Two wires as shown in the figure below, made of steel and have...

Let the maximum mass that can be added to the pan be $m$.

The breaking stress for both wires is \sigma_\max = 12 \times 10^8 \text{ N/m}^2.

For the upper wire:
Area of cross-section, $A_u = 0.008 \text{ cm}^2 = 8 \times 10^{-7} \text{ m}^2$
Maximum tension the upper wire can withstand is:
T_{u,max} = \sigma_\max \times A_u = (12 \times 10^8) \times (8 \times 10^{-7}) = 960 \text{ N}
The total mass supported by the upper wire is $(30 + 10 + m) = (40 + m) \text{ kg}$.
Tension in the upper wire, $T_u = (40 + m)g = (40 + m) \times 10$
To avoid breaking the upper wire:
$(40 + m) \times 10 \le 960$
$40 + m \le 96 \implies m \le 56 \text{ kg}$

For the lower wire:
Area of cross-section, $A_l = 0.004 \text{ cm}^2 = 4 \times 10^{-7} \text{ m}^2$
Maximum tension the lower wire can withstand is:
T_{l,max} = \sigma_\max \times A_l = (12 \times 10^8) \times (4 \times 10^{-7}) = 480 \text{ N}
The total mass supported by the lower wire is $(10 + m) \text{ kg}$.
Tension in the lower wire, $T_l = (10 + m)g = (10 + m) \times 10$
To avoid breaking the lower wire:
$(10 + m) \times 10 \le 480$
$10 + m \le 48 \implies m \le 38 \text{ kg}$

To ensure neither wire breaks, the added mass $m$ must satisfy both conditions. Therefore, the maximum mass that can be added is the smaller of the two values:
$m = 38 \text{ kg}$

Answer: $38$