Physics · Newton's Laws of Motion

JEE Main 2026 — 5 April, Morning Shift — Question 4

Three masses $m_1 = 4\mathrm{kg}$ , $m_2 = 4\mathrm{kg}$ and $m_3 = 6\mathrm{kg}$ are suspended from a fixed smooth frictionless pulley as shown in the figure below. The value of $T_1/T_2$ is (take $g = 10\mathrm{m/s}^2$ ).

Option A:
$\frac{5}{3}$
Correct
Option B:
$\frac{2}{3}$
Option C:
$\frac{3}{5}$
Option D:
$\frac{2}{5}$

Answer: A

Step-by-step solution

From equations of motion: $60 - T_2 = 6a$ , $T_2 + 40 - T_1 = 4a$ , solving gives $a = 30/7$ , $T_1 = 400/7$ , $T_2 = 240/7$ , ratio = 5/3.

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Exam: JEE Main 2026
Paper: 5 April, Morning Shift
Subject: Physics
Chapter: Newton's Laws of Motion
Topic: Pulley - Block Problems

A wedge Y with mass of 10 kg and all frictionless surfaces and the inclined surface making 37^circ with horizontal. A block X with mass 2…

Answer: A

Step-by-step solution

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