h`

Explanation

To solve the problem of how high water will rise in a capillary tube with a cross-sectional area of \(4A\), we can follow these steps:

### Step 1: Understand the relationship between capillary rise and cross-sectional area
The height to which a liquid rises in a capillary tube is given by the formula:
\[ h = \frac{2T \cos \theta}{\rho r g} \]
where:
- \(T\) is the surface tension of the liquid,
- \(\theta\) is the angle of contact,
- \(\rho\) is the density of the liquid,
- \(r\) is the radius of the capillary tube,
- \(g\) is the acceleration due to gravity.

From this formula, we can see that the height \(h\) is inversely proportional to the radius \(r\) of the capillary tube.

### Step 2: Relate the radius to the cross-sectional area
The cross-sectional area \(A\) of a circular tube is given by:
\[ A = \pi r^2 \]
From this, we can express the radius \(r\) in terms of the area \(A\):
\[ r = \sqrt{\frac{A}{\pi}} \]

### Step 3: Establish the relationship between heights for different areas
Since the height \(h\) is inversely proportional to the radius, we can say:
\[ h \propto \frac{1}{r} \]
Substituting \(r\) in terms of \(A\):
\[ h \propto \frac{1}{\sqrt{A}} \]
This means that the height \(h\) is inversely proportional to the square root of the cross-sectional area \(A\).

### Step 4: Set up the ratio of heights for the two cases
Let \(h_1\) be the height of water in the first capillary tube with area \(A\) and \(h_2\) be the height in the second capillary tube with area \(4A\). From our relationship:
\[ \frac{h_2}{h_1} = \frac{\sqrt{A}}{\sqrt{4A}} = \frac{\sqrt{A}}{2\sqrt{A}} = \frac{1}{2} \]
Thus, we can express \(h_2\) in terms of \(h_1\):
\[ h_2 = \frac{1}{2} h_1 \]

### Step 5: Substitute the known height
Given that water rises to a height \(h\) in the first capillary tube, we can substitute \(h_1 = h\):
\[ h_2 = \frac{1}{2} h \]

### Final Answer
The height to which water will rise in a capillary tube of cross-sectional area \(4A\) is:
\[ h_2 = \frac{h}{2} \]