The period of oscillation of a simple pendulum is given by `T = 2pisqrt((L)/(g))`, where L is the length of the pendulum and g is the acceleration due to gravity. The length is measured using a meter scale which has 500 divisions. If the measured value L is 50 cm, the accuracy in the determination of g is `1.1%` and the time taken for 100 oscillations is 100 seconds, what should be the possible error in measurement of the clock in one minute (in milliseconds) ?

Options

A. 1
B. 2
C. 5
D. 0.25

Answer

To solve the problem, we need to determine the possible error in the measurement of the clock in one minute, given the parameters of the pendulum and the accuracy of the measurements.

### Step-by-Step Solution:

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

2. **Given values**:
- Length \( L = 50 \) cm = 0.5 m
- Accuracy in \( g \) is \( 1.1\% \)
- Time for 100 oscillations = 100 seconds

3. **Calculate the time period \( T \)**:
\[ T = \frac{100 \text{ seconds}}{100} = 1 \text{ second} \]

4. **Calculate the least count of the meter scale**:
The least count is given by:
\[ \text{Least Count} = \frac{1 \text{ meter}}{500} = \frac{100 \text{ cm}}{500} = 0.2 \text{ cm} = 0.002 \text{ m} \]
This means the possible error in measuring \( L \) is \( \Delta L = 0.002 \text{ m} \).

5. **Convert the percentage accuracy of \( g \) into a fraction**:
\[ \frac{\Delta g}{g} = \frac{1.1}{100} = 0.011 \]

6. **Calculate the relative error in \( T \)**:
The formula for the relative error in \( T \) is:
\[ \frac{\Delta T}{T} = \frac{1}{2} \left( \frac{\Delta L}{L} + \frac{\Delta g}{g} \right) \]

7. **Substituting values**:
- \( \Delta L = 0.002 \text{ m} \)
- \( L = 0.5 \text{ m} \)
- \( \Delta g/g = 0.011 \)

So,
\[ \frac{\Delta L}{L} = \frac{0.002}{0.5} = 0.004 \]

Now substituting into the error formula:
\[ \frac{\Delta T}{T} = \frac{1}{2} \left( 0.004 + 0.011 \right) = \frac{1}{2} \times 0.015 = 0.0075 \]

8. **Calculate \( \Delta T \)**:
Since \( T = 1 \text{ second} \):
\[ \Delta T = 0.0075 \times 1 = 0.0075 \text{ seconds} = 7.5 \text{ milliseconds} \]

9. **Determine the error in one minute**:
In 100 seconds, the error is \( 7.5 \text{ milliseconds} \). To find the error in 60 seconds:
\[ \text{Error in 60 seconds} = \frac{7.5 \text{ ms}}{100 \text{ seconds}} \times 60 \text{ seconds} = 4.5 \text{ milliseconds} \]

### Final Answer:
The possible error in the measurement of the clock in one minute is approximately **4.5 milliseconds**.