As shown in the figure, an aluminium wire of length 60 cm is...

The linear mass density of the steel wire is given by:

$\mu_s = \rho_s A_s = (7.8 \times 10^3 \text{ kg m}^{-3})(1.0 \times 10^{-6} \text{ m}^2) = 7.8 \times 10^{-3} \text{ kg m}^{-1}$

The linear mass density of the aluminium wire is:

$\mu_a = \rho_a A_a = (2.6 \times 10^3 \text{ kg m}^{-3})(3.0 \times 10^{-6} \text{ m}^2) = 7.8 \times 10^{-3} \text{ kg m}^{-1}$

Since $\mu_s = \mu_a = \mu$, the wave speed $v$ is the same in both wires:

$v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{40}{7.8 \times 10^{-3}}} \approx 71.6 \text{ m s}^{-1}$

For the joint to act as a node, both wires must form standing waves with nodes at their ends. The frequency $f$ must be the same for both wires:

$f = \dfrac{n_s v}{2L_s} = \dfrac{n_a v}{2L_a}$

where $n_s$ and $n_a$ are the number of loops in the steel and aluminium wires, respectively. Equating the two expressions gives:

$\dfrac{n_s}{L_s} = \dfrac{n_a}{L_a} \Rightarrow \dfrac{n_s}{0.8} = \dfrac{n_a}{0.6} \Rightarrow \dfrac{n_s}{n_a} = \dfrac{4}{3}$

For the minimum frequency, $n_s$ and $n_a$ must be the smallest possible integers. Thus, $n_s = 4$ and $n_a = 3$.

Substituting $n_s = 4$ into the frequency formula yields:

$f = \dfrac{4 \times 71.6}{2 \times 0.8} = \dfrac{71.6}{0.4} = 179 \text{ Hz}$

The closest value among the given options is $180 \text{ Hz}$.

Answer: $180 \text{ Hz}$