In a vertical `U` - tube a column of mercury oscillates simple harmonically. If the tube contains `1kg` of mercury and `1cm` of mercury column weighs `20 g`, then the period of oscillation is

Options

A. `1s`
B. `2 s`
C. `sqrt (2) s`
D. Insufficient data

Explanation

To find the period of oscillation of the mercury column in a vertical U-tube, we can follow these steps:

### Step 1: Understand the problem
We have a U-tube containing mercury, and we need to determine the period of oscillation of the mercury column. The problem states that the tube contains 1 kg of mercury, and 1 cm of the mercury column weighs 20 g.

### Step 2: Calculate the total length of the mercury column
Since 1 cm of mercury weighs 20 g, we can find out how many centimeters correspond to 1 kg (1000 g) of mercury.

\[ \text{Total length of mercury column} = \frac{1000 \text{ g}}{20 \text{ g/cm}} = 50 \text{ cm} \]

### Step 3: Determine the length of one side of the U-tube
In a U-tube, the total length of the mercury column is divided equally between the two sides. Therefore, the length \( L \) of the mercury column on one side is:

\[ L = \frac{50 \text{ cm}}{2} = 25 \text{ cm} \]

### Step 4: Convert the length to meters
To use standard units, we convert 25 cm to meters:

\[ L = 25 \text{ cm} = 0.25 \text{ m} \]

### Step 5: Use the formula for the period of oscillation
The formula for the period \( T \) of a simple harmonic oscillator is given by:

\[ T = 2\pi \sqrt{\frac{L}{g}} \]

where \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \).

### Step 6: Substitute the values into the formula
Now we can substitute \( L = 0.25 \, \text{m} \) and \( g = 9.8 \, \text{m/s}^2 \) into the formula:

\[ T = 2\pi \sqrt{\frac{0.25}{9.8}} \]

### Step 7: Calculate the value inside the square root
Calculating the fraction:

\[ \frac{0.25}{9.8} \approx 0.02551 \]

### Step 8: Calculate the square root
Now, we find the square root:

\[ \sqrt{0.02551} \approx 0.1597 \]

### Step 9: Calculate the period \( T \)
Now we can calculate \( T \):

\[ T = 2\pi \times 0.1597 \approx 1.003 \, \text{s} \]

### Final Answer
Thus, the period of oscillation \( T \) is approximately \( 1 \, \text{s} \).

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