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4 votes

Computation at AND-1 takes 10+ 9 =19ns   and  AND-2 takes =12ns

So, at OR gate output from AND-2 will be there at 12th and output from AND-1 gate comes at 19th second 

Hence For 19-12 = 7ns the output will be spurious

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1 votes
1 votes
Nothing is given about the OR gate so, lets consider its delay as 0.

And at time=0 ns,    a,b and c are applied to the input.

Now at time t=12ns (delay of AND 2 gate), We will have output bc from AND2 gate, and some output (does not matter) from final OR gate.

Complement of b will be input of AND 1 gate at time =9 nsand b'a comes as output of AND1 gate at t=19ns.

At t=19 ns, output bc+b'a comes from final OR gate.

So, time difference between correct output and first output is 19ns-12ns=7ns

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