The least count of a stop watch is 1/5 second. The time of 20 oscillations of a pendulum is measured to be 25 seconds. The maximum percentage error in the measurement of time will be

Options

A. `0.1`%
B. 0.008
C. 0.018
D. 0.08

Explanation

To solve the problem of finding the maximum percentage error in the measurement of time for 20 oscillations of a pendulum, we will follow these steps:

### Step 1: Identify the given values
- Least count of the stopwatch (Δt) = 1/5 seconds = 0.2 seconds
- Time for 20 oscillations (T) = 25 seconds

### Step 2: Calculate the error per oscillation
To find the error per oscillation, we divide the least count by the number of oscillations:
\[ \text{Error per oscillation} = \frac{\Delta t}{n} = \frac{1/5}{20} = \frac{0.2}{20} = 0.01 \text{ seconds} \]

### Step 3: Calculate the time period for one oscillation
The time period (t) for one oscillation can be calculated by dividing the total time for 20 oscillations by 20:
\[ t = \frac{T}{n} = \frac{25 \text{ seconds}}{20} = 1.25 \text{ seconds} \]

### Step 4: Calculate the percentage error
The percentage error in the measurement can be calculated using the formula:
\[ \text{Percentage error} = \left( \frac{\text{Error per oscillation}}{t} \right) \times 100 \]
Substituting the values we calculated:
\[ \text{Percentage error} = \left( \frac{0.01}{1.25} \right) \times 100 \]
Calculating this gives:
\[ \text{Percentage error} = \left( 0.008 \right) \times 100 = 0.8\% \]

### Conclusion
The maximum percentage error in the measurement of time is **0.8%**.