A narrow capillary tube when dipped in beaker containing water, the rise is 20 cm . If the area of cross section of the bore is reduced to `(1//4)^(th)` calue , water will rise to a height of

Options

A. 10 cm
B. 20 cm
C. 40 cm
D. 80 cm

Answer

To solve the problem, we need to analyze the relationship between the height of water rise in a capillary tube and the radius of the tube. Let's break it down step by step.

### Step 1: Understand the initial conditions
Let the initial height of water rise in the capillary tube be \( h_1 = 20 \, \text{cm} \). Let the radius of the capillary tube be \( r_1 \).

### Step 2: Determine the new radius after reducing the cross-sectional area
The problem states that the area of cross-section of the bore is reduced to \( \frac{1}{4} \) of its original value. The area \( A \) of a circular cross-section is given by:
\[ A = \pi r^2 \]
If the new area \( A_2 \) is \( \frac{1}{4} A_1 \), we can write:
\[ \pi r_2^2 = \frac{1}{4} \pi r_1^2 \]
Cancelling \( \pi \) from both sides gives:
\[ r_2^2 = \frac{1}{4} r_1^2 \]
Taking the square root of both sides:
\[ r_2 = \frac{1}{2} r_1 \]

### Step 3: Relate the height of water rise to the radius
The height of water rise in a capillary tube is given by the formula:
\[ h = \frac{2T \cos \theta}{\rho g r} \]
where:
- \( T \) is the surface tension,
- \( \theta \) is the angle of contact,
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity,
- \( r \) is the radius of the capillary tube.

From this formula, we can see that the height \( h \) is inversely proportional to the radius \( r \):
\[ h \propto \frac{1}{r} \]

### Step 4: Set up the ratio of heights
Using the relationship of heights and radii, we can write:
\[ \frac{h_1}{h_2} = \frac{r_2}{r_1} \]
Substituting \( r_2 = \frac{1}{2} r_1 \):
\[ \frac{h_1}{h_2} = \frac{\frac{1}{2} r_1}{r_1} = \frac{1}{2} \]
This implies:
\[ h_2 = 2 h_1 \]

### Step 5: Calculate the new height
Substituting the value of \( h_1 \):
\[ h_2 = 2 \times 20 \, \text{cm} = 40 \, \text{cm} \]

### Final Answer
The new height of water rise in the capillary tube when the area of cross-section is reduced to \( \frac{1}{4} \) of its original value is \( \boxed{40 \, \text{cm}} \).