The time period of a simple pendulum is given by `T = 2 pi sqrt(L//g)`, where `L` is length and `g` acceleration due to gravity. Measured value of length is `10 cm` known to `1 mm` accuracy and time for 50 oscillations of the pendulum is 80 s using a wrist watch of 1 s resloution. What is the accuracy in the determination of `g`?

Options

A. 0.02
B. 0.03
C. 0.04
D. 0.05

Explanation

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

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

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

### Step 3: Identify the measured values and their uncertainties
- Length \( L = 10 \, \text{cm} = 0.1 \, \text{m} \) with an uncertainty of \( \Delta L = 1 \, \text{mm} = 0.001 \, \text{m} \)
- Time for 50 oscillations \( T_{50} = 80 \, \text{s} \)
- Time period \( T = \frac{T_{50}}{50} = \frac{80 \, \text{s}}{50} = 1.6 \, \text{s} \)
- Uncertainty in time \( \Delta T = 1 \, \text{s} \) (resolution of the wristwatch)

### Step 4: Calculate the uncertainty in the time period `T`
The uncertainty in the time period can be calculated as:
\[ \Delta T = \frac{\Delta T_{50}}{50} = \frac{1 \, \text{s}}{50} = 0.02 \, \text{s} \]

### Step 5: Calculate the uncertainty in `g`
To find the uncertainty in `g`, we need to use the propagation of uncertainties formula. The relative uncertainty in `g` can be calculated as follows:
\[ \frac{\Delta g}{g} = \sqrt{\left(\frac{\Delta L}{L}\right)^2 + \left(2 \frac{\Delta T}{T}\right)^2} \]

Substituting the values:
1. Calculate \( \frac{\Delta L}{L} \):
\[ \frac{\Delta L}{L} = \frac{0.001 \, \text{m}}{0.1 \, \text{m}} = 0.01 \]

2. Calculate \( \frac{\Delta T}{T} \):
\[ \frac{\Delta T}{T} = \frac{0.02 \, \text{s}}{1.6 \, \text{s}} = 0.0125 \]

3. Now substitute these values into the uncertainty formula:
\[ \frac{\Delta g}{g} = \sqrt{(0.01)^2 + (2 \times 0.0125)^2} \]
\[ = \sqrt{0.0001 + 0.000625} \]
\[ = \sqrt{0.000725} \]
\[ \approx 0.0269 \]

### Step 6: Calculate the value of `g`
Now we can calculate the value of `g` using the formula:
\[ g = \frac{4\pi^2 L}{T^2} \]
Substituting \( L = 0.1 \, \text{m} \) and \( T = 1.6 \, \text{s} \):
\[ g = \frac{4 \times (3.14)^2 \times 0.1}{(1.6)^2} \]
\[ \approx \frac{4 \times 9.86 \times 0.1}{2.56} \]
\[ \approx \frac{3.944}{2.56} \]
\[ \approx 1.54 \, \text{m/s}^2 \]

### Step 7: Calculate the absolute uncertainty in `g`
Now, we can calculate the absolute uncertainty in `g`:
\[ \Delta g = g \times \frac{\Delta g}{g} \]
\[ \Delta g = 1.54 \times 0.0269 \]
\[ \approx 0.0414 \, \text{m/s}^2 \]

### Final Result
Thus, the value of `g` with its uncertainty is:
\[ g = 1.54 \pm 0.0414 \, \text{m/s}^2 \]