Two copper wires of radii r_1 and r_2 (r_1 > r_2 ) are subjected to...

The velocity of a transverse wave in a stretched wire is given by $v = \sqrt{\dfrac{T}{\mu}}$, where $T$ is the tension and $\mu$ is the linear mass density.

The linear mass density $\mu$ can be expressed as $\mu = \pi r^2 \rho$, where $r$ is the radius of the wire and $\rho$ is the density of the material.

Substituting $\mu$ into the velocity formula gives $v = \sqrt{\dfrac{T}{\pi r^2 \rho}} = \dfrac{1}{r} \sqrt{\dfrac{T}{\pi \rho}}$.

Since both wires are made of copper, their densities $\rho$ are the same. The tension $T$ is also the same for both wires.

Therefore, the velocity $v$ is inversely proportional to the radius $r$, which means $v \propto \dfrac{1}{r}$.

Since the thinner wire has a smaller radius, the transverse wave will travel faster in the thinner wire.

Answer: travel faster in the thinner wire.