The two wires shown in figure are made of the same material which has a breaking stress of 8 x ${10}^8$ $\mathrm{N}/{\mathrm{m}}^2$. The area of the cross-section of the upper wire is 0.006 ${\mathrm{cm}}^2$ and that of the lower wire is 0.003 ${\mathrm{cm}}^2$. The mass ${\mathrm{m}}_{1 }= 10 \mathrm{kg}, {\mathrm{m}}_2 =$20 kg and the hanger is light. The maximum load that can be put on the hanger without breaking a wire is

 

Let load put on the hanger is F, then stress in lower wire

${\mathrm{S}}_1 = \frac{{\mathrm{m}}_1\mathrm{g}+\mathrm{F}}{0.003\times {10}^{-4}}$

Let ${\mathrm{S}}_1 = 8 \times 108 \mathrm{N}/{\mathrm{m}}^2, \mathrm{then}$

  $8\times {10}^8 = \frac{10\times 10+\mathrm{F}}{3\times {10}^{-7}} \Rightarrow \mathrm{F} =140 \mathrm{N}$

Let stress developed in upper wire is ${\mathrm{S}}_2$, then

${\mathrm{S}}_2 = \frac{({\mathrm{m}}_1+{\mathrm{m}}_2)\mathrm{g}+\mathrm{F}}{0.006\times {10}^{-4}}$

Let ${\mathrm{S}}_2 = 8 \times {10}^8 \mathrm{N}/{\mathrm{m}}^2$

$8\times {10}^8 = \frac{(10+20)10+\mathrm{F}}{6\times {10}^{-7}} \Rightarrow \mathrm{F} = 180 \mathrm{N}$

We have to select the lower value of F. Hence maximum load that can be suspended is $\frac{140}{\mathrm{g}}$ = 14 kg .