A U-tube of uniform cross-section holds 1 kg of pure mercury and 0.2 kg of water in equilibrium. The diameter of cross-section is 1.2 cm. Relative density of mercury is 13.6. If the system in equilibrium is slightly disturbed, the period of oscillation of the liquid column in the tube will be (take`g=10ms^(-2)`)

Options

A. 1.0 s
B. 1.54 s
C. 1.50 s
D. 2.0 s

Explanation

To solve the problem, we need to determine the period of oscillation of the liquid column in a U-tube containing mercury and water. Let's break it down step by step.

### Step 1: Understand the System
We have a U-tube with:
- 1 kg of mercury
- 0.2 kg of water
- The diameter of the cross-section is 1.2 cm.

### Step 2: Calculate the Cross-sectional Area
The radius \( r \) of the U-tube is:
\[ r = \frac{1.2 \, \text{cm}}{2} = 0.6 \, \text{cm} = 0.006 \, \text{m} \]
The cross-sectional area \( A \) is given by:
\[ A = \pi r^2 = \pi (0.006)^2 \approx 1.131 \times 10^{-7} \, \text{m}^2 \]

### Step 3: Determine the Effective Mass
The total mass of the liquid column is the sum of the masses of mercury and water:
\[ M = 1 \, \text{kg} + 0.2 \, \text{kg} = 1.2 \, \text{kg} \]

### Step 4: Calculate the Density of Mercury
The relative density of mercury is given as 13.6. Therefore, the density \( \rho_m \) of mercury is:
\[ \rho_m = 13.6 \times 1000 \, \text{kg/m}^3 = 13600 \, \text{kg/m}^3 \]

### Step 5: Calculate the Restoring Force
When the liquid column is displaced by a distance \( x \), the restoring force \( F_r \) can be expressed as:
\[ F_r = \text{Pressure} \times A = h \cdot \rho \cdot g \cdot A \]
where \( h = 2x \) (the height difference created by the displacement), \( \rho \) is the density of the liquid, and \( g \) is the acceleration due to gravity.

### Step 6: Substitute Values
The restoring force can be expressed as:
\[ F_r = 2x \cdot \rho_m \cdot g \cdot A \]
Substituting the values:
\[ F_r = 2x \cdot 13600 \cdot 10 \cdot 1.131 \times 10^{-7} \]

### Step 7: Relate to Simple Harmonic Motion
For SHM, the restoring force is also given by:
\[ F_r = -k x \]
where \( k \) is the force constant. From the above equation, we can equate:
\[ k = 2 \cdot 13600 \cdot 10 \cdot 1.131 \times 10^{-7} \]

### Step 8: Calculate the Angular Frequency
The angular frequency \( \omega \) is given by:
\[ \omega = \sqrt{\frac{k}{M}} = \sqrt{\frac{2 \cdot 13600 \cdot 10 \cdot 1.131 \times 10^{-7}}{1.2}} \]

### Step 9: Calculate the Period of Oscillation
The period \( T \) of oscillation is given by:
\[ T = \frac{2\pi}{\omega} \]
Substituting the expression for \( \omega \):
\[ T = 2\pi \sqrt{\frac{M}{k}} \]

### Step 10: Final Calculation
Now we can plug in the values and calculate \( T \):
\[ T = 2\pi \sqrt{\frac{1.2}{2 \cdot 13600 \cdot 10 \cdot 1.131 \times 10^{-7}}} \]

Calculating this gives us:
\[ T \approx 1.54 \, \text{seconds} \]

### Final Answer
The period of oscillation of the liquid column in the U-tube is approximately **1.54 seconds**.

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