In the arrangement shown, block $A$ of mass $2$ kg lies on a smooth horizontal table. The string from $A$ passes over a fixed pulley, goes round a movable pulley that carries block $B$ of mass $2$ kg, and its other end is fixed to a rigid support. The strings and pulleys are light and frictionless. The tension in the string is ($g = 10\ \text{m/s}^2$)
Answer: (C) $8$ N
Concept. Write the string constraint first, then Newton's second law for each block. One string means one tension $T$ throughout.
Step 1. Constraint. If $B$ goes down by $y$, both vertical segments lengthen by $y$, so $A$ moves $2y$:
$$a_A = 2a_B$$
Step 2. Block A. $T = 2a_A = 4a_B$.
Step 3. Block B. Two segments hold the movable pulley:
$$2g - 2T = 2a_B \;\Rightarrow\; 20 - 8a_B = 2a_B \;\Rightarrow\; a_B = 2\ \text{m/s}^2$$
Step 4. Tension. $T = 4 \times 2 = 8$ N (and $a_A = 4\ \text{m/s}^2$).
Answer. $8$ N (Option C). Ignoring the constraint ($a_A = a_B$) gives the trap answers.
Solution by Sreeraj P, M.Sc Physics