Effective length of a seconds pendulum is about.

Options

A. `0.5m`
B. `1m`
C. `1.5m`
D. 2m

Answer

To find the effective length of a seconds pendulum, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: A seconds pendulum has a time period of 2 seconds. We need to find its effective length.

2. **Use the Formula for Time Period**: The formula for the time period (T) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \(L\) is the length of the pendulum and \(g\) is the acceleration due to gravity.

3. **Substitute the Known Values**: For a seconds pendulum, the time period \(T\) is 2 seconds. We can substitute this into the formula:
\[ 2 = 2\pi \sqrt{\frac{L}{g}} \]

4. **Simplify the Equation**: Divide both sides by 2:
\[ 1 = \pi \sqrt{\frac{L}{g}} \]

5. **Isolate the Square Root**: To isolate \(\sqrt{\frac{L}{g}}\), divide both sides by \(\pi\):
\[ \sqrt{\frac{L}{g}} = \frac{1}{\pi} \]

6. **Square Both Sides**: Square both sides to eliminate the square root:
\[ \frac{L}{g} = \left(\frac{1}{\pi}\right)^2 \]

7. **Multiply by g**: Rearranging gives us:
\[ L = g \left(\frac{1}{\pi}\right)^2 \]

8. **Substitute the Value of g**: We know \(g \approx 9.8 \, \text{m/s}^2\):
\[ L = 9.8 \left(\frac{1}{\pi}\right)^2 \]

9. **Calculate \(\frac{1}{\pi}\)**: The value of \(\pi\) is approximately 3.14, so:
\[ \frac{1}{\pi} \approx 0.318 \]
Therefore,
\[ \left(\frac{1}{\pi}\right)^2 \approx 0.101 \]

10. **Final Calculation**: Now substituting back:
\[ L \approx 9.8 \times 0.101 \approx 0.992 \]

11. **Conclusion**: Rounding this value gives us approximately 1 meter. Thus, the effective length of a seconds pendulum is about 1 meter.

### Final Answer:
The effective length of a seconds pendulum is approximately **1 meter**.