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

Options

A. 10 cm
B. 20 cm
C. 28.28 cm
D. 14.14 cm

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Understand the Problem
We know that a capillary tube initially causes water to rise to a height of 20 cm. We need to find out how high the water will rise when the area of cross-section of the tube is reduced to half.

### Step 2: Relate the Area Change to Radius
The area of a circle is given by \( A = \pi r^2 \). If the area is reduced to half, we can express this as:
\[ A' = \frac{1}{2} A \]
This implies:
\[ \pi r'^2 = \frac{1}{2} \pi r^2 \]
Dividing both sides by \( \pi \):
\[ r'^2 = \frac{1}{2} r^2 \]
Taking the square root of both sides gives:
\[ r' = \frac{r}{\sqrt{2}} \]

### Step 3: Use the Capillary Rise Formula
The height of the liquid column in a capillary tube is given by the formula:
\[ h = \frac{2S \cos \theta}{\rho g r} \]
Where:
- \( h \) is the height of the liquid column,
- \( S \) is the surface tension,
- \( \theta \) 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.

### Step 4: Relate the Heights
From the formula, we can see that the height \( h \) is inversely proportional to the radius \( r \):
\[ h \propto \frac{1}{r} \]
Thus, we can write:
\[ \frac{h'}{h} = \frac{r}{r'} \]
Where \( h' \) is the new height and \( h \) is the original height.

### Step 5: Substitute the Values
We know:
- \( h = 20 \) cm,
- \( r' = \frac{r}{\sqrt{2}} \).

Substituting this into the ratio gives:
\[ \frac{h'}{20} = \frac{r}{\frac{r}{\sqrt{2}}} = \sqrt{2} \]
Thus:
\[ h' = 20 \sqrt{2} \]

### Step 6: Calculate the New Height
Now we calculate \( h' \):
\[ h' = 20 \times 1.414 \approx 28.28 \text{ cm} \]

### Final Answer
The new height to which the water will rise is approximately **28.28 cm**.

---