In a simple pendulum experiment for the determination of acceleration due to gravity time period is measured with an accuracy of `0.1 %` while length was measured with an accuracy of `0.3 %`. the percentage accuracy in the value of g so obtained is

Options

A. `0.8%`
B. `0.7%`
C. `0.5%`
D. `0.6%`

Explanation

To find the percentage accuracy in the value of acceleration due to gravity (g) obtained from the simple pendulum experiment, we can follow these steps:

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

### Step 2: Rearrange the formula to express g
We can rearrange the formula to express \( g \) in terms of \( T \) and \( L \):
\[ g = \frac{4\pi^2 L}{T^2} \]

### Step 3: Differentiate to find the relationship between accuracies
To find the percentage accuracy in \( g \), we differentiate the equation:
\[ \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T} \]
Where:
- \( \Delta g \) is the uncertainty in \( g \),
- \( \Delta L \) is the uncertainty in \( L \),
- \( \Delta T \) is the uncertainty in \( T \).

### Step 4: Convert the uncertainties to percentage
The percentage uncertainties are given as:
- For time period \( T \): \( \frac{\Delta T}{T} \times 100 = 0.1\% \)
- For length \( L \): \( \frac{\Delta L}{L} \times 100 = 0.3\% \)

### Step 5: Substitute the values into the differentiated equation
Now, substituting the values into the equation:
\[ \frac{\Delta g}{g} \times 100 = \frac{\Delta L}{L} \times 100 + 2 \left(\frac{\Delta T}{T} \times 100\right) \]
\[ \frac{\Delta g}{g} \times 100 = 0.3 + 2 \times 0.1 \]
\[ \frac{\Delta g}{g} \times 100 = 0.3 + 0.2 \]
\[ \frac{\Delta g}{g} \times 100 = 0.5\% \]

### Conclusion
Thus, the percentage accuracy in the value of \( g \) obtained from the experiment is:
\[ \boxed{0.5\%} \]