Transverse waves are generated in two uniform steel wires `A and B` by attaching their free ends to a fork of frequency `500 Hz`. The diameter of wire `A` is half that `B` and tension in wire `A` is half the tension in wire `B`. What is the ratio of velocities of waves in `A and B`?

Options

A. `1:2`
B. `sqrt2:1`
C. `2:1`
D. `1:sqrt2`

Explanation

To solve the problem, we need to find the ratio of the velocities of waves in two steel wires A and B, given the conditions about their diameters and tensions.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- Frequency of the fork: \( f = 500 \, \text{Hz} \)
- Diameter of wire A: \( D_A = \frac{1}{2} D_B \)
- Tension in wire A: \( T_A = \frac{1}{2} T_B \)

2. **Formula for Wave Velocity**:
The velocity \( v \) of a wave in a string is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension and \( \mu \) is the mass per unit length of the wire.

3. **Calculate Mass per Unit Length**:
The mass per unit length \( \mu \) can be expressed as:
\[ \mu = \rho \cdot A \]
where \( \rho \) is the density of the material and \( A \) is the cross-sectional area of the wire. For a circular wire, the area \( A \) is given by:
\[ A = \frac{\pi D^2}{4} \]

4. **Cross-Sectional Areas**:
For wire A:
\[ A_A = \frac{\pi (D_A)^2}{4} = \frac{\pi \left(\frac{1}{2} D_B\right)^2}{4} = \frac{\pi D_B^2}{16} \]
For wire B:
\[ A_B = \frac{\pi (D_B)^2}{4} \]

5. **Mass per Unit Length for Each Wire**:
For wire A:
\[ \mu_A = \rho \cdot A_A = \rho \cdot \frac{\pi D_B^2}{16} \]
For wire B:
\[ \mu_B = \rho \cdot A_B = \rho \cdot \frac{\pi D_B^2}{4} \]

6. **Ratio of Mass per Unit Length**:
\[ \frac{\mu_A}{\mu_B} = \frac{\frac{\rho \cdot \frac{\pi D_B^2}{16}}{\rho \cdot \frac{\pi D_B^2}{4}} = \frac{1/16}{1/4} = \frac{1}{4} \]

7. **Substituting into the Velocity Formula**:
The velocities can be expressed as:
\[ v_A = \sqrt{\frac{T_A}{\mu_A}} \quad \text{and} \quad v_B = \sqrt{\frac{T_B}{\mu_B}} \]

8. **Substituting Tensions**:
Since \( T_A = \frac{1}{2} T_B \):
\[ v_A = \sqrt{\frac{\frac{1}{2} T_B}{\mu_A}} \quad \text{and} \quad v_B = \sqrt{\frac{T_B}{\mu_B}} \]

9. **Finding the Ratio of Velocities**:
\[ \frac{v_A}{v_B} = \frac{\sqrt{\frac{\frac{1}{2} T_B}{\mu_A}}}{\sqrt{\frac{T_B}{\mu_B}}} = \sqrt{\frac{\frac{1}{2}}{\frac{\mu_A}{\mu_B}}} \]
Substituting \( \frac{\mu_A}{\mu_B} = \frac{1}{4} \):
\[ \frac{v_A}{v_B} = \sqrt{\frac{\frac{1}{2}}{\frac{1}{4}}} = \sqrt{2} \]

10. **Final Answer**:
The ratio of the velocities of waves in wires A and B is:
\[ \frac{v_A}{v_B} = \sqrt{2} : 1 \]