What is the percentage error in the measurement of time period of a pendulum if maximum errors in the measurement of `l` ands `g` are 2% and 4% respectively?

Options

A. `6%`
B. `4%`
C. `3%`
D. `5%`

Answer

To find the percentage error in the measurement of the time period of a pendulum given the maximum errors in the measurement of length (l) and acceleration due to gravity (g), 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:
\[ T = 2\pi \sqrt{\frac{l}{g}} \]

2. **Identify the Variables**:
- \( l \) = length of the pendulum
- \( g \) = acceleration due to gravity
- Given maximum percentage errors:
- Maximum error in \( l \) = 2%
- Maximum error in \( g \) = 4%

3. **Express the Percentage Errors**:
The percentage error in \( l \) is given as:
\[ \frac{\Delta l}{l} \times 100 = 2\% \]
The percentage error in \( g \) is given as:
\[ \frac{\Delta g}{g} \times 100 = 4\% \]

4. **Determine the Formula for Percentage Error in Time Period**:
To find the percentage error in \( T \), we can use the formula for the propagation of errors. For the formula \( T = 2\pi \sqrt{\frac{l}{g}} \), we can derive the percentage error as follows:
\[ \frac{\Delta T}{T} = \frac{1}{2} \left( \frac{\Delta l}{l} + \frac{\Delta g}{g} \right) \]
Here, \( \Delta T \) is the error in the time period, and \( T \) is the time period itself.

5. **Substitute the Known Values**:
Substitute the percentage errors into the equation:
\[ \frac{\Delta T}{T} = \frac{1}{2} \left( \frac{2}{100} + \frac{4}{100} \right) \]

6. **Calculate the Result**:
\[ \frac{\Delta T}{T} = \frac{1}{2} \left( \frac{2 + 4}{100} \right) = \frac{1}{2} \left( \frac{6}{100} \right) = \frac{3}{100} \]
To express this as a percentage:
\[ \text{Percentage error in } T = \frac{3}{100} \times 100 = 3\% \]

7. **Final Answer**:
The percentage error in the measurement of the time period of the pendulum is **3%**.