Q 11-04-055JEE MainJEE Main 2026 (5 Apr, Shift 1)Medium
Three masses $m_1 = 4$ kg, $m_2 = 4$ kg and $m_3 = 6$ kg are suspended from a fixed smooth frictionless pully as shown in the figure below. The value of $T_1/T_2$ is ______ (take $g = 10\ \text{m/s}^2$)
Answer: (A) $5/3$
The right side ($m_2 + m_3 = 10$ kg) goes down:
$$a = \frac{(10 - 4)g}{14} = \frac{30}{7}\ \text{m/s}^2$$
$m_1$ accelerates upward: $T_1 = m_1(g + a) = 4 \times \dfrac{100}{7} = \dfrac{400}{7}$ N.
$m_3$ accelerates downward: $T_2 = m_3(g - a) = 6 \times \dfrac{40}{7} = \dfrac{240}{7}$ N.
$\dfrac{T_1}{T_2} = \dfrac{400}{240} = \dfrac{5}{3}$.
Solution by Sreeraj P, M.Sc Physics