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Physics - NLM Question with Solution | TestHub

PhysicsNLMChallenging application of NLMMedium2 minQB

PhysicsMediumsingle choice

In the system of pulleys shown what should be the value of $m_{1}$ such that 100gm remains at rest w.r.t. ground :

Options:

Answer:

Solution:

$T = 100 g$

$a_{1} + a_{2} + 2 a_{3} = 0 a_{1} = 0 T - 200 g = 200 a_{2} 2 T - m_{1} g = m_{1} a_{3}$

Solving equation we get

$200 g - m_{1} g = \frac{- M _{1} a _{2}}{2} \Rightarrow 200 g - m_{1} g = - \frac{m _{1}}{2} (- g /2) \Rightarrow 200 g = \frac{5}{4} m_{1} g \Rightarrow m_{1} = \frac{800}{5} = 160 gm$

Stream:JEESubject:PhysicsTopic:NLMSubtopic:Challenging application of NLM

⏱ 2mℹ️ Source: QB

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