The length of a simple pendulum is about 100 cm known to have an accuracy of 1mm. Its period of oscillation is 2 s determined by measuring the time for 100 oscillations using a clock of 0.1s resolution. What is the accuracy in the determined value of g?

Options

A. 0.002
B. 0.005
C. 0.001
D. 0.02

Explanation

To find the accuracy in the determined value of \( g \) for a simple pendulum, we will follow these steps:

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

### Step 2: Identify the known values and their accuracies
- Length \( L = 100 \, \text{cm} = 1.00 \, \text{m} \) with an accuracy of \( \Delta L = 1 \, \text{mm} = 0.001 \, \text{m} \).
- Period \( T = 2 \, \text{s} \) determined by measuring the time for 100 oscillations, with a clock resolution of \( 0.1 \, \text{s} \).

### Step 3: Calculate the accuracy in the period \( T \)
Since the time for 100 oscillations is measured, the period \( T \) can be calculated as:
\[ T = \frac{\text{Total time for 100 oscillations}}{100} \]
The accuracy in the total time measurement is \( 0.1 \, \text{s} \), so the accuracy in the period \( T \) is:
\[ \Delta T = \frac{0.1 \, \text{s}}{100} = 0.001 \, \text{s} \]

### Step 4: Calculate the relative uncertainties
The relative uncertainty in length \( \frac{\Delta L}{L} \) is:
\[ \frac{\Delta L}{L} = \frac{0.001 \, \text{m}}{1.00 \, \text{m}} = 0.001 \]
The relative uncertainty in period \( \frac{\Delta T}{T} \) is:
\[ \frac{\Delta T}{T} = \frac{0.001 \, \text{s}}{2 \, \text{s}} = 0.0005 \]

### Step 5: Use the formula for uncertainty in \( g \)
The formula for the uncertainty in \( g \) based on the uncertainties in \( L \) and \( T \) is given by:
\[ \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T} \]
Substituting the values we calculated:
\[ \frac{\Delta g}{g} = 0.001 + 2 \times 0.0005 = 0.001 + 0.001 = 0.002 \]

### Step 6: Calculate the absolute uncertainty in \( g \)
Now, to find the absolute uncertainty \( \Delta g \):
\[ \Delta g = g \times \frac{\Delta g}{g} \]
However, since we are interested in the relative uncertainty, we can directly state:
\[ \Delta g = 0.002 \times g \]
Since we do not need the exact value of \( g \) for this problem, we conclude that the accuracy in the determined value of \( g \) is \( 0.002 \).

### Final Answer
The accuracy in the determined value of \( g \) is \( 0.002 \).

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