As student performs an experiment for determine of `g[=(4pi^(2)L)/(T^(2))]. L approx 1m`, and has commits an error of `Delta L` for `T` he tajes the teime of `n` osciollations wityh the stop watch of least count `Delta T`. For which of the following data the measurement of `g` will be most accurate?

Options

A. ` Delta L = 0.5 , Delta T = 0.1 , n = 20`
B. ` Delta L = 0.5 , Delta T = 0.1 , n = 50`
C. ` Delta L = 0.5 , Delta T = 0.01 , n = 20`
D. ` Delta L = 0.5 , Delta T = 0.05 , n = 50`

Explanation

To determine the most accurate measurement of \( g \) in the given experiment, we need to analyze the formula and the errors associated with the measurements. The formula for \( g \) is given by:

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

where \( L \) is the length (approximately 1 m) and \( T \) is the time period of oscillation. The student has committed an error \( \Delta L \) in measuring \( L \) and takes the time of \( n \) oscillations with a stopwatch that has a least count of \( \Delta T \).

### Step-by-Step Solution:

1. **Understanding the Formula**:
The formula for \( g \) shows that it is directly proportional to \( L \) and inversely proportional to the square of \( T \). Thus, any error in \( T \) will have a significant effect on the value of \( g \).

2. **Error Propagation**:
To analyze the errors, we can use the concept of propagation of errors. The relative error in \( g \) can be expressed as:
\[ \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T} \]
Here, \( \Delta g \) is the error in \( g \), \( \Delta L \) is the error in \( L \), and \( \Delta T \) is the error in \( T \).

3. **Calculating \( T \)**:
The time period \( T \) for \( n \) oscillations is given by:
\[ T = \frac{t}{n} \]
where \( t \) is the total time measured for \( n \) oscillations. The error in \( T \) can be derived from the error in \( t \):
\[ \Delta T = \frac{\Delta t}{n} \]
where \( \Delta t \) is the least count of the stopwatch.

4. **Substituting into the Error Formula**:
Substituting \( \Delta T \) into the error propagation equation:
\[ \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta t}{nT} \]
This shows that increasing \( n \) will decrease the term \( \frac{\Delta t}{nT} \), thereby reducing the overall error in \( g \).

5. **Choosing the Best Option**:
To minimize the error in \( g \), we should maximize \( n \) (the number of oscillations). The more oscillations measured, the smaller the relative error in the time period \( T \) becomes. Therefore, the option with the highest \( n \) will yield the most accurate measurement of \( g \).

### Conclusion:
The most accurate measurement of \( g \) will occur when the number of oscillations \( n \) is maximized, as this reduces the relative error associated with the time measurement.