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the given answer will be satisfied only if we considere 5 NOT gates here.. but why not to consider 6 NOT gate?? how is D answer?

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every odd no of Not gate always complement the input(1->0, 0->1)
u can also think this problem as fo 6th not gate i/p is complemented after 2*5=10 ns so for 6th not gate it is in 1 for 10 ns nd in 0 for 10 ns hence time period =10+10=20 ns

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