Given

Breaking stress:

$\sigma = 12\times10^8\ \text{N/m}^2$

Area of upper wire:

$A_1=0.008\ \text{cm}^2$

Area of lower wire:

$A_2=0.004\ \text{cm}^2$

Masses attached:

• Upper block $=30\ \text{kg}$
• Lower block $=10\ \text{kg}$

Let maximum additional mass in pan be $m$ kg.

Take

$g=10\ \text{m/s}^2$

Step 1 : Maximum tension in each wire

Using

$F=\sigma A$

Convert areas into $\text{m}^2$:

$A_1=0.008\times10^{-4}=8\times10^{-7}\ \text{m}^2$

$A_2=0.004\times10^{-4}=4\times10^{-7}\ \text{m}^2$

Upper wire

$T_1=(12\times10^8)(8\times10^{-7})$

$T_1=960\ \text{N}$

Equivalent supported mass:

$\dfrac{960}{10}=96\ \text{kg}$

Lower wire

$T_2=(12\times10^8)(4\times10^{-7})$

$T_2=480\ \text{N}$

Equivalent supported mass:

$\dfrac{480}{10}=48\ \text{kg}$

Step 2 : Condition for lower wire

Lower wire supports:

$(10+m)\ \text{kg}$

So,

$10+m\leq 48$

$m\leq 38$

Step 3 : Condition for upper wire

Upper wire supports:

$(30+10+m)\ \text{kg}$

So,

$40+m\leq 96$

$m\leq 56$

Step 4 : Maximum possible mass

Smaller value is allowed.

Hence,

$m_{\max}=38\ \text{kg}$

Final Answer

$\boxed{38\ \text{kg}}$