(i) In equilibrium, the forces on each particle must balance. For particle A, the tension in S₁ must balance the weight of A and the tension in S₂. Therefore, we have:

\(T_1 = W_A + T_2\)

For particle B, the tension in S₂ must balance the weight of B:

\(T_2 = W_B\)

Given that the weight of each particle is \(W = mg = 0.2 \times 9.8 = 1.96 \text{ N}\), we find:

\(T_2 = 1.96 \approx 2 \text{ N}\)

Substituting back, we find:

\(T_1 = 1.96 + 1.96 = 3.92 \approx 4 \text{ N}\)

(ii) When S₁ is cut, both particles fall with acceleration. Using Newton's second law for particle A:

\(T_2 + 2 - 0.4 = 0.2a\)

For particle B:

\(2 - T_2 - 0.2 = 0.2a\)

Solving these equations simultaneously, we find:

\(a = 8.5 \text{ m/s}^2\)

\(T_2 = 0.1 \text{ N}\)