A ring whose diameter is 1 meter, oscillates simple harmonically in a vertical plane about a nail fixed at its circumference. The time period will be

Options

A. 1/4 sec
B. 1/2sec
C. 1 sec
D. 2sec

Explanation

To find the time period of a ring oscillating in a vertical plane about a nail fixed at its circumference, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Diameter and Radius of the Ring**:
- Given that the diameter of the ring is 1 meter, we can calculate the radius \( r \) as:
\[ r = \frac{\text{Diameter}}{2} = \frac{1 \, \text{m}}{2} = 0.5 \, \text{m} \]

2. **Understand the Formula for Time Period**:
- The time period \( T \) for a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{l}{g}} \]
- In our case, the effective length \( l \) is equal to the distance from the pivot point (the nail) to the center of the ring, which is equal to the radius of the ring.

3. **Calculate the Effective Length**:
- Since the pivot is at the circumference of the ring, the effective length \( l \) for our calculation is:
\[ l = 2r = 2 \times 0.5 \, \text{m} = 1 \, \text{m} \]

4. **Substitute Values into the Time Period Formula**:
- Now substituting \( l \) and the acceleration due to gravity \( g \approx 9.8 \, \text{m/s}^2 \) into the time period formula:
\[ T = 2\pi \sqrt{\frac{1 \, \text{m}}{9.8 \, \text{m/s}^2}} \]

5. **Calculate the Square Root**:
- First, calculate the square root:
\[ \sqrt{\frac{1}{9.8}} \approx 0.319 \]

6. **Calculate the Time Period**:
- Now substitute this back into the time period formula:
\[ T \approx 2\pi \times 0.319 \approx 2 \times 3.14 \times 0.319 \approx 2.006 \, \text{s} \]

7. **Final Result**:
- Therefore, the time period of the oscillation is approximately:
\[ T \approx 2 \, \text{s} \]

### Summary:
The time period of the ring oscillating in a vertical plane about a nail fixed at its circumference is approximately **2 seconds**.

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