When water rises in a capillary tube of radius `r` to height h, then its potential energy `U_(1)` if capillary tube of radius 2r is dipped in same water then potential energy of water is `U_(2)` then `U_(1):U_(2)` will be

Options

A. `2:1`
B. `1:2`
C. `4:1`
D. `1:1`

Answer

To solve the problem, we need to find the ratio of potential energies \( U_1 \) and \( U_2 \) for two capillary tubes with different radii. Let's break down the solution step by step.

### Step 1: Understand the formula for potential energy
The potential energy \( U \) of a column of liquid is given by the formula:
\[ U = mgh \]
where \( m \) is the mass of the liquid, \( g \) is the acceleration due to gravity, and \( h \) is the height of the liquid column.

### Step 2: Express mass in terms of density and volume
The mass \( m \) can be expressed as:
\[ m = \text{density} \times \text{volume} \]
For a cylindrical capillary tube, the volume \( V \) of the water column is given by:
\[ V = \pi r^2 h \]
Thus, we can write:
\[ m = \rho \cdot V = \rho \cdot (\pi r^2 h) \]
where \( \rho \) is the density of water.

### Step 3: Substitute mass into the potential energy formula
Now substituting the expression for mass into the potential energy formula, we get:
\[ U = \rho \cdot (\pi r^2 h) \cdot g \cdot h = \pi \rho g r^2 h^2 \]

### Step 4: Calculate potential energy for both tubes
1. For the first capillary tube with radius \( r \) and height \( h_1 \):
\[ U_1 = \pi \rho g r^2 h_1^2 \]

2. For the second capillary tube with radius \( 2r \) and height \( h_2 \):
\[ U_2 = \pi \rho g (2r)^2 h_2^2 = \pi \rho g \cdot 4r^2 h_2^2 \]

### Step 5: Find the ratio \( U_1 : U_2 \)
Now, we can find the ratio of the potential energies:
\[ \frac{U_1}{U_2} = \frac{\pi \rho g r^2 h_1^2}{\pi \rho g \cdot 4r^2 h_2^2} \]
The terms \( \pi \), \( \rho \), \( g \), and \( r^2 \) cancel out:
\[ \frac{U_1}{U_2} = \frac{h_1^2}{4h_2^2} \]

### Step 6: Relate heights \( h_1 \) and \( h_2 \)
Using the capillary rise formula:
\[ h = \frac{2T \cos \theta}{\rho g r} \]
we see that height \( h \) is inversely proportional to the radius \( r \). Therefore, for the two tubes:
\[ \frac{h_1}{h_2} = \frac{r_2}{r_1} = \frac{2r}{r} = 2 \]
Squaring both sides gives:
\[ \frac{h_1^2}{h_2^2} = 4 \]

### Step 7: Substitute back into the ratio of potential energies
Substituting this back into our ratio:
\[ \frac{U_1}{U_2} = \frac{h_1^2}{4h_2^2} = \frac{4}{4} = 1 \]

### Final Result
Thus, the ratio of potential energies \( U_1 : U_2 \) is:
\[ U_1 : U_2 = 1 : 1 \]