Student `I ,II , and III` perform an experiment for measuring the acceleration due to gravity `(g) ` usinf a simple pendulum. They use lengths of the pendulum and // or record time for different number of oscillations . The observations are shown in the following table . Least count for length ` = 0.1 cm`
`{:(underset(III)underset(II)underset(I)underset()underset()("Student"),underset(20.0)underset(64.0)underset(64.0)underset((cm))underset("Pendulam")("Length of "),underset(4)underset(4)underset(8)underset((n))underset("n Oscillation")("Number of"),underset(9.0)underset(16.0)underset(16.0)underset((s))underset("Period")("Time")):}`
Least count for time `= 0.1 s`.
If `E_(I), E_(II)` , and `E_(III)` are the percentage errors in `g` , i.,e., `((Delta g)/( g) xx 100)` for students I,II , and III, respectively , then

Options

A. `E_(I) = 0`
B. ` E_(I) is minimum`
C. `E_(I) = E_(II)`
D. `E_(II) is maximum`

Explanation

To solve the problem of calculating the percentage errors in the acceleration due to gravity (g) for students I, II, and III, we can follow these steps:

### Step 1: Understand the formula for g
The formula for calculating the acceleration due to gravity (g) using the period (T) of a simple pendulum is given by:
\[ g = \frac{4\pi^2 L}{T^2} \]
where \( L \) is the length of the pendulum and \( T \) is the period of oscillation.

### Step 2: Calculate g for each student
We need to calculate \( g \) for each student using their respective lengths and periods.

1. **For Student I:**
- Length \( L_I = 64.0 \, \text{cm} = 0.64 \, \text{m} \)
- Period \( T_I = 9.0 \, \text{s} \)
\[ g_I = \frac{4\pi^2 \times 0.64}{9.0^2} \]

2. **For Student II:**
- Length \( L_{II} = 64.0 \, \text{cm} = 0.64 \, \text{m} \)
- Period \( T_{II} = 16.0 \, \text{s} \)
\[ g_{II} = \frac{4\pi^2 \times 0.64}{16.0^2} \]

3. **For Student III:**
- Length \( L_{III} = 20.0 \, \text{cm} = 0.20 \, \text{m} \)
- Period \( T_{III} = 4.0 \, \text{s} \)
\[ g_{III} = \frac{4\pi^2 \times 0.20}{4.0^2} \]

### Step 3: Calculate the percentage error in g
The formula for the percentage error in \( g \) is given by:
\[ E = \frac{\Delta g}{g} \times 100 = \frac{\Delta L}{L} \times 100 + 2 \times \frac{\Delta T}{T} \times 100 \]
Where:
- \( \Delta L \) is the least count for length = 0.1 cm = 0.001 m
- \( \Delta T \) is the least count for time = 0.1 s

Now we can calculate the percentage errors for each student.

1. **For Student I:**
\[ E_I = \frac{0.1}{0.64} \times 100 + 2 \times \frac{0.1}{9.0} \times 100 \]

2. **For Student II:**
\[ E_{II} = \frac{0.1}{0.64} \times 100 + 2 \times \frac{0.1}{16.0} \times 100 \]

3. **For Student III:**
\[ E_{III} = \frac{0.1}{0.20} \times 100 + 2 \times \frac{0.1}{4.0} \times 100 \]

### Step 4: Calculate the values
Now we can compute the actual values for \( E_I, E_{II}, \) and \( E_{III} \).

1. **For Student I:**
\[ E_I = \frac{0.1}{0.64} \times 100 + 2 \times \frac{0.1}{9.0} \times 100 = 15.625 + 2.222 = 17.847 \% \]

2. **For Student II:**
\[ E_{II} = \frac{0.1}{0.64} \times 100 + 2 \times \frac{0.1}{16.0} \times 100 = 15.625 + 1.25 = 16.875 \% \]

3. **For Student III:**
\[ E_{III} = \frac{0.1}{0.20} \times 100 + 2 \times \frac{0.1}{4.0} \times 100 = 50 + 5 = 55 \% \]

### Final Results
- \( E_I \approx 17.85\% \)
- \( E_{II} \approx 16.88\% \)
- \( E_{III} \approx 55\% \)