If a pendulum has a period of exactly `1.00sec` at the equator what would be its period at the south pole? Assume the earth to be spherical and rotational effect of the Earth is to be taken .

Answer

To find the period of a pendulum at the South Pole given that it has a period of 1.00 seconds at the equator, we can follow these steps:

### Step-by-Step Solution:

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

2. **Recognize that the length of the pendulum remains constant**:
Since the problem states that the pendulum's length does not change, we can focus on how the acceleration due to gravity \( g \) varies between the equator and the pole.

3. **Identify the values of \( g \)**:
- At the equator, the acceleration due to gravity \( g_e \) is approximately \( 9.78 \, \text{m/s}^2 \).
- At the South Pole, the acceleration due to gravity \( g_p \) is approximately \( 9.834 \, \text{m/s}^2 \).

4. **Relate the periods at the equator and the pole**:
Since the period is inversely proportional to the square root of \( g \), we can express the relationship between the periods at the equator \( T_e \) and at the pole \( T_p \) as:
\[ \frac{T_p}{T_e} = \sqrt{\frac{g_e}{g_p}} \]

5. **Substitute the known values**:
Given that \( T_e = 1.00 \, \text{s} \):
\[ T_p = T_e \cdot \sqrt{\frac{g_e}{g_p}} = 1.00 \cdot \sqrt{\frac{9.78}{9.834}} \]

6. **Calculate the square root**:
First, calculate the fraction:
\[ \frac{9.78}{9.834} \approx 0.9943 \]
Now, take the square root:
\[ \sqrt{0.9943} \approx 0.9971 \]

7. **Find the period at the South Pole**:
Now, substitute back to find \( T_p \):
\[ T_p \approx 1.00 \cdot 0.9971 \approx 0.997 \, \text{s} \]

### Final Answer:
The period of the pendulum at the South Pole is approximately \( 0.997 \, \text{seconds} \).

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