Three masses m₁ = 4 kg, m₂ = 4 kg and m₃ = 6 kg are suspended from a… | JEE Main 2026 (5 Apr, Shift 1)

Q4JEE Main 2026 · 5 APR · SHIFT 1Medium

Connected Bodies & Pulleys

Three masses $m_1 = 4\text{ kg}, m_2 = 4\text{ kg}$ and $m_3 = 6\text{ kg}$ are suspended from a fixed smooth frictionless pulley as shown in the figure below. The value of $T_1/T_2$ is

______

(take $g = 10\text{ m/s}^{2}$ )

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Answer: (a)

Left side: $m_1 = 4$ kg; right side: $m_2 + m_3 = 10$ kg, so the right side goes down.
$a = \dfrac{(10-4)g}{14} = \dfrac{30}{7}\text{ m/s}^2$
$T_1 = m_1(g+a) = 4\left(10 + \dfrac{30}{7}\right) = \dfrac{400}{7}$ N
$T_2 = m_3(g-a) = 6\left(10 - \dfrac{30}{7}\right) = \dfrac{240}{7}$ N
$\dfrac{T_1}{T_2} = \dfrac{400}{240} = \dfrac{5}{3}$ , so the answer is (a) 5/3.

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