1. Use the wave speed formula on a stretched string

For a transverse wave on a wire,

$v = \sqrt{\frac{T}{\mu}}$

where $T$ is the tension and $\mu$ is the linear mass density.

Since the same wire is used, $\mu$ remains constant.

2. Relate the two situations

Initially,

$v = \sqrt{\frac{T_1}{\mu}}$

with

$T_1 = 2.06 \times 10^4\,\text{N}$

After changing the tension to $T$, the speed becomes $\dfrac{v}{2}$:

$\frac{v}{2} = \sqrt{\frac{T}{\mu}}$

3. Square both equations and compare

From the first,

$v^2 = \frac{T_1}{\mu}$

From the second,

$\frac{v^2}{4} = \frac{T}{\mu}$

Now substitute $v^2 = \dfrac{T_1}{\mu}$:

$\frac{1}{4}\cdot \frac{T_1}{\mu} = \frac{T}{\mu}$

So,

$T = \frac{T_1}{4}$

4. Calculate $T$

$T = \frac{2.06 \times 10^4}{4} = 0.515 \times 10^4 = 5.15 \times 10^3\,\text{N}$

5. Match with the options

$T = 5.15 \times 10^3\,\text{N}$

This corresponds to Option D.

6. Check all options briefly

• A: $30.5 \times 10^4$ N — much too large
• B: $2.50 \times 10^4$ N — larger than initial tension, impossible if speed is halved
• C: $10.2 \times 10^2 = 1.02 \times 10^3$ N — too small
• D: $5.15 \times 10^3$ N — correct

Therefore, the correct answer is D.