A vertical `U-tube` contains a liquid. The total length of the liquid column inside the tube is `1`. When the liquid is in equilibrium, the liquid surface in one of the arms of the `U-tube` is pushed down slightly and released. The entire liquid column will undergo a periodic motion.

Options

A. The motion is not simple harmonic motion.
B. the motion is simple harmonic motion
C. if it undergoes simple harmonic motion the time period will be `2pisqrt((l)/(g))`
D. It is undergoes simple harmonic motion the time period will be `2pisqrt((l)/(2g))`

Answer

To solve the problem, we need to analyze the motion of the liquid in the U-tube when it is disturbed slightly from its equilibrium position. Here’s a step-by-step solution:

### Step 1: Understanding the Setup
We have a U-tube filled with a liquid of density \( \rho \) and total length \( L \). Initially, the liquid levels in both arms of the U-tube are equal. When one side is pushed down by a small distance \( y \), the liquid in the other arm rises by the same distance \( y \).

**Hint:** Visualize the U-tube and the liquid levels before and after the disturbance.

### Step 2: Determine the Mass of the Liquid Column
The mass of the liquid in the U-tube can be calculated using the formula:
\[ m_i = \rho \cdot V = \rho \cdot A \cdot L \]
where \( A \) is the cross-sectional area of the tube.

**Hint:** Remember that the volume of the liquid is the product of the cross-sectional area and the height of the liquid column.

### Step 3: Analyzing the Displacement
When the liquid is pushed down by \( y \) in one arm, the height of the liquid in that arm becomes \( L + y \), while the height in the other arm becomes \( L - y \). The effective height of the liquid column that contributes to the restoring force is \( 2y \).

**Hint:** Consider how the liquid levels change in both arms when one side is pushed down.

### Step 4: Calculating the Restoring Force
The restoring force \( F \) acting on the liquid column due to the weight of the displaced liquid can be expressed as:
\[ F = -\text{Weight of the liquid column} = -\rho \cdot A \cdot (2y) \cdot g \]
This can be simplified to:
\[ F = -2 \rho A g y \]

**Hint:** The negative sign indicates that the force acts in the opposite direction to the displacement.

### Step 5: Relating Restoring Force to Simple Harmonic Motion
According to Hooke's law, the restoring force is proportional to the displacement:
\[ F = -k y \]
where \( k \) is the spring constant. From our expression for \( F \):
\[ k = 2 \rho A g \]

**Hint:** Recognize that the form of the restoring force indicates that the motion is simple harmonic.

### Step 6: Finding the Time Period
The time period \( T \) of simple harmonic motion is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Substituting \( m = \rho A L \) and \( k = 2 \rho A g \):
\[ T = 2\pi \sqrt{\frac{\rho A L}{2 \rho A g}} = 2\pi \sqrt{\frac{L}{2g}} \]

**Hint:** Simplify the expression by canceling out common terms.

### Final Result
The time period of the oscillation of the liquid column in the U-tube is:
\[ T = 2\pi \sqrt{\frac{L}{2g}} \]

### Summary
The liquid in the U-tube undergoes simple harmonic motion when disturbed, and the time period of this motion is \( T = 2\pi \sqrt{\frac{L}{2g}} \).

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