Water rises to a height of `10cm` in a glass capillary tube. If the area of cross section of the tube is reduced to one fourth of the former value what is the height of water rise now?

Options

A. 20 cm
B. 5 cm
C. 2.5 cm
D. 7 cm

Explanation

To solve the problem of how the height of water rise in a capillary tube changes when the cross-sectional area is reduced, we can follow these steps:

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

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

### Step 2: Determine the initial conditions
We know from the problem that the initial height of water rise in the capillary tube is:
\[ h_1 = 10 \, \text{cm} \]
Let the initial radius of the tube be \( r_1 \).

### Step 3: Determine the new radius
The area of cross-section of the tube is reduced to one fourth of its former value. The area \( A \) of a circle is given by:
\[ A = \pi r^2 \]
If the area is reduced to one fourth, we have:
\[ A_2 = \frac{1}{4} A_1 \]
This implies:
\[ \pi r_2^2 = \frac{1}{4} \pi r_1^2 \]
From this, we can solve for the new radius \( r_2 \):
\[ r_2^2 = \frac{1}{4} r_1^2 \]
Taking the square root of both sides gives:
\[ r_2 = \frac{1}{2} r_1 \]

### Step 4: Relate the new height to the new radius
Since the height is inversely proportional to the radius, we can express the new height \( h_2 \) in terms of the old height \( h_1 \):
\[ h_2 = \frac{2s \cos \theta}{\rho g r_2} \]
Substituting \( r_2 = \frac{1}{2} r_1 \) into the equation gives:
\[ h_2 = \frac{2s \cos \theta}{\rho g \left(\frac{1}{2} r_1\right)} = \frac{4s \cos \theta}{\rho g r_1} \]
This shows that:
\[ h_2 = 4 \cdot \frac{2s \cos \theta}{\rho g r_1} = 4h_1 \]

### Step 5: Calculate the new height
Now substituting \( h_1 = 10 \, \text{cm} \):
\[ h_2 = 4 \cdot 10 \, \text{cm} = 40 \, \text{cm} \]

### Final Answer
Thus, the height of water rise in the capillary tube after reducing the area of cross-section to one fourth is:
\[ h_2 = 40 \, \text{cm} \]