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?

Options

A. `3%`
B. `1%`
C. `5%`
D. `2%`

Answer

To determine the accuracy in the determination of \( g \) based on the given information, we will follow these steps:

### Step 1: Understand the relationship between \( g \), \( L \), and \( T \)
The formula for the period of oscillation 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 uncertainties in \( L \) and \( T \)
- The measured value of \( L \) is \( 20.0 \, \text{cm} \) with an accuracy of \( 1 \, \text{mm} \) (which is \( 0.1 \, \text{cm} \)).
- 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 uncertainty in \( T \) is \( 1 \, \text{s} \) (the resolution of the wristwatch).

### Step 3: Calculate the relative uncertainties
The relative uncertainty in \( L \) is given by:
\[ \frac{\Delta L}{L} = \frac{0.1 \, \text{cm}}{20.0 \, \text{cm}} = 0.005 \]

The relative uncertainty in \( T \) is:
\[ \frac{\Delta T}{T} = \frac{1 \, \text{s}}{90 \, \text{s}} \approx 0.0111 \]

### Step 4: Determine the relative uncertainty in \( g \)
Using the formula for \( g \):
\[ g = \frac{4\pi^2 L}{T^2} \]
The relative uncertainty in \( g \) can be calculated using the formula:
\[ \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T} \]
Substituting the values we found:
\[ \frac{\Delta g}{g} = 0.005 + 2 \times 0.0111 \]
\[ \frac{\Delta g}{g} = 0.005 + 0.0222 = 0.0272 \]

### Step 5: Convert to percentage
To express the relative uncertainty in percentage:
\[ \text{Percentage uncertainty} = \frac{\Delta g}{g} \times 100 = 0.0272 \times 100 \approx 2.72\% \]

### Step 6: Round to the nearest whole number
Rounding \( 2.72\% \) gives approximately \( 3\% \).

### Final Answer
The accuracy in the determination of \( g \) is approximately \( 3\% \).

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