Time period of simple pendulum of length l at a place, where acceleration due to gravity is g and period T. Then period of simple pendulum of the same length at a place where the acceleration due to gravity 1.02 g will be

Options

A. T
B. 1.02 T
C. 0.99 T
D. 1.01 T

Explanation

To find the time period of a simple pendulum at a place where the acceleration due to gravity is 1.02g, 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. **Initial Conditions**:
We are given that at a place where the acceleration due to gravity is \( g \), the time period is \( T \). Therefore, we can write:
\[ T = 2\pi \sqrt{\frac{l}{g}} \quad \text{(Equation 1)} \]

3. **New Conditions**:
Now, we need to find the time period \( T' \) at a place where the acceleration due to gravity is \( 1.02g \). Using the same formula for the time period, we have:
\[ T' = 2\pi \sqrt{\frac{l}{1.02g}} \quad \text{(Equation 2)} \]

4. **Relate the Two Time Periods**:
To find the relationship between \( T' \) and \( T \), we can divide Equation 2 by Equation 1:
\[ \frac{T'}{T} = \frac{2\pi \sqrt{\frac{l}{1.02g}}}{2\pi \sqrt{\frac{l}{g}}} \]

5. **Simplifying the Expression**:
The \( 2\pi \) and \( \sqrt{l} \) terms cancel out:
\[ \frac{T'}{T} = \sqrt{\frac{g}{1.02g}} = \sqrt{\frac{1}{1.02}} = \frac{1}{\sqrt{1.02}} \]

6. **Calculating the New Time Period**:
Now, we can express \( T' \) in terms of \( T \):
\[ T' = T \cdot \frac{1}{\sqrt{1.02}} \]

7. **Approximate the Value**:
We can approximate \( \sqrt{1.02} \):
\[ \sqrt{1.02} \approx 1.01 \quad \text{(using a calculator or binomial approximation)} \]
Therefore:
\[ T' \approx T \cdot \frac{1}{1.01} \approx 0.99T \]

### Final Answer:
Thus, the time period of the simple pendulum at a place where the acceleration due to gravity is \( 1.02g \) will be approximately:
\[ T' \approx 0.99T \]