A capillary glass tube records a rise of `20cm` when dipped in water. When the area of cross-section of the tube is reduced to half of the former value, water will rise to a height of
(A)`10sqrt(2)cm` (B)`10cm` (C)`20cm` (D)`20sqrt(2)cm`

Options

A. `10sqrt(2)cm`
B. `10cm`
C. `20cm`
D. `20sqrt(2)cm`

Explanation

To solve the problem step-by-step, we will analyze the situation using the principles of capillarity.

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

### Step 2: Identify the initial conditions
From the problem, we know that when the radius of the tube is \( r \), the height of rise \( h \) is \( 20 \, \text{cm} \).

### Step 3: Determine the new radius when the area is halved
The area \( A \) of the cross-section of the tube is given by:
\[ A = \pi r^2 \]
If the area is reduced to half, we have:
\[ \frac{A}{2} = \frac{\pi r^2}{2} \]
Let the new radius be \( r' \). Thus,
\[ \frac{\pi (r')^2}{2} = \frac{\pi r^2}{2} \]
This implies:
\[ (r')^2 = \frac{r^2}{2} \]
Taking the square root:
\[ r' = \frac{r}{\sqrt{2}} \]

### Step 4: Substitute the new radius into the height formula
Now, we need to find the new height \( h' \) when the radius is \( r' \):
\[ h' = \frac{2T \cos \theta}{\rho g r'} \]
Substituting \( r' = \frac{r}{\sqrt{2}} \):
\[ h' = \frac{2T \cos \theta}{\rho g \left(\frac{r}{\sqrt{2}}\right)} \]
This simplifies to:
\[ h' = \frac{2T \cos \theta \cdot \sqrt{2}}{\rho g r} \]
Notice that the term \( \frac{2T \cos \theta}{\rho g r} \) is the original height \( h \), which is \( 20 \, \text{cm} \):
\[ h' = \sqrt{2} \cdot h \]

### Step 5: Calculate the new height
Substituting the value of \( h \):
\[ h' = \sqrt{2} \cdot 20 \, \text{cm} = 20\sqrt{2} \, \text{cm} \]

### Final Answer
Thus, the height of water when the area of cross-section is reduced to half is:
\[ \boxed{20\sqrt{2} \, \text{cm}} \]