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. `(rhoghr_(1)r_(2))/(2(r_(2)-r_(1)))`
B. `(rhogh(r_(1)-r_(2)))/(2r_(1)r_(2))`
C. `(2(r_(2)-r_(1)))/(rhoghr_(1)r_(2))`
D. `(rhogh)/(2(r_(2)-r_(1)))`

Explanation

To solve the problem, we will derive the expression for the surface tension of the liquid in the U-tube based on the given parameters. Here’s a step-by-step solution:

### Step 1: Understand the pressure balance
In a U-tube, the pressure at the same horizontal level in both arms must be equal. Therefore, we can set up the pressure balance equation at the level of the liquid in both arms.

### Step 2: Write the pressure equations
Let \( P_1 \) be the pressure at the level in the left arm and \( P_2 \) be the pressure at the level in the right arm. The pressures can be expressed as:
- For the left arm:
\[ P_1 = P_0 - \frac{2T}{r_1} + \rho g h \]
- For the right arm:
\[ P_2 = P_0 - \frac{2T}{r_2} \]

### Step 3: Set the pressures equal
Since the pressures at the same level must be equal:
\[ P_1 = P_2 \]
Substituting the expressions from Step 2:
\[ P_0 - \frac{2T}{r_1} + \rho g h = P_0 - \frac{2T}{r_2} \]

### Step 4: Simplify the equation
Cancel \( P_0 \) from both sides:
\[ -\frac{2T}{r_1} + \rho g h = -\frac{2T}{r_2} \]
Rearranging gives:
\[ \rho g h = \frac{2T}{r_2} - \frac{2T}{r_1} \]
Factoring out \( 2T \):
\[ \rho g h = 2T \left( \frac{1}{r_2} - \frac{1}{r_1} \right) \]

### Step 5: Solve for surface tension \( T \)
Now, we can isolate \( T \):
\[ T = \frac{\rho g h}{2 \left( \frac{1}{r_2} - \frac{1}{r_1} \right)} \]
This can be rewritten as:
\[ T = \frac{\rho g h \cdot r_1 r_2}{2(r_2 - r_1)} \]

### Final Expression
Thus, the expression for the surface tension \( T \) is:
\[ T = \frac{\rho g h r_1 r_2}{2(r_2 - r_1)} \]