JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Laws of Motion

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

m/s

^{2}

)

Answer: (A)

Solution

Let

a

be the common acceleration, with the heavier right side moving down. The

4

kg mass

m_{1}

on the left rises:

T_{1} - m_{1} g = m_{1} a

, i.e.

T_{1} - 40 = 4 a

. On the right the over-pulley string

T_{1}

holds

m_{2}

, and

m_{2}

carries

m_{3}

below it through the lower string

T_{2}

. For

m_{3}

m_{3} g - T_{2} = m_{3} a

, i.e.

60 - T_{2} = 6 a

. For

m_{2}

(moving down):

m_{2} g + T_{2} - T_{1} = m_{2} a

, i.e.

40 + T_{2} - T_{1} = 4 a

. Adding the three equations,

60 = 14 a

, so

a = \frac{30}{7} m/s^{2}

. Then

T_{1} = 40 + 4 a = \frac{400}{7} N

and

T_{2} = 60 - 6 a = \frac{240}{7} N

, giving

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

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Single correct option