The two wires depicted in the figure are constructed from the same...

Let $M$ be the mass of the load placed on the hanger.

The tension in the lower wire is given by $T_1 = (m_1 + M)g = (10 + M) \times 10$.

The tension in the upper wire is given by $T_2 = (m_1 + m_2 + M)g = (10 + 20 + M) \times 10 = (30 + M) \times 10$.

The cross-sectional area of the lower wire is $A_1 = 0.003 \text{ cm}^2 = 3 \times 10^{-7} \text{ m}^2$.

The cross-sectional area of the upper wire is $A_2 = 0.006 \text{ cm}^2 = 6 \times 10^{-7} \text{ m}^2$.

The breaking stress for both wires is $\sigma_{\text{max}} = 8 \times 10^8 \text{ N/m}^2$.

For the lower wire not to break, the stress must be less than or equal to the breaking stress:

$\dfrac{T_1}{A_1} \le \sigma_{\text{max}}$

$\dfrac{(10 + M) \times 10}{3 \times 10^{-7}} \le 8 \times 10^8$

$100 + 10M \le 240$

$10M \le 140 \Rightarrow M \le 14 \text{ kg}$

For the upper wire not to break, the stress must be less than or equal to the breaking stress:

$\dfrac{T_2}{A_2} \le \sigma_{\text{max}}$

$\dfrac{(30 + M) \times 10}{6 \times 10^{-7}} \le 8 \times 10^8$

$300 + 10M \le 480$

$10M \le 180 \Rightarrow M \le 18 \text{ kg}$

The maximum load that can be placed without breaking either wire is the minimum of the two values, which is $14 \text{ kg}$.

Since the maximum load limit for the lower wire is reached first, the lower wire will break first if the load is increased beyond $14 \text{ kg}$.