The least count of a stop watch is `(1//5)s`. The time 20 oscillations of a pendulum is measured to be `25 s`. The maximum percentage error in this measurement is

Options

A. `8%`
B. `1%`
C. `0.8%`
D. `16%`

Explanation

To solve the problem, we need to calculate the maximum percentage error in the measurement of the time taken for 20 oscillations of a pendulum. Here’s a step-by-step breakdown:

### Step 1: Identify the given values
- Least count of the stopwatch (LC) = \( \frac{1}{5} \) seconds = 0.2 seconds
- Total time for 20 oscillations (T) = 25 seconds

### Step 2: Calculate the error for one oscillation
The least count gives us the maximum possible error in the measurement. Since the time is measured for 20 oscillations, we need to find the error per oscillation.

\[ \text{Error per oscillation} = \frac{\text{Least Count}}{\text{Number of oscillations}} = \frac{LC}{20} = \frac{0.2 \text{ seconds}}{20} = 0.01 \text{ seconds} \]

### Step 3: Calculate the time period for one oscillation
The time period (T₀) for one oscillation can be calculated by dividing the total time by the number of oscillations.

\[ T_0 = \frac{T}{20} = \frac{25 \text{ seconds}}{20} = 1.25 \text{ seconds} \]

### Step 4: Calculate the maximum percentage error
The percentage error can be calculated using the formula:

\[ \text{Percentage Error} = \left( \frac{\text{Error per oscillation}}{\text{Time period for one oscillation}} \right) \times 100 \]

Substituting the values we have:

\[ \text{Percentage Error} = \left( \frac{0.01 \text{ seconds}}{1.25 \text{ seconds}} \right) \times 100 \]

Calculating this gives:

\[ \text{Percentage Error} = \left( \frac{0.01}{1.25} \right) \times 100 = 0.008 \times 100 = 0.8\% \]

### Final Answer
The maximum percentage error in this measurement is **0.8%**.

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