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PhysicsNLMMedium2 minQB

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Three blocks are connected as shown in the figure on a horizontal frictionless table if $m_{1} = 1 kg, m_{2} = 8 kg, m_{3} = 27 kg$ and $T_{3} = 36 N, T_{2}$ will be

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Answer:

Solution:

F.B.D. of the whole system

$T_{3} = (m_{1} + m_{2} + m_{3}) a = (1 + 8 + 27) a = 36$

$a = 1$

F.B.D. of $m_{3}$

$T_{3} - T_{2} = m_{3} \times a$

$36 - 27 \times 1 = T_{2} = 9 N$

Stream:JEESubject:PhysicsTopic:NLM

⏱ 2mℹ️ Source: QB

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