In an experiment for determining the value of acceleration due to gravity `(g)` using a simpe pendulum , the following observations were recorded:
Length of the string `(l) = 98.0 cm`
Diameter of the bob `(d) = 2.56 cm`
Time for `10 oscillations (T) = 20.0 s`
Calculate the value of `g` with maximum permissible absolute error and the percentage relative error.

Explanation

To calculate the value of acceleration due to gravity `(g)` using the given data from the simple pendulum experiment, we can follow these steps:

### Step 1: Calculate the Effective Length of the Pendulum
The effective length of the pendulum `(L_eff)` is the length of the string plus the radius of the bob. The radius of the bob can be calculated from its diameter.

Given:
- Length of the string, \( l = 98.0 \, \text{cm} \)
- Diameter of the bob, \( d = 2.56 \, \text{cm} \)

Calculate the radius of the bob:
\[ \text{Radius} = \frac{d}{2} = \frac{2.56 \, \text{cm}}{2} = 1.28 \, \text{cm} \]

Now, calculate the effective length:
\[ L_{eff} = l + \text{Radius} = 98.0 \, \text{cm} + 1.28 \, \text{cm} = 99.28 \, \text{cm} = 0.9928 \, \text{m} \]

### Step 2: Calculate the Time Period of the Pendulum
The time period \( T \) for one oscillation can be calculated from the total time for 10 oscillations.

Given:
- Time for 10 oscillations, \( T_{10} = 20.0 \, \text{s} \)

Calculate the time period for one oscillation:
\[ T = \frac{T_{10}}{10} = \frac{20.0 \, \text{s}}{10} = 2.0 \, \text{s} \]

### Step 3: Calculate the Value of \( g \)
Using the formula for the acceleration due to gravity derived from the time period of a simple pendulum:
\[ g = \frac{4\pi^2 L_{eff}}{T^2} \]

Substituting the values:
\[ g = \frac{4 \cdot \pi^2 \cdot 0.9928 \, \text{m}}{(2.0 \, \text{s})^2} \]
\[ g = \frac{4 \cdot 9.8696 \cdot 0.9928}{4} = 9.8696 \cdot 0.9928 \approx 9.8 \, \text{m/s}^2 \]

### Step 4: Calculate the Maximum Permissible Absolute Error in \( g \)
To find the maximum permissible absolute error in \( g \), we need to consider the errors in the measurements of \( L \) and \( T \).

1. **Error in Length**:
- The least count for length measurement is \( 0.1 \, \text{cm} = 0.001 \, \text{m} \).
- So, \( \Delta L = 0.001 \, \text{m} \).

2. **Error in Time**:
- The least count for time measurement is \( 0.1 \, \text{s} \).
- So, for 10 oscillations, \( \Delta T_{10} = 0.1 \, \text{s} \).
- Therefore, for 1 oscillation, \( \Delta T = \frac{0.1}{10} = 0.01 \, \text{s} \).

Using the formula for percentage error:
\[ \text{Percentage error in } g = 2 \left( \frac{\Delta L}{L_{eff}} \right) + 2 \left( \frac{\Delta T}{T} \right) \]

Calculating the individual errors:
\[ \frac{\Delta L}{L_{eff}} = \frac{0.001}{0.9928} \approx 0.00101 \]
\[ \frac{\Delta T}{T} = \frac{0.01}{2.0} = 0.005 \]

Now substituting into the percentage error formula:
\[ \text{Percentage error in } g = 2(0.00101) + 2(0.005) \approx 0.00202 + 0.01 = 0.01202 \]
\[ \text{Percentage error in } g \approx 1.2\% \]

### Step 5: Calculate the Absolute Error in \( g \)
The absolute error in \( g \) can be calculated as:
\[ \Delta g = \text{Percentage error} \times g = 0.01202 \times 9.8 \approx 0.118 \, \text{m/s}^2 \]

### Final Result
Thus, the value of \( g \) with the maximum permissible absolute error is:
\[ g = 9.8 \pm 0.12 \, \text{m/s}^2 \]