The period of oscillation of a simple pendulum is given by `T=2pisqrt((l)/(g))` where l is about 100 cm and is known to have 1 mm accuracy. The period is about 2 s. The time of 100 oscillation is measrued by a stop watch of least count 0.1 s. The percentage error is g is

Options

A. `0.1%`
B. `1%`
C. `0.2%`
D. `0.8%`

Explanation

To find the percentage error in \( g \) based on the given parameters, we will follow these steps:

### Step-by-Step Solution

1. **Identify the given values:**
- Length \( l = 100 \) cm (which is \( 1 \) m)
- Accuracy in length \( \Delta l = 1 \) mm = \( 0.1 \) cm
- Period \( T \approx 2 \) s
- Least count of stopwatch \( \Delta T = 0.1 \) s
- Number of oscillations \( n = 100 \)

2. **Calculate the period of 100 oscillations:**
\[ T_{100} = n \cdot T = 100 \cdot 2 = 200 \text{ s} \]

3. **Calculate the percentage error in time \( T \):**
\[ \frac{\Delta T}{T} = \frac{0.1}{200} = 0.0005 \]
To convert this to a percentage:
\[ \text{Percentage error in } T = \frac{0.1}{200} \times 100 = 0.05\% \]

4. **Use the formula for the period of a simple pendulum:**
\[ T = 2\pi \sqrt{\frac{l}{g}} \]
Rearranging gives:
\[ g = \frac{4\pi^2 l}{T^2} \]

5. **Calculate the percentage error in \( g \):**
Using the formula for error propagation:
\[ \frac{\Delta g}{g} = \frac{\Delta l}{l} + 2 \frac{\Delta T}{T} \]
Substitute the known values:
\[ \frac{\Delta g}{g} = \frac{0.1}{100} + 2 \cdot \frac{0.1}{200} \]

6. **Calculate each term:**
- First term:
\[ \frac{0.1}{100} = 0.001 \]
- Second term:
\[ 2 \cdot \frac{0.1}{200} = 2 \cdot 0.0005 = 0.001 \]

7. **Combine the errors:**
\[ \frac{\Delta g}{g} = 0.001 + 0.001 = 0.002 \]

8. **Convert to percentage:**
\[ \text{Percentage error in } g = 0.002 \times 100 = 0.2\% \]

### Final Answer
The percentage error in \( g \) is \( \pm 0.2\% \).

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