answer with correct explanation....
irrelevant answer will b reported...

Answers
let the tension in the block be t
then
mgsin53°-T=ma......1)
T-mgsin37°=ma.......2)
adding both
mg(sin53°-sin 37°)=2ma
mg(4/5-3/5)=2ma
g/5=2a
a=g/10
a=1m/s
this is the acceleration of both blocks...
Now let's find the acceleration of centre of mass of both the blocks, using the vector
for block first
acceleration=acos53°i+asin53°j
acceleration=a(3/5i+4/5j)
for block second
acceleration=acos53°i+asin53°j
acceleration=a(4/5i +3/5 j)
A(cm)= (ma1+ma2)/(m1+m2)
A(cm)=ma(7/5i+7/5j)/2m
put a=1
A(cm)=(7/5i+7/5j)/2
A(cm)={(98/25)^1/2}/2
A(cm)={(3.9)^1/2}/2
3.9=4 (approx)
A(cm)={(4)^1/2}/2
A(cm)=2/2=1
this shows that acceleration of centre of mass of both the blocks is same as acceleration of each block separately......
so option1),will be the correct option
keep asking ur doubts......


