Q 11-04-139JEE MainJEE Main 2022 (26 Jun, Shift 2)Medium
In the arrangement shown in figure, $a_1, a_2, a_3$ and $a_4$ are the accelerations of masses $m_1, m_2, m_3$ and $m_4$ respectively. Which of the following relations is true for this arrangement?
Answer: (A) $4a_1 + 2a_2 + a_3 + a_4 = 0$
Take downward accelerations as positive, and let $b_2$, $b_3$ be the accelerations of the second and third pulleys.
String over the fixed pulley: $a_1 + b_2 = 0$.
String over pulley 2 (its length relative to pulley 2 is constant): $(a_2 - b_2) + (b_3 - b_2) = 0\Rightarrow a_2 + b_3 = 2b_2$.
String over pulley 3: $(a_3 - b_3) + (a_4 - b_3) = 0\Rightarrow b_3 = \dfrac{a_3 + a_4}{2}$.
Substituting: $a_2 + \dfrac{a_3 + a_4}{2} = -2a_1\Rightarrow 4a_1 + 2a_2 + a_3 + a_4 = 0$.
Solution by Sreeraj P, M.Sc Physics