A student performs exeriment with simple pendulum and measures time for `10` vibrations. If the measures the time for `100` vibrations, the error in measurement of time period will be reduced by a factor of......

Options

A. `1000`
B. `100`
C. `90`
D. `10`

Answer

To solve the problem, we need to understand how the error in measurement changes when the number of vibrations measured increases.

### Step-by-Step Solution:

1. **Understanding the Relationship Between Measurements and Error**:
- The error in measurement is inversely related to the number of readings taken. This means that if you take more readings, the error decreases.

2. **Initial Measurement**:
- The student initially measures the time for `10` vibrations. Let's denote the error in this measurement as \( e_1 \).

3. **New Measurement**:
- The student then measures the time for `100` vibrations. We denote the error in this measurement as \( e_2 \).

4. **Establishing the Relationship**:
- Since the error is inversely proportional to the number of readings, we can express this relationship mathematically:
\[ e_1 \propto \frac{1}{n_1} \quad \text{and} \quad e_2 \propto \frac{1}{n_2} \]
- Here, \( n_1 = 10 \) and \( n_2 = 100 \).

5. **Calculating the Errors**:
- We can express the errors as:
\[ e_1 = k \cdot \frac{1}{10} \quad \text{and} \quad e_2 = k \cdot \frac{1}{100} \]
- Where \( k \) is a proportionality constant.

6. **Finding the Factor of Reduction**:
- To find the factor by which the error is reduced, we can take the ratio of the two errors:
\[ \text{Factor of reduction} = \frac{e_1}{e_2} = \frac{k \cdot \frac{1}{10}}{k \cdot \frac{1}{100}} = \frac{100}{10} = 10 \]

7. **Conclusion**:
- Therefore, the error in the measurement of the time period will be reduced by a factor of **10**.

### Final Answer:
The error in measurement of the time period will be reduced by a factor of **10**.