In a simple pendulum experiment for determination of acceleration due to gravity `(g)`, time taken for `20` oscillations is measure by using a watch of `1` second least count. The means value of time taken comes out to be 30s. The length pf pendulum is measured by using a meter scale of least count 1mm and the value obtined is `55.0 cm` The percentage error in the determination of `g` is close to :

Options

A. `0.7%`
B. `0.2%`
C. `3.5%`
D. `6.8%`

Answer

To solve the problem of determining the percentage error in the calculation of acceleration due to gravity \( g \) from a simple pendulum experiment, we will follow these steps:

### Step 1: Understand the given data
- Time taken for 20 oscillations, \( T_{20} = 30 \, \text{s} \)
- Length of the pendulum, \( L = 55.0 \, \text{cm} = 0.55 \, \text{m} \)
- Least count of the watch (for time measurement), \( \delta t = 1 \, \text{s} \)
- Least count of the meter scale (for length measurement), \( \delta L = 1 \, \text{mm} = 0.001 \, \text{m} \)

### Step 2: Calculate the time period \( T \)
The time period \( T \) for one oscillation is given by:
\[ T = \frac{T_{20}}{20} = \frac{30 \, \text{s}}{20} = 1.5 \, \text{s} \]

### Step 3: Calculate \( g \) using the formula
The formula for the acceleration due to gravity \( g \) in terms of the length \( L \) and the time period \( T \) is:
\[ g = \frac{4\pi^2 L}{T^2} \]
Substituting the values:
\[ g = \frac{4\pi^2 \times 0.55}{(1.5)^2} \]

### Step 4: Calculate \( g \)
Calculating \( g \):
\[ g = \frac{4 \times (3.14)^2 \times 0.55}{2.25} \approx \frac{4 \times 9.86 \times 0.55}{2.25} \approx \frac{21.688}{2.25} \approx 9.65 \, \text{m/s}^2 \]

### Step 5: Calculate the percentage error in \( g \)
The percentage error in \( g \) can be calculated using the formula:
\[ \text{Percentage error} = \left( \frac{\Delta g}{g} \right) \times 100 \]
Where \( \Delta g \) is the sum of the percentage errors due to the length and time period measurements.

### Step 6: Calculate the errors in \( L \) and \( T \)
1. **Error in length \( L \)**:
\[ \Delta L = \delta L = 0.001 \, \text{m} \]
The percentage error in \( L \):
\[ \frac{\Delta L}{L} \times 100 = \frac{0.001}{0.55} \times 100 \approx 0.182 \% \]

2. **Error in time period \( T \)**:
The least count for time measurement gives:
\[ \Delta t = \frac{\delta t}{20} = \frac{1}{20} = 0.05 \, \text{s} \]
The percentage error in \( T \):
\[ \frac{\Delta T}{T} \times 100 = \frac{0.05}{1.5} \times 100 \approx 3.33 \% \]

### Step 7: Combine the errors
The total percentage error in \( g \) is given by:
\[ \text{Total percentage error} = \text{Percentage error in } L + 2 \times \text{Percentage error in } T \]
Substituting the values:
\[ \text{Total percentage error} = 0.182 + 2 \times 3.33 \approx 0.182 + 6.66 \approx 6.84 \% \]

### Final Answer
Thus, the percentage error in the determination of \( g \) is approximately \( 6.8\% \).