A transverse wave travels on a taut steel wire with a velocity of when tension in it is `2.06xx10^4 N` . When the tension is changed to T, the velocity changed to 3v. The value of T is close to :

Options

A. `10.2 xx10^2 N`
B. `5.15 xx10^3 N`
C. `2.50xx10^4 N`
D. `18.54 xx10^4 N`

Explanation

To solve the problem, we need to understand the relationship between the velocity of a wave on a string and the tension in the string. The formula that relates these quantities is:

\[ v = \sqrt{\frac{T}{\mu}} \]

where:
- \( v \) is the velocity of the wave,
- \( T \) is the tension in the string,
- \( \mu \) is the linear mass density of the string.

From this formula, we can derive that the square of the velocity is proportional to the tension:

\[ v^2 \propto T \]

This means that if we change the velocity, the tension will change according to the square of the velocity.

### Step-by-step Solution:

1. **Initial Conditions**:
- The initial tension is \( T_1 = 2.06 \times 10^4 \, \text{N} \).
- The initial velocity is \( v \).

2. **Final Conditions**:
- The final velocity is \( v_2 = 3v \).

3. **Using the Proportionality**:
- Since \( v^2 \propto T \), we can write:
\[ \frac{T_2}{T_1} = \left(\frac{v_2}{v_1}\right)^2 \]
- Substituting the values:
\[ \frac{T_2}{T_1} = \left(\frac{3v}{v}\right)^2 = 3^2 = 9 \]

4. **Calculating the New Tension**:
- Rearranging gives:
\[ T_2 = 9 T_1 \]
- Substituting the value of \( T_1 \):
\[ T_2 = 9 \times (2.06 \times 10^4 \, \text{N}) = 18.54 \times 10^4 \, \text{N} \]

5. **Final Answer**:
- Therefore, the value of \( T \) is approximately \( 1.854 \times 10^5 \, \text{N} \).