Equation of travelling wave on a stretched string of linear density 5 g/m is $$y = 0.03 \sin(450t - 9x)$$ where distance and time are measured in SI units. The tension in the string is:

The given equation of the transverse wave is $$y = 0.03 \sin(450t - 9x)$$.

We compare it with the standard form of a travelling wave, $$y = A \sin(\omega t - kx)$$, where $$\omega$$ is the angular frequency and $$k$$ is the angular wave number.

From inspection we have $$\omega = 450 \; \text{rad s}^{-1}$$ and $$k = 9 \; \text{rad m}^{-1}$$.

For any wave, the linear speed $$v$$ is related to $$\omega$$ and $$k$$ by the formula $$v = \dfrac{\omega}{k}$$.

Substituting the identified values,

$$v = \dfrac{450}{9} = 50 \; \text{m s}^{-1}.$$

The speed of a transverse wave on a stretched string is also given by the relation $$v = \sqrt{\dfrac{T}{\mu}},$$ where $$T$$ is the tension and $$\mu$$ is the linear mass density.

The linear density is provided as $$5 \; \text{g m}^{-1}$$. Converting grams to kilograms (since SI units require kilograms), $$\mu = 5 \; \text{g m}^{-1} = 5 \times 10^{-3} \; \text{kg m}^{-1} = 0.005 \; \text{kg m}^{-1}.$$

We now square both sides of the speed-tension formula to isolate $$T$$:

$$v^{2} = \dfrac{T}{\mu} \quad \Longrightarrow \quad T = \mu v^{2}.$$

Substituting $$\mu = 0.005 \; \text{kg m}^{-1}$$ and $$v = 50 \; \text{m s}^{-1}$$,

$$T = 0.005 \times (50)^{2} = 0.005 \times 2500 = 12.5 \; \text{N}.$$

Hence, the correct answer is Option C.