A boy performs an experiment in which he uses a simple pendulum to find the value of 'g' using the formula `g=(4pi^(2)l)/(T^(2))=` Where l = 1m. Error in measurements of length is `Delta l`, human error in time measurement is 0.15 sec and least count of stop watch `Delta T`, for what value of amplitude (A). `Delta l` and `Delta T` error in calculation of 'g' is maximum.

Options

A. `A=1.4, Delta l = 0.5m, Delta T = 0.1 sec`
B. `A=0.4, Delta l = 0.1 m, Delta T = 0.5 sec`
C. `A=0.5, Delta l =0.5m, Delta T = 0.2 sec`
D. `A=0.05, Delta l = 0.1 m, Delta T = 0.5 sec`

Answer

To solve the problem, we need to analyze the formula given for calculating the acceleration due to gravity (g) using a simple pendulum. The formula is:

\[ g = \frac{4\pi^2 l}{T^2} \]

Where:
- \( l \) = length of the pendulum (1 m)
- \( T \) = time period of the pendulum

We are given:
- Error in measurements of length = \( \Delta l \)
- Human error in time measurement = \( \Delta T = 0.15 \) sec
- Least count of stopwatch = \( \Delta T \)

We need to find the value of the amplitude (A) for which the errors \( \Delta l \) and \( \Delta T \) in the calculation of \( g \) are maximum.

### Step 1: Understand the relationship between errors and \( g \)

From the formula for \( g \), we can express the relative error in \( g \) as:

\[ \frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T} \]

### Step 2: Substitute known values

Since \( l = 1 \) m, we can substitute this into the equation:

\[ \frac{\Delta g}{g} = \Delta l + 2\frac{\Delta T}{T} \]

### Step 3: Analyze the time period \( T \)

The time period \( T \) for a simple pendulum is given by:

\[ T = 2\pi\sqrt{\frac{l}{g}} \]

For \( l = 1 \) m and assuming \( g \approx 9.81 \) m/s², we can calculate \( T \):

\[ T = 2\pi\sqrt{\frac{1}{9.81}} \approx 0.63 \text{ sec} \]

### Step 4: Calculate \( \Delta T \)

Given \( \Delta T = 0.15 \) sec, we can now find the relative error in \( T \):

\[ \frac{\Delta T}{T} = \frac{0.15}{0.63} \approx 0.238 \]

### Step 5: Substitute \( \Delta T \) into the error formula

Now substituting this back into the error formula for \( g \):

\[ \frac{\Delta g}{g} = \Delta l + 2(0.238) \]

### Step 6: Maximize the error

To maximize \( \Delta g \), we need to consider the values of \( \Delta l \) and \( \Delta T \). Since \( \Delta l \) is a variable that we can control, we can set it to a value that maximizes the overall error.

### Step 7: Find the value of amplitude \( A \)

The amplitude \( A \) of the pendulum affects the time period \( T \) and thus the error in \( g \). For small angles, the amplitude can be approximated as:

\[ A \approx \Delta l \]

To maximize the error in \( g \), we can set \( \Delta l \) to a value that balances the errors.

### Conclusion

Thus, the value of amplitude \( A \) that maximizes the errors \( \Delta l \) and \( \Delta T \) in the calculation of \( g \) is approximately:

\[ A \approx 1.4 \text{ m} \]