A `U` tube pf uniform born of cross sectional area `A` has been set up vertically with open ends facing up Now `m gm` of a liquid of density `d` is poured into it. The column of liquid in this tube will oscillation with a period `T` such that

Options

A. `T = 2pi sqrt((M)/(g))`
B. `T = 2pi sqrt((MA)/(gd))`
C. `T = 2pi sqrt((M)/(gdA))`
D. `T = 2pi sqrt((M)/(2Adg))`

Explanation

To find the period of oscillation \( T \) of the liquid column in a U-tube, we can follow these steps:

### Step 1: Understand the System
We have a U-tube of uniform cross-sectional area \( A \) with a liquid of density \( d \) poured into one side. When the liquid is displaced by a height \( y \), it creates a restoring force due to the weight of the liquid column.

### Step 2: Calculate the Force Acting on the Liquid
The weight of the liquid column that is causing the oscillation can be expressed as:
\[ F = mg \]
where \( m \) is the mass of the liquid. The mass can be expressed in terms of density and volume:
\[ m = d \cdot V \]
The volume \( V \) of the liquid in the U-tube can be expressed as:
\[ V = A \cdot h \]
where \( h \) is the height of the liquid column.

### Step 3: Determine the Displacement
When the liquid is displaced by a height \( y \), the height of the liquid in one side of the U-tube decreases by \( y \) and increases by \( y \) in the other side. The effective height difference that contributes to the restoring force is \( 2y \).

### Step 4: Calculate the Restoring Force
The restoring force \( F \) due to the height difference can be calculated as:
\[ F = \text{Weight of the liquid column} = \text{Density} \times \text{Volume} \times g = d \cdot (A \cdot 2y) \cdot g = 2A d g y \]

### Step 5: Relate Force to Acceleration
According to Newton's second law, the acceleration \( a \) can be expressed as:
\[ F = m \cdot a \]
where \( m \) is the mass of the liquid column. The mass \( m \) can be expressed as:
\[ m = d \cdot (A \cdot h) \]
For small oscillations, we can assume that the total mass of the liquid in the U-tube is \( M \).

### Step 6: Write the Equation of Motion
The acceleration \( a \) can be expressed as:
\[ a = \frac{F}{M} = \frac{2A d g y}{M} \]
This indicates that the acceleration is proportional to the displacement \( y \), which is characteristic of simple harmonic motion (SHM).

### Step 7: Determine the Time Period
The time period \( T \) of SHM is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
where \( k \) is the effective spring constant. In our case, we can relate it to the restoring force:
\[ k = \frac{F}{y} = \frac{2A d g}{M} \]
Thus, the time period can be expressed as:
\[ T = 2\pi \sqrt{\frac{M}{2A d g}} \]

### Final Expression
The final expression for the period of oscillation \( T \) is:
\[ T = 2\pi \sqrt{\frac{M}{2A d g}} \]