Three masses M = 100 kg, m₁ = 10 kg and m₂ = 20 kg are arranged in a… | JEE Main 2022 (26 Jul, Shift 1)

Q26JEE Main 2022 · 26 JUL · SHIFT 1Medium

Connected Bodies & Pulleys

Three masses $M = 100\text{ kg},\ m_1 = 10\text{ kg}$ and $m_2 = 20\text{ kg}$ are arranged in a system as shown in figure. All the surfaces are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force $F$ is applied on the system so that the mass $m_2$ moves upward with an acceleration of $2\text{ ms}^{-2}$ . The value of $F$ is:

( Take $g = 10\text{ ms}^{-2}$ )

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Answer: (a)

Let $M$ accelerate right with $A$ . $m_2$ moves with $M$ horizontally and goes up at $2\text{ m/s}^2$ : $T - m_2 g = m_2(2) \Rightarrow T = 20(12) = 240$ N.
As $m_2$ rises, $m_1$ slides left relative to $M$ at $2\text{ m/s}^2$ , so its ground acceleration is $A - 2$ (rightward), caused only by $T$ .
$T = m_1(A - 2) \Rightarrow 240 = 10(A - 2) \Rightarrow A = 26\text{ m/s}^2$
Whole system: $F = MA + m_1(A-2) + m_2 A = 100(26) + 10(24) + 20(26)$
$F = 2600 + 240 + 520 = 3360$ N, so the answer is (a).

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