1. Use wave speed on a stretched wire

For a transverse wave on a wire, $v = \sqrt{\frac{T}{\mu}}$ where:

• $v = 90\,\text{m s}^{-1}$
• $T$ = tension in wire
• $\mu$ = mass per unit length

So, $T = \mu v^2$

2. Find linear mass density

Given:

• mass $m = 6.0\,\text{g} = 6.0 \times 10^{-3}\,\text{kg}$
• length $L = 60\,\text{cm} = 0.60\,\text{m}$

Thus, $\mu = \frac{m}{L} = \frac{6.0 \times 10^{-3}}{0.60} = 1.0 \times 10^{-2}\,\text{kg m}^{-1}$

3. Calculate tension

$T = \mu v^2 = (1.0 \times 10^{-2})(90)^2$ $T = 0.01 \times 8100 = 81\,\text{N}$

4. Relate tension to extension using Young's modulus

Young's modulus is $Y = \frac{\text{stress}}{\text{strain}} = \frac{T/A}{\Delta L/L}$

So, $Y = \frac{TL}{A\Delta L}$

Hence, $\Delta L = \frac{TL}{AY}$

Given:

• $A = 1.0\,\text{mm}^2 = 1.0 \times 10^{-6}\,\text{m}^2$
• $Y = 16 \times 10^{11}\,\text{N m}^{-2}$
• $T = 81\,\text{N}$
• $L = 0.60\,\text{m}$

Substitute: $\Delta L = \frac{81 \times 0.60}{(1.0 \times 10^{-6})(16 \times 10^{11})}$

$\Delta L = \frac{48.6}{16 \times 10^5}$ $\Delta L = 3.0375 \times 10^{-5}\,\text{m}$

5. Convert into mm

$\Delta L = 3.0375 \times 10^{-5}\,\text{m} = 0.030375\,\text{mm}$

So approximately, $\Delta L \approx 0.03\,\text{mm}$

6. Evaluate options

• A: $0.03\,\text{mm}$ ✅
• B: $0.04\,\text{mm}$ ❌
• C: $0.02\,\text{mm}$ ❌
• D: $0.01\,\text{mm}$ ❌

Therefore, the correct option is A.