Velocity of a wave in a wire is v when tension in it is `2.06 × 10^4 N`. Find value of tension in wire when velocity of wave become `v/2`.

Options

A. `5.15 xx 10^3 N`
B. `8.24 xx 10^4 N`
C. `6 xx 10^4`
D. `5.15 xx 10^4 N`

Explanation

To find the tension in the wire when the velocity of the wave becomes \( \frac{v}{2} \), we can use the relationship between the velocity of a wave in a wire and the tension in the wire.

### Step-by-step Solution:

1. **Understand the relationship**: The velocity \( v \) of a wave in a wire is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the wire and \( \mu \) is the linear mass density of the wire.

2. **Proportionality**: From the formula, we can see that the velocity \( v \) is directly proportional to the square root of the tension \( T \):
\[ v \propto \sqrt{T} \]

3. **Set up the ratio**: If the new velocity is \( v' = \frac{v}{2} \), we can write the ratio of the new velocity to the original velocity:
\[ \frac{v'}{v} = \frac{\frac{v}{2}}{v} = \frac{1}{2} \]

4. **Square the ratio**: Since \( v' \) is proportional to \( \sqrt{T'} \) and \( v \) is proportional to \( \sqrt{T} \), we can set up the following equation:
\[ \frac{v'}{v} = \sqrt{\frac{T'}{T}} \]
Squaring both sides gives:
\[ \left(\frac{1}{2}\right)^2 = \frac{T'}{T} \]
Simplifying this, we get:
\[ \frac{1}{4} = \frac{T'}{T} \]

5. **Solve for \( T' \)**: Rearranging the equation to find \( T' \):
\[ T' = \frac{T}{4} \]

6. **Substitute the given tension**: The original tension \( T \) is given as \( 2.06 \times 10^4 \, \text{N} \):
\[ T' = \frac{2.06 \times 10^4}{4} = 5.15 \times 10^3 \, \text{N} \]

7. **Final answer**: Therefore, the tension in the wire when the velocity of the wave becomes \( \frac{v}{2} \) is:
\[ T' = 5.15 \times 10^3 \, \text{N} \]