A transverse wave travels on a taut steel wire with a velocity of $$v$$ when tension in it is $$2.06 \times 10^4$$ N. When the tension is changed to T, the velocity changed to $$\frac{v}{2}$$. The value of T is close to:

We begin by recalling the basic formula for the speed of a transverse wave on a stretched string or wire. The speed $$v$$ is given by

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

where $$T$$ is the tension in the wire and $$\mu$$ is the linear mass density (mass per unit length) of the wire. Because the physical wire does not change, the value of $$\mu$$ remains the same in every part of the problem.

First, let us write the expression for the initial situation. When the tension is

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

the corresponding speed is given to be simply $$v$$, so

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

Now the tension is altered to a new value $$T_2$$, and under this new tension the speed is reported to become $$\dfrac{v}{2}$$. Therefore we have

$$\frac{v}{2} = \sqrt{\frac{T_2}{\mu}}.$$

Because the same wire is used, $$\mu$$ is identical in both expressions. We can now form a ratio of the two speed formulas in order to eliminate $$\mu$$:

$$\frac{\dfrac{v}{2}}{v} = \frac{\sqrt{\dfrac{T_2}{\mu}}}{\sqrt{\dfrac{T_1}{\mu}}}.$$

Simplifying the left-hand side first, we see that

$$\frac{\dfrac{v}{2}}{v} = \frac{1}{2}.$$

On the right-hand side, the square roots have the same denominator $$\sqrt{\mu}$$, which cancels. We are left with

$$\frac{1}{2} = \sqrt{\frac{T_2}{T_1}}.$$

To remove the square root, we square both sides:

$$\left(\frac{1}{2}\right)^2 = \left(\sqrt{\frac{T_2}{T_1}}\right)^2.$$

Hence,

$$\frac{1}{4} = \frac{T_2}{T_1}.$$

Now we solve for $$T_2$$ by multiplying both sides by $$T_1$$:

$$T_2 = \frac{T_1}{4}.$$

We substitute the numerical value of $$T_1$$:

$$T_2 = \frac{2.06 \times 10^{4}\ \text{N}}{4}.$$

Carrying out the division, we write $$2.06$$ as $$2.06 = 2.060$$ so that the arithmetic is transparent:

$$T_2 = 0.515 \times 10^{4}\ \text{N}.$$

Finally, we shift the decimal point one place to convert $$0.515 \times 10^{4}$$ into standard scientific notation:

$$T_2 = 5.15 \times 10^{3}\ \text{N}.$$

Therefore, the tension that gives a wave speed of $$\dfrac{v}{2}$$ is approximately $$5.15 \times 10^{3}\ \text{N}$$.

Hence, the correct answer is Option B.