Two strings of copper are stretched to the same tension. If their cross-section area are in the ratio 1 : 4 , then the respective wave velocities will be

Options

A. (a)`4:1`
B. (b)`2:1`
C. (c)`1:2`
D. (d)`1:4`

Answer

To solve the problem of finding the respective wave velocities of two copper strings stretched to the same tension with cross-sectional areas in the ratio of 1:4, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between wave velocity, tension, and mass per unit length**:
The velocity of a wave (v) on a string is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the string and \( \mu \) is the mass per unit length of the string.

2. **Express mass per unit length in terms of density and cross-sectional area**:
The mass per unit length (\( \mu \)) can be expressed as:
\[ \mu = \frac{m}{L} = \frac{\text{density} \times \text{volume}}{L} = \frac{\text{density} \times (A \times L)}{L} = \text{density} \times A \]
where \( A \) is the cross-sectional area and \( L \) is the length of the string.

3. **Substitute \( \mu \) in the wave velocity formula**:
Substituting \( \mu \) into the wave velocity formula gives:
\[ v = \sqrt{\frac{T}{\text{density} \times A}} \]

4. **Set up the ratio of velocities for the two strings**:
Let \( v_1 \) and \( v_2 \) be the velocities of the first and second strings, respectively, with cross-sectional areas \( A_1 \) and \( A_2 \). Since the tension is the same for both strings:
\[ \frac{v_1}{v_2} = \sqrt{\frac{A_2}{A_1}} \]

5. **Use the given ratio of areas**:
The areas are in the ratio \( A_1 : A_2 = 1 : 4 \). Therefore:
\[ A_1 = 1k \quad \text{and} \quad A_2 = 4k \quad \text{for some constant } k \]

6. **Substitute the area values into the velocity ratio**:
\[ \frac{v_1}{v_2} = \sqrt{\frac{4k}{1k}} = \sqrt{4} = 2 \]

7. **Conclude the ratio of velocities**:
Thus, the ratio of the velocities \( v_1 : v_2 = 2 : 1 \).

### Final Answer:
The respective wave velocities will be in the ratio \( 2 : 1 \).

---