The period of oscillation of a simple pendulum is `T = 2pisqrt(L//g)`. Measured value of L is `20.0 cm` known to `1mm` accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. What is the accuracy in the determination of g?

Explanation

To determine the accuracy in the measurement of the acceleration due to gravity \( g \) using the given data, we can follow these steps:

### Step 1: Understand the formula for the period of a simple pendulum
The period of oscillation \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

### Step 2: Rearrange the formula to express \( g \)
We can rearrange the formula to express \( g \) in terms of \( L \) and \( T \):
\[ g = \frac{4\pi^2 L}{T^2} \]

### Step 3: Identify the uncertainties in measurements
We are given:
- The length \( L = 20.0 \, \text{cm} \) with an accuracy of \( \Delta L = 1 \, \text{mm} = 0.1 \, \text{cm} \).
- The time for 100 oscillations is \( 90 \, \text{s} \), so the period \( T \) is:
\[ T = \frac{90 \, \text{s}}{100} = 0.9 \, \text{s} \]
- The resolution of the wristwatch gives an uncertainty in time \( \Delta T = 1 \, \text{s} \).

### Step 4: Calculate the percentage uncertainty in \( g \)
The percentage uncertainty in \( g \) can be calculated using the formula:
\[ \frac{\Delta g}{g} \times 100 = \frac{\Delta L}{L} \times 100 + 2 \times \frac{\Delta T}{T} \times 100 \]

### Step 5: Substitute the values into the formula
1. Calculate \( \frac{\Delta L}{L} \):
\[ \frac{\Delta L}{L} = \frac{0.1 \, \text{cm}}{20.0 \, \text{cm}} = 0.005 \]
Therefore,
\[ \frac{\Delta L}{L} \times 100 = 0.005 \times 100 = 0.5\% \]

2. Calculate \( \frac{\Delta T}{T} \):
\[ \frac{\Delta T}{T} = \frac{1 \, \text{s}}{90 \, \text{s}} \approx 0.0111 \]
Therefore,
\[ 2 \times \frac{\Delta T}{T} \times 100 = 2 \times 0.0111 \times 100 \approx 2.22\% \]

### Step 6: Combine the uncertainties
Now, we can combine the uncertainties:
\[ \frac{\Delta g}{g} \times 100 = 0.5\% + 2.22\% = 2.72\% \]

### Final Answer
The accuracy in the determination of \( g \) is approximately:
\[ \boxed{2.72\%} \]