The radii of the two columne is U-tube are `r_(1)` and `r_(2)(gtr_(1))`. When a liquid of density `rho` (angle of contact is `0^@))` is filled in it, the level different of liquid in two arms is h. The surface tension of liquid is
`(g=` acceleration due to gravity)

Options

A. `(rho g hr_(1)r_(2))/(2(r_(2)-r_(1)))`
B. `(rho g h(r_(1)r_(2)))/(2r_(2)-r_(1))`
C. `(2(r_(1)r_(2)))/(rho gh r_(2)-r_(1))`
D. `(rho g h)/(2(r_(2)-r_(1)))`

Explanation

To solve the problem, we need to find the surface tension (T) of the liquid in the U-tube based on the given parameters. Here’s the step-by-step solution:

### Step 1: Understand the setup
We have a U-tube with two columns of liquid. The radii of the two arms are \( r_1 \) and \( r_2 \) (where \( r_2 > r_1 \)). When a liquid of density \( \rho \) is filled in the U-tube, there is a height difference \( h \) between the two arms.

### Step 2: Write the pressure balance equation
At equilibrium, the pressure at the same horizontal level in both arms must be equal. We can express this as:
\[ P_1 = P_2 \]
Where:
- \( P_1 \) is the pressure at the lower level in the left arm (with radius \( r_1 \)).
- \( P_2 \) is the pressure at the lower level in the right arm (with radius \( r_2 \)).

### Step 3: Express the pressures
The pressure at point 1 can be expressed as:
\[ P_1 = P_0 - \frac{2T}{r_1} + \rho g h \]
And the pressure at point 2 can be expressed as:
\[ P_2 = P_0 - \frac{2T}{r_2} \]
Where \( P_0 \) is the atmospheric pressure, \( T \) is the surface tension, and \( g \) is the acceleration due to gravity.

### Step 4: Set the pressures equal
Setting \( P_1 \) equal to \( P_2 \):
\[ P_0 - \frac{2T}{r_1} + \rho g h = P_0 - \frac{2T}{r_2} \]

### Step 5: Simplify the equation
Cancel \( P_0 \) from both sides:
\[ -\frac{2T}{r_1} + \rho g h = -\frac{2T}{r_2} \]
Rearranging gives:
\[ \rho g h = \frac{2T}{r_2} - \frac{2T}{r_1} \]

### Step 6: Factor out the surface tension
Factoring out \( 2T \):
\[ \rho g h = 2T \left( \frac{1}{r_2} - \frac{1}{r_1} \right) \]

### Step 7: Solve for surface tension
Rearranging for \( T \):
\[ T = \frac{\rho g h}{2 \left( \frac{1}{r_2} - \frac{1}{r_1} \right)} \]
Finding a common denominator:
\[ T = \frac{\rho g h \cdot r_1 r_2}{2(r_2 - r_1)} \]

### Final Answer
Thus, the surface tension \( T \) is given by:
\[ T = \frac{\rho g h \cdot r_1 r_2}{2(r_2 - r_1)} \]