The acceleration due to gravity is measured on the surface of earth by using a simple pendulum. If `alphaandbeta` are relative errors in the measurement of length and time period respectively, then percentage error in the measurement of acceleration due to gravity is

Options

A. `(alpha+1/2beta)xx100`
B. `(alpha-2beta)`
C. `(2alpha+beta)xx100`
D. `(alpha+2beta)xx100`

Answer

To solve the problem of finding the percentage error in the measurement of acceleration due to gravity (g) using a simple pendulum, 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:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
Here, \(L\) is the length of the pendulum and \(g\) is the acceleration due to gravity.

2. **Express the Relative Errors**:
We are given:
- Relative error in length \(L\) is \(\alpha = \frac{dL}{L}\)
- Relative error in time period \(T\) is \(\beta = \frac{dT}{T}\)

3. **Differentiate the Time Period Formula**:
Taking the logarithm of both sides of the time period equation:
\[ \log T = \log(2\pi) + \frac{1}{2} \log L - \frac{1}{2} \log g \]
Differentiating this equation gives:
\[ \frac{dT}{T} = \frac{1}{2} \frac{dL}{L} - \frac{1}{2} \frac{dg}{g} \]

4. **Rearranging the Equation**:
Rearranging the above equation to express \(\frac{dg}{g}\):
\[ \frac{dg}{g} = \frac{dL}{L} - 2\frac{dT}{T} \]
Substituting the relative errors:
\[ \frac{dg}{g} = \alpha - 2\beta \]

5. **Calculate the Percentage Error**:
The percentage error in \(g\) is given by:
\[ \text{Percentage Error in } g = \left(\frac{dg}{g}\right) \times 100\% \]
Substituting the expression we found:
\[ \text{Percentage Error in } g = (\alpha - 2\beta) \times 100\% \]

### Final Answer:
The percentage error in the measurement of acceleration due to gravity \(g\) is:
\[ \text{Percentage Error in } g = (\alpha - 2\beta) \times 100\% \]