The correct option is D 4 H
Let, L1 and L2 be the individual inductors,
When, L1 and L2 are connected in parallel, their equivalent inductance is,
L1L2L1+L2=1.5 ........(1)
When, L1 and L2 are connected in series, their equivalent inductance is,
L1+L2=8 .........(2)
From (1) and (2) we get,
L1L2=12
And, we can write,
(L1−L2)2=(L1+L2)2−4L1L2
=(8)2−4×12=64−48=16
∴L1−L2=4 H
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Hence, (D) is the correct answer.