=c Free body diagram of the two blocks brass and steel are
Let young's modulus of steel is Y1 and of brass is Y2. ∴ Y1=
F1.l1
A1.Δl1
...(i) and Y2=
F2.l2
A2.Δl2
...(ii) Dividing Eq.(i) by (ii),
Y1
Y2
=
F1.l1
A1.Δl1
F2.l2
A2.Δl2
or
Y1
Y2
=
F1.A2.l1.Δl2
F2.A1.l2.Δl1
...(iii) Force on steel wire from free body diagram T=F1=(2g) newton Force on brass wire from free body diagram F2=T′=T+2g=(4g) newton Now, putting the value of F1,F2 in Eq. (iii), we get