Use the string-length constraints

This is a standard pulley-constraint question. Let us take downward displacement of each mass as positive and denote them by $x_1,x_2,x_3,x_4$.

Since the string lengths are constant, we write one constraint equation for each string and then differentiate twice to relate accelerations.

First movable pulley relation

For a pulley supported by two segments of the same string, if the pulley moves by $x$, the two supporting segments together change by $2x$.

Thus, for the pulley/string arrangement involving $m_1$ and $m_2$, the constraint gives a relation of the type $2x_1 + x_2 = \text{constant}$ and hence 2a_1 + a_2 = 0. \tag{1}

Second pulley relation

Similarly, for the next stage involving the previous pulley motion and mass $m_3$, we get $2x_2 + x_3 = \text{constant}$ so 2a_2 + a_3 = 0. \tag{2}

Third pulley relation

For the final stage involving the motion of the second movable pulley and mass $m_4$, we get $2x_3 + x_4 = \text{constant}$ so 2a_3 + a_4 = 0. \tag{3}

Eliminate intermediate accelerations

From Eq. (1): $a_2 = -2a_1.$

From Eq. (2): $a_3 = -2a_2 = 4a_1.$

From Eq. (3): $a_4 = -2a_3 = -8a_1.$

Now test option A: $4a_1 + 2a_2 + a_3 + a_4$ $= 4a_1 + 2(-2a_1) + 4a_1 - 8a_1$ $= 4a_1 - 4a_1 + 4a_1 - 8a_1 = 0.$

So option A is satisfied.

Check the others quickly

• Option B: $a_1 + 4a_2 + 3a_3 + a_4$ $= a_1 + 4(-2a_1) + 3(4a_1) - 8a_1 = -3a_1 \neq 0$

• Option C: $a_1 + 4a_2 + 3a_3 + 2a_4$ $= a_1 - 8a_1 + 12a_1 - 16a_1 = -11a_1 \neq 0$

• Option D: $2a_1 + 2a_2 + 3a_3 + a_4$ $= 2a_1 - 4a_1 + 12a_1 - 8a_1 = 2a_1 \neq 0$

Final answer

The correct relation is $\boxed{4a_1 + 2a_2 + a_3 + a_4 = 0}$ so the correct option is A.