1. Given data

• Breaking stress of both wires: $\sigma_b = 1.25 \times 10^9\ \text{N/m}^2$
• Cross-sectional areas: $A_1 = 8 \times 10^{-7}\ \text{m}^2, \qquad A_2 = 4 \times 10^{-7}\ \text{m}^2$
• Hanging masses already attached: $m_1 = 20\ \text{kg}, \qquad m_2 = 10\ \text{kg}$
• $g = 10\ \text{m/s}^2$

We need the maximum additional mass that can be placed in the pan so that neither wire breaks.

2. Maximum tension each wire can sustain

Breaking stress is $\sigma_b = \frac{T_{\max}}{A}$ So, $T_{\max} = \sigma_b A$

For wire $W_1$

$T_{1,\max} = 1.25 \times 10^9 \times 8 \times 10^{-7}$ $= 10 \times 10^2 = 1000\ \text{N}$

For wire $W_2$

$T_{2,\max} = 1.25 \times 10^9 \times 4 \times 10^{-7}$ $= 5 \times 10^2 = 500\ \text{N}$

3. Extra load each wire can still support

The already hanging masses produce tensions:

For $W_1$

$T_1 = 20 \times 10 = 200\ \text{N}$ Remaining allowable tension: $1000 - 200 = 800\ \text{N}$ Equivalent additional mass: $\frac{800}{10} = 80\ \text{kg}$

For $W_2$

$T_2 = 10 \times 10 = 100\ \text{N}$ Remaining allowable tension: $500 - 100 = 400\ \text{N}$ Equivalent additional mass: $\frac{400}{10} = 40\ \text{kg}$

4. Condition for the pan

The pan is supported by both wires, so the same additional mass placed in the pan must be safely supported by each relevant wire according to the setup. Therefore, the limiting wire is the weaker one in terms of remaining capacity.

Thus, maximum allowable mass in the pan is $m_{\max} = \min(80, 40) = 40\ \text{kg}$

5. Final answer

$\boxed{40}$ kg