The speed of a transverse wave passing through a string of length 50 cm and mass 10 g is 60 m s$$^{-1}$$. The area of cross-section of the wire is 2.0 mm$$^2$$ and its Young's modulus is $$1.2 \times 10^{11}$$ N m$$^{-2}$$. The extension of the wire over its natural length due to its tension will be $$x \times 10^{-5}$$ m. The value of x is _____

We have a string of length $$L = 50$$ cm $$= 0.5$$ m, mass $$m = 10$$ g $$= 0.01$$ kg, cross-sectional area $$A = 2.0$$ mm$$^2 = 2.0 \times 10^{-6}$$ m$$^2$$, and Young's modulus $$Y = 1.2 \times 10^{11}$$ N/m$$^2$$. The speed of a transverse wave on the string is $$v = 60$$ m/s.

The linear mass density is $$\mu = \frac{m}{L} = \frac{0.01}{0.5} = 0.02$$ kg/m. The wave speed on a string is $$v = \sqrt{\frac{T}{\mu}}$$, so the tension is $$T = \mu v^2 = 0.02 \times 60^2 = 0.02 \times 3600 = 72$$ N.

Now, using the definition of Young's modulus, $$Y = \frac{T/A}{\Delta L / L}$$, we can find the extension: $$\Delta L = \frac{TL}{YA} = \frac{72 \times 0.5}{1.2 \times 10^{11} \times 2.0 \times 10^{-6}} = \frac{36}{2.4 \times 10^{5}} = 1.5 \times 10^{-4}$$ m $$= 15 \times 10^{-5}$$ m.

Hence $$x = 15$$.

Hence, the correct answer is 15.