An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be:

Options

A. `1.7%`
B. `4.4%`
C. `2.7%`
D. `2.27%`

Answer

To find the error in the determination of the acceleration due to gravity \( g \) using the simple pendulum method, we will follow these steps:

### Step 1: Understand the formula for \( g \)
The formula for \( g \) derived from the time period \( T \) of a simple pendulum is given by:
\[ g = \frac{4\pi^2 L}{T^2} \]
where:
- \( L \) is the length of the pendulum,
- \( T \) is the time period of one oscillation.

### Step 2: Determine the values and their uncertainties
From the problem:
- Length \( L = 20.0 \, \text{cm} = 0.20 \, \text{m} \)
- Time for 100 oscillations \( T_{100} = 90.0 \, \text{s} \)
- Time period \( T = \frac{T_{100}}{100} = \frac{90.0}{100} = 0.90 \, \text{s} \)

The least count of the meter scale is \( 1 \, \text{mm} = 0.001 \, \text{m} \), so:
- Uncertainty in length \( \Delta L = 0.001 \, \text{m} \)

The least count of the watch is \( 1 \, \text{s} \), so:
- Uncertainty in time \( \Delta T_{100} = 1 \, \text{s} \)
- Therefore, uncertainty in time period \( \Delta T = \frac{\Delta T_{100}}{100} = \frac{1}{100} = 0.01 \, \text{s} \)

### Step 3: Calculate the percentage error in \( g \)
The formula for the percentage error in \( g \) is given by:
\[ \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \cdot \frac{\Delta T}{T} \]

### Step 4: Substitute the values into the error formula
1. Calculate \( \frac{\Delta L}{L} \):
\[ \frac{\Delta L}{L} = \frac{0.001 \, \text{m}}{0.20 \, \text{m}} = 0.005 = 0.5\% \]

2. Calculate \( \frac{\Delta T}{T} \):
\[ \frac{\Delta T}{T} = \frac{0.01 \, \text{s}}{0.90 \, \text{s}} \approx 0.0111 \approx 1.11\% \]

3. Now substitute these values into the error formula:
\[ \frac{\Delta g}{g} = 0.005 + 2 \cdot 0.0111 = 0.005 + 0.0222 = 0.0272 \]

### Step 5: Convert to percentage
To find the percentage error:
\[ \text{Percentage error in } g = 0.0272 \times 100 \approx 2.72\% \]

### Conclusion
The error in the determination of \( g \) is approximately \( 2.72\% \).