Given information:
- Two identical sonometer wires have a fundamental frequency of 500 vibrations per second when kept under the same tension.

To find:
- The fractional increase in the tension of the one wire that would cause an occurrence of 5 beats/sec when both wires vibrate together.

Let's break down the problem step by step:

1. Understanding beats:
- When two sound waves of slightly different frequencies are superimposed, they produce a phenomenon known as beats.
- Beats are periodic variations in the loudness or intensity of sound when two waves interfere constructively and destructively.
- The number of beats produced per second is equal to the difference in frequency between the two waves.

2. Relationship between frequency and tension:
- In the case of a stretched wire, the frequency of vibration is directly proportional to the square root of tension.
- Mathematically, f ∝ √T, where f is the frequency and T is the tension.

3. Finding the frequency difference:
- The two wires have a fundamental frequency of 500 vibrations per second when kept under the same tension.
- Therefore, the frequency of each wire is 500 Hz.
- For beats to occur, the frequency of one wire needs to be slightly different from the other.
- Let's assume the frequency of the second wire is (500 + Δf) Hz, where Δf is the frequency difference that causes 5 beats/sec.

4. Calculating the tension difference:
- From the relationship f ∝ √T, we can write the equation as f = k√T, where k is a constant.
- For the first wire: 500 = k√T1
- For the second wire: (500 + Δf) = k√T2

5. Finding the tension difference:
- Taking the ratio of the two equations: (500 + Δf)/500 = √(T2/T1)
- Squaring both sides: ((500 + Δf)/500)^2 = T2/T1
- Since the wires are identical, T2/T1 = 1
- Simplifying the equation: ((500 + Δf)/500)^2 = 1
- Solving for Δf: Δf = 500(√2 - 1) ≈ 207.11 Hz

6. Calculating the fractional increase in tension:
- Now, we need to find the fractional increase in tension that causes Δf = 207.11 Hz.
- Let's assume the tension of the first wire is T, and the tension of the second wire is T + ΔT.
- From the relationship f ∝ √T, we can write the equation as f = k√(T + ΔT)
- For the first wire: 500 = k√T
- For the second wire: (500 + Δf) = k√(T + ΔT)

7. Simplifying the equation:
- Taking the ratio of the two equations: (500 + Δf)/500 = √((T + ΔT)/T)
- Squaring both sides: ((500 + Δf)/500)^2 = (T + ΔT)/T
- Rearranging the equation: ΔT = T(((500 + Δf)/500)^2 - 1)

8. Calculating the fractional increase in tension:
- Substituting the values: ΔT = T(((500