Two wires shown in figure are made of the same material which has a breaking stress of `8xx10^(8) N/m^(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) = 20 kg ` and the hanger is light . 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 ? ( Take `g = 10 m//s^(2))`

Options

A. 14 kg
B. 1.4 kg
C. 1.2 kg
D. 12 kg

Answer

A

Explanation

Let load put on the hanger is F then stress in lower wire
`S_1=(m_1g+F)/(0.003xx10^-4)`
Let `S_1=8xx108(N)/(m^2)`, then
`8xx10^8=(10xx10+F)/(3xx10^-7)impliesF=140N`
Let stress developed in upper wire is `S_2`, then `S_2=((m_1+m_2)g+F)/(0.006xx10^-4)`
Let `S_2=8xx10^9(N)/(m^2)`
`8xx10^8=((10+20)10+f)/(6xx10^-7)impliesF=180N`
We have to select the lower value of F. Hence maximum load that can be suspended is `(140)/(g)=14kg`.