The two wires shown in figure are made of the same material which has a breaking stress of `8xx10^8Nm^-2`. The area of cross section of the upper wire is `0.006 cm^2` and that of the lower wire is `0.003 cm^2`. The mass `m_1=10 kg`, `m_2=20kg` and the hanger is light. a. Find the maximum load that can be put on the hanger without breaking a wire. Which wire will break first if the load is increased? b. Repeat the above part `m_1=10 kg and m_2=36kg`.

Answer

`2kg`

Explanation

Let load put on the hanger be, then stress in lower wire
`S_(1)=(m_(1)g+F)/(0.003x10^(-4))`
Let `S_(1)=8xx10^(8)N//m^(2)`, then
`impliesF=140N`
Let stress developed in upper wire be `S_(2)` then
`S_(2)=[(m_(1)+m_(2))g+F]/(0.006xx10^(-4))`
Let `S_(2)=8xx10^(8)N//m^(2)`
`implies F-=180N`
We have to select the lower vaue of `F`. Hence, maximum load that can be suspended is `140//g=14kg`
b. `S_(1)=(m_(1)g+F)/(0.003xx10^(-4))=8xx10^(8)`
`implies 10x10+F=240impliesF=140N`
`S_(2)=[(m_(1)+m_(2))g+F]//(0.006xx10^(-4))=8xx10^(8)`
`implies 46g+F=-480impliesF=20N`
Hence the maximum load that can be suspended is `20//g=2kg`