Two wires as shown in the figure below, made of steel and have breaking stress of \(12 \times 10^8\) N/m\(^2\). Area of cross-section of upper wire is \(0.008\) cm\(^2\) and of lower wire is \(0.004\) cm\(^2\). The maximum mass that can be added to pan without breaking any wire is _____ kg.
(take \(g = 10\) m/s\(^2\))

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: \(38\)