The two wires shown in figure are made of the same material which has a breaking stress of $8\times {10}^{8}N/m^2$. The area of the cross-section of the upper wire is $0.006c{m}^2$ and that of the lower wire is $=0.003c{m}^2$. The mass $m_1=10kg$ and $m_2=20kg$ the hanger is light. The maximum load (in $kg)$ that can be put on the hanger without breaking a wire is

Draw a free body diagram


Calculate load put on hanger without breaking lower wire

Let load put on hanger will be $F$. Then stress in lower wire will be,
$S_1=\frac{m_{1}g+F}{0.003\times {10}^{-4}}$
$S_1=8\times {10}^{8}N/m^2$,then
$8\times {10}^8=\frac{10\times 10+F}{3\times {10}^{-7}}$
$F=140N$

Calculate load put on hanger without breaking upper wire

let the stress developed in upper wire is $S_2$, then
$S_2=\frac{(m_1+m_2)g+F}{0.006\times {10}^{-4}}$
Let , $S_2=8\times {10}^{8}N/m^2$,therefore,
$8\times {10}^8=\frac{(10+20)10+F}{6\times {10}^{-7}}$
$F=180N$

Hence, maximum load that can we suspended will be $14kg$.