The period of oscillation of a simple pendulum is `T = 2 pi sqrt((L)/(g)) .L` is about `10 cm` and is known to `1mm` accuracy . The period of oscillation is about `0.5 s`. The time of 100 oscillation is measured with a wrist watch of `1 s` resolution . What is the accuracy in the determination of `g` ?

Explanation

To determine the accuracy in the determination of \( g \) based on the provided information, we can follow these steps:

### Step 1: Identify the given values
- Length \( L = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Accuracy in length \( \Delta L = 1 \, \text{mm} = 0.001 \, \text{m} \)
- Period of oscillation \( T = 0.5 \, \text{s} \)
- Time for 100 oscillations measured with a wristwatch of \( 1 \, \text{s} \) resolution.

### Step 2: Calculate the percentage error in length
The percentage error in length can be calculated using the formula:
\[ \frac{\Delta L}{L} \times 100 \]
Substituting the values:
\[ \frac{0.001 \, \text{m}}{0.1 \, \text{m}} \times 100 = 1\% \]

### Step 3: Calculate the period of oscillation for 100 oscillations
The total time for 100 oscillations is measured with a resolution of \( 1 \, \text{s} \). Thus, the time for one oscillation is:
\[ \Delta T = \frac{1 \, \text{s}}{100} = 0.01 \, \text{s} \]

### Step 4: Calculate the percentage error in time
The percentage error in the period of oscillation can be calculated as:
\[ \frac{\Delta T}{T} \times 100 \]
Substituting the values:
\[ \frac{0.01 \, \text{s}}{0.5 \, \text{s}} \times 100 = 2\% \]

### Step 5: Use the formula for the percentage error in \( g \)
The formula for the percentage error in \( g \) based on the errors in \( L \) and \( T \) is given by:
\[ \frac{\Delta g}{g} \times 100 = \frac{\Delta L}{L} \times 100 + 2 \times \frac{\Delta T}{T} \times 100 \]
Substituting the values:
\[ \frac{\Delta g}{g} \times 100 = 1\% + 2 \times 2\% \]
\[ \frac{\Delta g}{g} \times 100 = 1\% + 4\% = 5\% \]

### Step 6: Conclusion
The accuracy in the determination of \( g \) is \( 5\% \).

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