A pole is floating in a liquid with 80 cm of its length immersed. It is pushed doun a certain distance and then released. Time period of vertical oscillation is

Options

A. `(4pi)/7s`
B. `(3pi)/7s`
C. `(2pi)/7s`
D. `(pi)/7s`

Answer

To solve the problem of finding the time period of vertical oscillation of a pole floating in a liquid, we can follow these steps:

### Step 1: Understand the Problem
A pole is floating in a liquid, with 80 cm of its length immersed. When the pole is pushed down and released, it will oscillate vertically. We need to find the time period of these oscillations.

### Step 2: Convert Units
The length of the immersed part of the pole is given as 80 cm. We need to convert this into meters for consistency in SI units.
\[ \text{Length immersed} = 80 \, \text{cm} = \frac{80}{100} \, \text{m} = 0.8 \, \text{m} \]

### Step 3: Write the Formula for Time Period
The time period \( T \) of vertical oscillation for an object floating in a liquid can be calculated using the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
Where:
- \( L \) is the length of the immersed part of the pole (0.8 m)
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \))

### Step 4: Substitute Values into the Formula
Now we substitute the values of \( L \) and \( g \) into the formula:
\[ T = 2\pi \sqrt{\frac{0.8}{9.8}} \]

### Step 5: Calculate the Square Root
First, we calculate \( \frac{0.8}{9.8} \):
\[ \frac{0.8}{9.8} \approx 0.08163 \]
Now, take the square root:
\[ \sqrt{0.08163} \approx 0.2857 \]

### Step 6: Calculate the Time Period
Now, we can calculate \( T \):
\[ T \approx 2\pi \times 0.2857 \approx 1.795 \, \text{s} \]

### Step 7: Final Calculation
To express this in a more manageable form, we can simplify:
\[ T \approx \frac{4\pi}{7} \text{ seconds} \]

### Conclusion
Thus, the time period of the vertical oscillation of the pole is approximately:
\[ T \approx \frac{4\pi}{7} \text{ seconds} \]