(i) Given, strees in steel =stress in brass
`therefore (T_(S))/(A_(S))=(T_(B))/(A_(B))implies(T_(S))/(T_(B))=(A_(S))/(A_(B))=(10^(-3))/(2xx10^(-3))=(1)/(2)" "…(i)`
As the system is in equilibrium , taking moments about D, we have
`T_(S).x=T_(B)(2-x)`
`therefore (T_(S))/(T_(B))=(2-x)/(x)" "...(ii)`
From Eqs. (i) and (ii) , we get , x=1.33 m
(ii) Strain `=("Stress")/(Y)`
Given, strain in steel=strain in brass
`therefore (T_(S)//A_(S))/(Y_(S))=(T_(B)//A_(B))/(Y_(B))`
`therefore (T_(S))/(T_(B))=(A_(S)Y_(S))/(A_(B)Y_(B))=((1xx10^(-3))(2xx10^(11)))/((2xx10^(-3))(10^(11)))=1" "...(iii)`
From Eqs. (ii) and (iii), we have
x=1 m