In Melde's experiment, 8 loops are formed with a tension of a `0.75 N`. If the tension is increased to four times then the number of loops produced will be -

Options

A. 4
B. 8
C. 2
D. 12

Explanation

To solve the problem step by step, we can use the relationship between tension and the number of loops in Melde's experiment. The relationship can be expressed as:

\[ T_1 \cdot P_1^2 = T_2 \cdot P_2^2 \]

Where:
- \( T_1 \) is the initial tension,
- \( P_1 \) is the initial number of loops,
- \( T_2 \) is the final tension,
- \( P_2 \) is the final number of loops.

### Step 1: Identify the given values
From the problem, we know:
- \( T_1 = 0.75 \, \text{N} \)
- \( P_1 = 8 \)
- \( T_2 = 4 \times T_1 = 4 \times 0.75 = 3.0 \, \text{N} \)

### Step 2: Substitute the values into the equation
Now we can substitute the known values into the equation:

\[ 0.75 \cdot 8^2 = 3.0 \cdot P_2^2 \]

### Step 3: Calculate \( P_1^2 \)
Calculate \( P_1^2 \):

\[ P_1^2 = 8^2 = 64 \]

### Step 4: Substitute \( P_1^2 \) into the equation
Now substitute \( P_1^2 \) back into the equation:

\[ 0.75 \cdot 64 = 3.0 \cdot P_2^2 \]

### Step 5: Simplify the equation
Calculating the left side:

\[ 48 = 3.0 \cdot P_2^2 \]

### Step 6: Solve for \( P_2^2 \)
Now, divide both sides by 3.0 to isolate \( P_2^2 \):

\[ P_2^2 = \frac{48}{3.0} = 16 \]

### Step 7: Take the square root to find \( P_2 \)
Now take the square root of both sides to find \( P_2 \):

\[ P_2 = \sqrt{16} = 4 \]

### Conclusion
Thus, the number of loops produced when the tension is increased to four times is:

\[ \boxed{4} \]