some discrepancy in the answer !

let the upper spring be S1 and the lower one be S2..since both the springs are identical so their spring constant will be same = k
since the system is kept vertically and the sring constants are same so S2 will be compressed by length x2 and S1 will be elongated by length x1
now balance the force by drawing the FBD
kx1 upwards
kx2 upwards
mg downwards
so mg = kx1 + kx2 .....1
therefore mg> kx1 and mg>kx2
so no matter which spring is cut the block will move downwards
case I when S1 is cut
mg - kx2 = 6m .....2
case II when S2 is cut
mg - kx1 = ma ...3 (here a is the acceleration which we are required to find in the question)
Adding equation 2 and 3
2mg - (kx1 +kx2) = 6m + ma
2mg - mg = m(6+a) from 1
g=6+a
so a = 4

hence the answer is b

@govind..
why should the elongation of upper spring and the compression of lower one be different?

this sum is form dc pandey and in the first attempt i had also done it wrong but if i remeber the answer is right (though not sure abt it)

lets first assume the force by the upper stirng is F1(upward) and by downwards is F2(upward)

mg=F1+f2
mg-F1=6m
therfore F1 4m

now the force F2=6m

now the force on the spring is 10m-6m =4m therofe acc is 4

i believe here identical springs refers to equal relaxed lengths and not spring constants .

thus different accelerations are produced .

btw 4m where u do u get the ques debotosh?is the correct ans 4m/s?

ya i too agree no matter which spring u cut the initial acceleration will be same!!!

thanks all,,,its done !