The time period of oscillation of a simple pendulum is given by `T=2pisqrt(l//g)`
The length of the pendulum is measured as `1=10+-0.1` cm and the time period as `T=0.5+-0.02s`. Determine percentage error in te value of g.

Options

A. `5%`
B. `8%`
C. `7%`
D. None of these

Answer

To determine the percentage error in the value of \( g \) based on the given measurements of the pendulum's length and time period, we can follow these steps:

### Step 1: Understand the relationship
The time period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
From this, we can derive \( g \):
\[ g = \frac{4\pi^2 L}{T^2} \]

### Step 2: Identify the given values
We have the following measurements:
- Length of the pendulum: \( L = 10 \, \text{cm} \) with an uncertainty of \( \Delta L = 0.1 \, \text{cm} \)
- Time period: \( T = 0.5 \, \text{s} \) with an uncertainty of \( \Delta T = 0.02 \, \text{s} \)

### Step 3: Differentiate the equation for \( g \)
To find the percentage error in \( g \), we differentiate the equation:
\[ g = \frac{4\pi^2 L}{T^2} \]
Using the formula for relative errors:
\[ \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T} \]

### Step 4: Substitute the values
Now, we substitute the values into the equation:
- For \( \Delta L \):
\[ \frac{\Delta L}{L} = \frac{0.1}{10} = 0.01 \]
- For \( \Delta T \):
\[ \frac{\Delta T}{T} = \frac{0.02}{0.5} = 0.04 \]
Now, substituting these into the error equation:
\[ \frac{\Delta g}{g} = 0.01 + 2 \times 0.04 = 0.01 + 0.08 = 0.09 \]

### Step 5: Calculate the percentage error
To find the percentage error, we multiply by 100:
\[ \text{Percentage error in } g = 0.09 \times 100 = 9\% \]

### Final Answer
The percentage error in the value of \( g \) is **9%**.

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