Tension in string B is due to the block of mass 2M:
TB=2Mg
Tension in string A is due to both blocks of mass M and 2M:
TA=Mg+2Mg=3Mg
The elongation in a string is given by Hooke's law as ΔL=AYTL.
The ratio of elongations in string A to string B is:
ΔLBΔLA=(TBTA)(LBLA)(AAAB)(YAYB)
Given that LBLA=2, AAAB=1, and YBYA=0.5⇒YAYB=2.
Substituting the values into the ratio equation:
ΔLBΔLA=(2Mg3Mg)×(2)×(1)×(2)
ΔLBΔLA=23×4=6
Answer: 6