Correct Answer - A
`A_(S) = 16cm^(2), A_(B) = 2xx10 = 20cm^(2)`
Load, `F = 5000kg`
`Y_(S) = 2.0xx10^(6)kg//cm^(2), Y_(B) = 1.0xx10^(6) kg//cm^(2)`
`l_(S) = 30cm` and `l_(B) = 20cm`
Let `sigma_(S)` = stress in steel and `sigma_(B)` = stress in brass decreases in length of steel rod = decreases in length of brass rod
`(sigma_(S))/(Y_(S)) xx l_(S) = (sigma_(B))/(Y_(B)) xx l_(B)`
`rArr sigma_(S) = (Y_(S))/(Y_(B)) xx (l_(B))/(l_(S)) xx sigma_(B).....(i)`
now using the relation `F = sigma_(S) A_(S) + sigma_(B) A_(B)`
`5000 = sigma_(S) xx 16 + sigma_(B) xx20.........(ii)`
solving equ.(i) and (ii) and we get
`sigma_(B) = 120.9kg//cm^(2)` and `sigma_(S) = 161.2kg//cm^(2)`