Mercury is poured in a U-tube. Temperature of one side is `50^(@)C` and level of mercury on this side is `h_(2)`. Temperature of the other side of `100^(@) C` and level of mercury on this side is `h_(2)`. Then

Options

A. `h_(1) = h_(2)`
B. `h_(2) lt h_(1)`
C. `h_(2) gt h_(1)`
D. `h_(2) = 2h_(1)`

Answer

To solve the problem, we need to analyze the behavior of mercury in a U-tube when subjected to different temperatures on either side. Let's break down the steps:

### Step-by-Step Solution:

1. **Understanding the U-tube Setup**:
- We have a U-tube filled with mercury.
- The left side is at a temperature of \(50^\circ C\) and has a mercury level denoted as \(h_1\).
- The right side is at a temperature of \(100^\circ C\) and has a mercury level denoted as \(h_2\).

2. **Effect of Temperature on Density**:
- The density of mercury decreases as the temperature increases. Therefore, the density at \(50^\circ C\) (denoted as \(\rho_1\)) is greater than the density at \(100^\circ C\) (denoted as \(\rho_2\)).
- Mathematically, we can express this as:
\[ \rho_1 > \rho_2 \]

3. **Applying Hydrostatic Pressure Principle**:
- At the same horizontal level in the U-tube, the pressure exerted by the mercury column on both sides must be equal.
- The pressure exerted by the mercury on the left side can be expressed as:
\[ P_1 = h_1 \cdot \rho_1 \cdot g \]
- The pressure exerted by the mercury on the right side can be expressed as:
\[ P_2 = h_2 \cdot \rho_2 \cdot g \]
- Here, \(g\) is the acceleration due to gravity.

4. **Setting the Pressures Equal**:
- Since the pressures at the same level must be equal, we can set the two pressure equations equal to each other:
\[ h_1 \cdot \rho_1 \cdot g = h_2 \cdot \rho_2 \cdot g \]
- The \(g\) cancels out from both sides:
\[ h_1 \cdot \rho_1 = h_2 \cdot \rho_2 \]

5. **Rearranging the Equation**:
- We can rearrange the equation to find the relationship between \(h_1\) and \(h_2\):
\[ \frac{h_1}{h_2} = \frac{\rho_2}{\rho_1} \]

6. **Interpreting the Result**:
- Since \(\rho_1 > \rho_2\), it follows that:
\[ \frac{\rho_2}{\rho_1} < 1 \]
- Therefore, we conclude that:
\[ h_1 < h_2 \]

### Final Conclusion:
The height of mercury on the side with \(50^\circ C\) (\(h_1\)) is less than the height on the side with \(100^\circ C\) (\(h_2\)).