JEE Main 2026 — Apr 5, Shift 1 — Physics solution

JEE Main2026Apr 5, Shift 1Physics

Three masses $m_{1} = 4$ kg, $m_{2} = 4$ kg and $m_{3} = 6$ kg are suspended from a fixed smooth frictionless pulley as shown in the figure below. The value of $T_{1} / T_{2}$ is __________. (take $g = 10$ m/s $^{2}$ )

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

Solution

Let $a$ be the magnitude of acceleration: $m_{1}$ moves up, while the connected system $(m_{2} + m_{3})$ moves down.

Newton's second law for each block:

$m_{1}$ : $T_{1} - m_{1} g = m_{1} a \Rightarrow T_{1} - 40 = 4 a$ ,

$m_{2}$ : $m_{2} g + T_{2} - T_{1} = m_{2} a \Rightarrow 40 + T_{2} - T_{1} = 4 a$ ,

$m_{3}$ : $m_{3} g - T_{2} = m_{3} a \Rightarrow 60 - T_{2} = 6 a$ .

Adding all three: $60 = 14 a$ , so $a = \frac{30}{7}$ m/s $^{2}$ .

Then $T_{1} = 40 + 4 \cdot \frac{30}{7} = \frac{400}{7}$ N and $T_{2} = 60 - 6 \cdot \frac{30}{7} = \frac{240}{7}$ N.

Therefore $\frac{T _{1}}{T _{2}} = \frac{400}{240} = \frac{5}{3}$ .

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