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_1r_2)/(2(r_2-r_1))`
B. `(rhogh(r_2-r_1))/(2r_1r_2)`
C. `(2(r_2-r_1))/(rhoghr_1r_2)`
D. `(rhogh)/(2(r_2-r_1))`

Answer

To solve the problem, we need to analyze the U-tube with two different radii and the liquid filled in it. Here’s a step-by-step solution:

### Step 1: Understand the Setup
We have a U-tube with two arms. The radius of one arm is \( r_1 \) and the radius of the other arm is \( r_2 \) (where \( r_2 > r_1 \)). The liquid has a density \( \rho \) and the angle of contact is \( 0^\circ \). The height difference between the two arms is \( h \).

### Step 2: Pressure at Points A, B, and C
Let’s denote:
- \( P_A \): Pressure at point A (atmospheric pressure, \( P_0 \))
- \( P_B \): Pressure at point B (in the arm with radius \( r_2 \))
- \( P_C \): Pressure at point C (in the arm with radius \( r_1 \))

Since the liquid is in equilibrium, we can express the pressures at points B and C in terms of surface tension \( T \).

### Step 3: Express Pressures Using Surface Tension
For point B (in the arm with radius \( r_2 \)):
\[ P_B = P_0 - \frac{2T}{r_2} \]

For point C (in the arm with radius \( r_1 \)):
\[ P_C = P_0 - \frac{2T}{r_1} \]

### Step 4: Relate Pressures Using Height Difference
The pressure difference between points B and C can be expressed in terms of the height difference \( h \):
\[ P_B - P_C = \rho g h \]

### Step 5: Substitute Pressures into the Equation
Substituting the expressions for \( P_B \) and \( P_C \) into the pressure difference equation:
\[ \left(P_0 - \frac{2T}{r_2}\right) - \left(P_0 - \frac{2T}{r_1}\right) = \rho g h \]

### Step 6: Simplify the Equation
The \( P_0 \) terms cancel out:
\[ -\frac{2T}{r_2} + \frac{2T}{r_1} = \rho g h \]

Rearranging gives:
\[ \frac{2T}{r_1} - \frac{2T}{r_2} = \rho g h \]

### Step 7: Factor Out Surface Tension
Factoring out \( 2T \):
\[ 2T \left(\frac{1}{r_1} - \frac{1}{r_2}\right) = \rho g h \]

### Step 8: Solve for Surface Tension \( T \)
Now, we can solve for \( T \):
\[ T = \frac{\rho g h}{2 \left(\frac{1}{r_1} - \frac{1}{r_2}\right)} \]

This can be rewritten as:
\[ T = \frac{\rho g h \cdot r_1 r_2}{2(r_2 - r_1)} \]

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

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