Let the original length of the wire be 'L' now, we have in the first case L1= L + dl1 or [from Hooke's Law; x = T/k] L1= L + (T1/k) .................(1) similarly, for the second case we will have L1= L + dl2 or L2= L + (T2/k) .................(2) we can now use (1) and (2) to solve for original length 'L' so, we get L = (L1T2- L2T1) / (T2- T1) we have L1=10 L2=12 T1=20N T2=40N putting the values we get L=8cm