If a simple pendulum is taken to place where g decreases by 2%, then the time period

Options

A. increase by 5%
B. increase by 1%
C. increase by 2%
D. decrease by 5 %

Explanation

To solve the problem of how the time period of a simple pendulum changes when the acceleration due to gravity (g) decreases by 2%, we can follow these steps:

### Step-by-Step Solution

1. **Understand the Formula for Time Period**:
The time 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. **Determine the New Value of g**:
If g decreases by 2%, the new value of g (denoted as \(g'\)) can be calculated as:
\[ g' = g - 0.02g = 0.98g \]

3. **Calculate the New Time Period (T')**:
Substitute \(g'\) into the time period formula:
\[ T' = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{0.98g}} = 2\pi \sqrt{\frac{L}{g}} \cdot \sqrt{\frac{1}{0.98}} \]
This can be simplified to:
\[ T' = T \cdot \sqrt{\frac{1}{0.98}} = T \cdot \frac{1}{\sqrt{0.98}} \]

4. **Calculate the Percentage Change in Time Period**:
To find the percentage change in the time period, we can use the formula:
\[ \text{Percentage Change} = \frac{T' - T}{T} \times 100 \]
Substituting \(T'\):
\[ \text{Percentage Change} = \left(\frac{T \cdot \frac{1}{\sqrt{0.98}} - T}{T}\right) \times 100 = \left(\frac{1}{\sqrt{0.98}} - 1\right) \times 100 \]

5. **Calculate \(\sqrt{0.98}\)**:
Using a calculator, we find:
\[ \sqrt{0.98} \approx 0.9899 \]
Thus,
\[ \frac{1}{\sqrt{0.98}} \approx 1.0101 \]

6. **Final Calculation**:
Now, substituting this back into the percentage change formula:
\[ \text{Percentage Change} = (1.0101 - 1) \times 100 \approx 1.01\% \]
This indicates an increase of approximately 1%.

### Conclusion
The time period of the pendulum increases by approximately 1% when g decreases by 2%.