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 step by step, we will follow the given information and derive the possible error in the measurement of the clock in one minute.

### Step 1: Understanding the Given Information
We are given:
- The formula for the period of oscillation of a simple pendulum:
\[ T = 2\pi\sqrt{\frac{L}{g}} \]
- The length \( L = 50 \) cm.
- The accuracy in the determination of \( g \) is \( 1.1\% \).
- The time taken for 100 oscillations is 100 seconds.

### Step 2: Calculate the Time Period \( T \)
The time period \( T \) for one oscillation can be calculated as:
\[ T = \frac{\text{Total time for 100 oscillations}}{\text{Number of oscillations}} = \frac{100 \text{ seconds}}{100} = 1 \text{ second} \]

### Step 3: Calculate the Least Count of the Meter Scale
The meter scale has 500 divisions and measures up to 100 cm:
\[ \text{Least Count} = \frac{100 \text{ cm}}{500} = 0.2 \text{ cm} \]
Thus, the least count error in length \( \Delta L \) is 0.2 cm.

### Step 4: Calculate the Relative Error in Length
The relative error in length can be calculated as:
\[ \frac{\Delta L}{L} = \frac{0.2 \text{ cm}}{50 \text{ cm}} = \frac{0.2}{50} = \frac{2}{500} = \frac{4}{1000} \]

### Step 5: Convert the Accuracy of \( g \) to a Fraction
The accuracy in the determination of \( g \) is given as \( 1.1\% \):
\[ \frac{\Delta g}{g} = \frac{1.1}{100} = \frac{11}{1000} \]

### Step 6: Use the Formula for Error Propagation
From the formula for the period \( T \):
\[ T^2 = \frac{4\pi^2 L}{g} \]
The error propagation formula gives us:
\[ \frac{2\Delta T}{T} = \frac{\Delta L}{L} + \frac{\Delta g}{g} \]
Substituting the values:
\[ \frac{2\Delta T}{T} = \frac{4}{1000} + \frac{11}{1000} = \frac{15}{1000} \]
Thus, we have:
\[ \Delta T = \frac{15}{2000} T \]

### Step 7: Substitute \( T \) to Find \( \Delta T \)
Since \( T = 1 \) second:
\[ \Delta T = \frac{15}{2000} \times 1 = \frac{15}{2000} \text{ seconds} \]

### Step 8: Calculate the Error for One Minute
Now, we need to find the error in measurement for one minute (60 seconds):
\[ \Delta T_{\text{one minute}} = \Delta T \times 60 = \frac{15}{2000} \times 60 = \frac{15 \times 60}{2000} = \frac{900}{2000} = 0.45 \text{ seconds} \]

### Step 9: Convert Seconds to Milliseconds
To convert this to milliseconds:
\[ 0.45 \text{ seconds} = 0.45 \times 1000 = 450 \text{ milliseconds} \]

### Final Answer
The possible error in measurement of the clock in one minute is **450 milliseconds**.