A steel rod of cross-sectional area `16 cm^(2)` and two brass rods each of cross-sectional area `10 cm^(2)` together support a load of `5000 kg ` as shown in the figure. ( Given, `Y_(steel) = 2xx10^(6) kg cm^(-2) and Y_(brass) = 10 ^(6) kg cm^(-2))`. Choose the correct option(s).

Explanation

Given, area of steel rod, `A_(S)=16 cm^(2)`
Area of two brass rods, `A_(B)=2xx10=20 cm^(2)`
Load, F=5000 kg
Young's modulus for the steel , `Y_(S)=2xx10^(6) kg cm^(-2)`
Young's modulus for the brass, `Y_(B)=1xx10^(6) kg cm^(-2)`
Length of the steel rod , `I_(S)=30 cm`
Length of the brass rod, `I_(B)=20 cm`
Let `" "sigma` =stress in steel and `sigma_(B)` =stress in brass
Decrease in length of the steel rod = decrease in length of the brass rod
or `(sigma_(S))/(Y_(S))xxl_(S)=(sigma_(B))/(Y_(B))xxl_(B)`
or `sigma_(S)=(Y_(S))/(Y_(B))xx(l_(B))/(l_(S))xxsigma_(B)=(2xx10^(6))/(1xx10^(6))xx(20)/(30)xxsigma_(B)`
`therefore sigma_(S)=(4)/(3)sigma_(B)" ...(i)`
Now, using the relation,
`F=sigma_(S)A_(S)+sigma_(B)A_(B)` or `5000=sigma_(S)xx16+sigma_(B)xx20" "...(ii)`
Solving Eqs. (i) and (ii) , we get
`sigma_(B)=120.9 kg cm^(-2)`