Problem Analysis: An inverted U-tube with unequal arms (radii $r_1$ and $r_2$, with $r_2 > r_1$) is dipped in water. The water level in the narrow arm ($r_1$) coincides with the outside water level. We need to find the depth $h$ of the water level in the wider arm ($r_2$).

Given Parameters:

• Surface Tension of water = $\sigma$
• Density of water = $\rho$
• Atmospheric Pressure = $p_0$
• Acceleration due to gravity = $g$

Step 1: Equation for the Narrow Arm ($r_1$)

The water level in the narrow arm coincides with the outside free surface. The pressure in the water just below the meniscus is therefore equal to the atmospheric pressure ($p_0$).

Since the meniscus is concave towards the air, the pressure inside the trapped air ($P$) is higher than the pressure in the water ($p_0$) by the excess pressure $\frac{2\sigma}{r_1}$.

$$ P – p_0 = \frac{2\sigma}{r_1} $$ $$ P = p_0 + \frac{2\sigma}{r_1} \quad \dots(1) $$

Step 2: Equation for the Wider Arm ($r_2$)

Let the water level in the wider arm be at a depth $h$ below the outside surface. The pressure in the water just below this meniscus ($P_{water}$) is given by the hydrostatic pressure equation:

$$ P_{water} = p_0 + \rho g h $$

Applying the excess pressure equation across this meniscus (air pressure $P$ vs water pressure $P_{water}$):

$$ P – P_{water} = \frac{2\sigma}{r_2} $$

Substituting the value of $P_{water}$:

$$ P – (p_0 + \rho g h) = \frac{2\sigma}{r_2} \quad \dots(2) $$

Step 3: Solving for h

We equate the value of trapped air pressure $P$ from equation (1) into equation (2).

$$ \left( p_0 + \frac{2\sigma}{r_1} \right) – (p_0 + \rho g h) = \frac{2\sigma}{r_2} $$

The atmospheric pressure $p_0$ cancels out:

$$ \frac{2\sigma}{r_1} – \rho g h = \frac{2\sigma}{r_2} $$

Rearranging to solve for the term with $h$:

$$ \rho g h = \frac{2\sigma}{r_1} – \frac{2\sigma}{r_2} $$ $$ \rho g h = 2\sigma \left( \frac{1}{r_1} – \frac{1}{r_2} \right) $$

Isolating $h$:

Physical Interpretation: Since $r_2 > r_1$, the capillary pressure $\frac{2\sigma}{r_1}$ is greater than $\frac{2\sigma}{r_2}$. This higher pressure in the narrow arm pushes the trapped air, which in turn pushes the water level down in the wider arm by height $h$.