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The above circuit produces the output sequence:

  1. $1111\quad 1111\quad0000\quad0000$
  2. $1111 \quad 0000\quad1111\quad0000$
  3. $1111 \quad 0001\quad0011\quad0101$
  4. $1010\quad1010\quad1010\quad1010$
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Let us suppose the initial output of all the JK flip flops is $1$

So we can draw the below table to get the output $Q_3$
$\small \begin{array} {c|c|c|c|c|c|c|c|c|c|c|c|c|c } \hline \text{Index}  & Q_3 & Q_2 & Q_1 &Q_0 & J_3 & K_3 & J_2 & K_2 & J_1 & K_1 & J_0 & K_0 \\\hline  & & & & & Q_2 & Q_2’ & Q_1 & Q_1’ & Q_0 & Q_0’ & Q_3 \oplus Q_2 & J_0’ \\\hline 0 & 1 & 1 & 1 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 1 \\\hline 1 & 1 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 1 \\\hline 2 & 1 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 1 \\\hline 3 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 1 & 0 \\\hline 4 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 1 & 0 & 0 & 1 \\\hline 5 & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 0 & 1 \\\hline 6 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 1 & 0 \\\hline 7 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 1 & 0 & 1 & 0 \\\hline 8 & 0 & 0 & 1 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 1 \\\hline 9 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 1 & 0 \\\hline 10 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 \\\hline 11 & 1 & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 \\\hline 12 & 0 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 1 & 0 \\\hline 13 & 1 & 0 & 1 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 0 \\\hline 14 & 0 & 1 & 1 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\\hline & 1 & 1 & 1 & 1  \\\hline \end{array}$

From the above table $Q_3$ that is output is $1111\; 0001\; 0011\; 0101$

So the answer should be C.
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