The period of oscillation of a simple pendulum is `T = 2pisqrt((L)/(g))`. Meaured value of `L` is `20.0 cm` know to `1mm` accuracy and time for `100` oscillation of the pendulum is found to be `90 s` using a wrist watch of `1 s` resolution. The accracy in the determinetion of `g` is :

Options

A. (a) `1%`
B. (b) `5%`
C. ( c ) `2%`
D. (d) `3%`

Explanation

To determine the accuracy in the determination of \( g \) using the formula for the period of a simple pendulum, we can follow these steps:

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

### Step 2: Identify the variables and their uncertainties
- The measured value of \( L \) is \( 20.0 \, \text{cm} \) with an accuracy of \( \delta L = 1 \, \text{mm} = 0.1 \, \text{cm} \).
- The total time for 100 oscillations is \( T_{100} = 90 \, \text{s} \). Therefore, the period \( T \) for one oscillation is:
\[ T = \frac{90 \, \text{s}}{100} = 0.9 \, \text{s} \]
- The accuracy in the measurement of \( T \) is \( \delta T = 1 \, \text{s} \) (the resolution of the wristwatch).

### Step 3: Calculate the percentage error in \( g \)
The percentage error in \( g \) can be calculated using the formula:
\[ \frac{\delta g}{g} \times 100 = \frac{\delta L}{L} \times 100 + 2 \frac{\delta T}{T} \times 100 \]

#### Step 3.1: Calculate \( \frac{\delta L}{L} \)
\[ \frac{\delta L}{L} = \frac{0.1 \, \text{cm}}{20.0 \, \text{cm}} = 0.005 \]
Thus, the percentage error due to \( L \) is:
\[ \frac{\delta L}{L} \times 100 = 0.005 \times 100 = 0.5\% \]

#### Step 3.2: Calculate \( \frac{\delta T}{T} \)
\[ \frac{\delta T}{T} = \frac{1 \, \text{s}}{90 \, \text{s}} \approx 0.0111 \]
Thus, the percentage error due to \( T \) is:
\[ \frac{\delta T}{T} \times 100 = 0.0111 \times 100 \approx 1.11\% \]

#### Step 3.3: Combine the errors
Now, we can combine the errors:
\[ \text{Total percentage error in } g = 0.5\% + 2 \times 1.11\% = 0.5\% + 2.22\% = 2.72\% \]

### Step 4: Final result
The accuracy in the determination of \( g \) is approximately \( 2.72\% \), which can be rounded to \( 3\% \).