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Physics - NLM Question with Solution | TestHub

PhysicsNLMChallenging application of NLMMedium2 minQB

PhysicsMediumsingle choice

The system is released from rest. 2 second after the release the bigger block is held for an instant and again released. Find the time elapsed after that before the string again become tight.

Options:

Answer:

Solution:

$4 mg - 2 T = 4 ma$

$\frac{T - mg = m ( 2 a )}{2 mg = 8 ma} a = \frac{g}{4} so, a_{1} = \frac{g}{2} and a_{4} = \frac{g}{4} Also, at t = 2, v_{1} = 0 + (\frac{g}{2}) 2 = g = 10 m / s So, S_{1} = 2 S_{4} 10 t - \frac{1}{2} gt^{2} = 2 (0 + \frac{1}{2} gt^{2}) \Rightarrow t = \frac{2}{3} s$

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

⏱ 2mℹ️ Source: QB

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