An open U-tube contains mercury. When 11.2 cm of water is poured into one of the arms of the tube, how high does the mercury rise in the other arm from its initial level ?

Options

A. 0.82 cm
B. 1.35 cm
C. 0.41 cm
D. 2.32 cm

Explanation

To solve the problem of how high the mercury rises in the other arm of an open U-tube when 11.2 cm of water is poured into one arm, we can follow these steps:

### Step 1: Understand the Pressure Equilibrium
When water is poured into one arm of the U-tube, it exerts pressure on the mercury. The pressure exerted by the water column must equal the pressure exerted by the mercury column in the other arm.

### Step 2: Write the Pressure Equations
Let:
- \( h \) = height the mercury rises in the other arm (in cm)
- \( \rho_w \) = density of water = \( 10^3 \, \text{kg/m}^3 \)
- \( \rho_m \) = density of mercury = \( 13.6 \times 10^3 \, \text{kg/m}^3 \)
- \( g \) = acceleration due to gravity = \( 9.81 \, \text{m/s}^2 \)

The pressure due to the water column is given by:
\[ P_{\text{water}} = \rho_w \cdot g \cdot h_w \]
where \( h_w = 11.2 \, \text{cm} = 0.112 \, \text{m} \).

The pressure due to the mercury column is given by:
\[ P_{\text{mercury}} = \rho_m \cdot g \cdot (2h) \]
because the mercury rises by \( h \) in one arm and falls by \( h \) in the other arm, making the effective height \( 2h \).

### Step 3: Set the Pressures Equal
Since the pressures must balance, we can set them equal to each other:
\[ \rho_w \cdot g \cdot h_w = \rho_m \cdot g \cdot (2h) \]

### Step 4: Cancel \( g \) from Both Sides
Since \( g \) appears on both sides, we can cancel it out:
\[ \rho_w \cdot h_w = \rho_m \cdot (2h) \]

### Step 5: Solve for \( h \)
Rearranging the equation to solve for \( h \):
\[ h = \frac{\rho_w \cdot h_w}{2 \cdot \rho_m} \]

### Step 6: Substitute Values
Now, substituting the known values:
\[ h = \frac{(10^3 \, \text{kg/m}^3) \cdot (0.112 \, \text{m})}{2 \cdot (13.6 \times 10^3 \, \text{kg/m}^3)} \]

### Step 7: Calculate \( h \)
Calculating the above expression:
\[ h = \frac{112}{2 \cdot 13600} = \frac{112}{27200} \approx 0.00412 \, \text{m} = 0.412 \, \text{cm} \]

### Final Answer
The height to which the mercury rises in the other arm is approximately:
\[ h \approx 0.41 \, \text{cm} \]

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