The least count of a stop watch is 0.2 second. The time of 20 oscillations of a pendulum is measured to be 25 seconds. The maximum percentage error in this measurement is found to be 0.x%. What is the value of x?

Answer

To solve the problem, we need to determine the maximum percentage error in the measurement of the time taken for 20 oscillations of a pendulum.

### Step-by-step Solution:

1. **Identify the given values:**
- Least count of the stopwatch (Δt) = 0.2 seconds
- Time for 20 oscillations (T) = 25 seconds
- Number of oscillations (n) = 20

2. **Calculate the time period (T) for one oscillation:**
\[ \text{Time period (t)} = \frac{\text{Total time for 20 oscillations}}{\text{Number of oscillations}} = \frac{25 \text{ seconds}}{20} = 1.25 \text{ seconds} \]

3. **Calculate the error in the time for one oscillation:**
The error in time for one oscillation (Δt) can be calculated as:
\[ \text{Error for one oscillation} = \frac{\text{Least count}}{n} = \frac{0.2 \text{ seconds}}{20} = 0.01 \text{ seconds} \]

4. **Calculate the percentage error:**
The percentage error can be calculated using the formula:
\[ \text{Percentage error} = \left(\frac{\Delta t}{t}\right) \times 100 \]
Substituting the values:
\[ \text{Percentage error} = \left(\frac{0.01}{1.25}\right) \times 100 \]
\[ = 0.008 \times 100 = 0.8\% \]

5. **Determine the value of x:**
The problem states that the maximum percentage error is found to be 0.x%. From our calculation, we found it to be 0.8%, which means:
\[ x = 8 \]

### Final Answer:
The value of x is **8**.

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