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Physics - NLM Question with Solution | TestHub

PhysicsNLMMedium2 minQB

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Fig shows two pulley arrangements for lifting a mass m. In (a) the mass is lifted by attaching a mass 2m while in (b) the mass is lifted by pulling the other end with a downward force $F = 2 mg$ , If $f_{a}$ and $f_{b}$ are the accelerations of the two masses then

Options:

Answer:

Solution:

$Case (a) T - mg = ma 2 mg - T = 2 ma} \Rightarrow f_{a} = g /3 = a$

Case (b)

$T = 2 mg \Rightarrow 2 mg - mg = mf_{b} \Rightarrow f_{b} = g \frac{f _{a}}{f _{b}} = \frac{g}{3 g} = \frac{1}{3} \Rightarrow f_{a} = f_{b} /3$

Stream:JEESubject:PhysicsTopic:NLM

⏱ 2mℹ️ Source: QB

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