1. For a cyclic process, the system returns to its initial state.

   Hence, the change in internal energy over one complete cycle is $\Delta U = 0.$

2. From the first law of thermodynamics, $Q = \Delta U + W.$ Since $\Delta U = 0$ for a cycle, $Q = W.$

   So, the magnitude of heat exchanged is equal to the magnitude of the net work done in the cycle, i.e. the area enclosed in the $P$-$V$ diagram.

3. From the figure, the path $ABCA$ encloses a semicircular area. The dimensions from the graph correspond to:

   • semi-major/semi-diameter scale giving radius along one axis as $5$
   • the other scale as $2$

   Therefore enclosed area is $\text{Area} = \frac{1}{2}\pi ab = \frac{1}{2}\pi(5)(2)=5\pi.$

4. Therefore, $|Q| = |W| = 5\pi \text{ J}.$

5. Checking options:

   • A: zero $\;\times$
   • B: $40\pi$ $\;\times$
   • C: $5\pi$ $\;\checkmark$
   • D: $10\pi$ $\;\times$

Thus the correct option is C.