The time period of a pendulum is given by `T = 2 pi sqrt((L)/(g))`. The length of pendulum is `20 cm` and is measured up to `1 mm` accuracy. The time period is about `0.6 s`. The time of `100` oscillations is measured with a watch of `1//10 s` resolution. What is the accuracy in the determination of `g`?

Explanation

To find the accuracy in the determination of \( g \) using the given formula for the time period of a pendulum, we will follow these steps:

### Step 1: Write down the formula for \( g \)
The time period \( T \) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
Squaring both sides, we get:
\[ T^2 = 4\pi^2 \frac{L}{g} \]
From this, we can express \( g \) as:
\[ g = \frac{4\pi^2 L}{T^2} \]

### Step 2: Identify the variables and their uncertainties
- Length of the pendulum \( L = 20 \, \text{cm} \) with an uncertainty \( \Delta L = 1 \, \text{mm} = 0.1 \, \text{cm} \).
- Time period \( T = 0.6 \, \text{s} \).
- The time for 100 oscillations is measured with a resolution of \( \Delta t = 0.1 \, \text{s} \).

### Step 3: Calculate the uncertainty in \( T \)
The time for 100 oscillations is:
\[ T_{100} = 100 \times T = 100 \times 0.6 = 60 \, \text{s} \]
The uncertainty in the time for 100 oscillations is:
\[ \Delta T_{100} = 0.1 \, \text{s} \]
Thus, the uncertainty in the time period \( T \) can be calculated as:
\[ \Delta T = \frac{\Delta T_{100}}{100} = \frac{0.1}{100} = 0.001 \, \text{s} \]

### Step 4: Calculate the relative uncertainties
The relative uncertainty in \( L \) is:
\[ \frac{\Delta L}{L} = \frac{0.1 \, \text{cm}}{20 \, \text{cm}} = \frac{1}{200} \]

The relative uncertainty in \( T \) is:
\[ \frac{\Delta T}{T} = \frac{0.001 \, \text{s}}{0.6 \, \text{s}} = \frac{1}{600} \]

### Step 5: Combine the uncertainties to find \( \Delta g \)
Using the formula for \( g \):
\[ g = \frac{4\pi^2 L}{T^2} \]
The relative uncertainty in \( g \) can be calculated using the following formula:
\[ \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T} \]
Substituting the values:
\[ \frac{\Delta g}{g} = \frac{1}{200} + 2 \times \frac{1}{600} \]
Calculating \( 2 \times \frac{1}{600} = \frac{2}{600} = \frac{1}{300} \):
\[ \frac{\Delta g}{g} = \frac{1}{200} + \frac{1}{300} \]
Finding a common denominator (600):
\[ \frac{1}{200} = \frac{3}{600}, \quad \frac{1}{300} = \frac{2}{600} \]
Thus:
\[ \frac{\Delta g}{g} = \frac{3}{600} + \frac{2}{600} = \frac{5}{600} = \frac{1}{120} \]

### Step 6: Convert relative uncertainty to percentage
To find the percentage uncertainty in \( g \):
\[ \Delta g \text{ (in percentage)} = \frac{\Delta g}{g} \times 100 = \frac{1}{120} \times 100 \approx 0.833\% \]

### Final Result
The accuracy in the determination of \( g \) is approximately \( 0.83\% \).

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