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Physics - NLM Question with Solution | TestHub

PhysicsNLMMedium2 minQB

PhysicsMediumsingle choice

In the arrangement as shown $M > 2 m$ . Find the extension in the spring of force constant $K_{1}$ . All pulleys are smooth and massless and the blocks do not oscillate.

Options:

Answer:

Solution:

$a = \frac{( 1 \times 1 - ∠1115 )}{( M + 2 m )}$

Now, $Mg - K_{1} x_{1} = Ma$

$x_{1} = \frac{M ( g - a )}{K _{1}} = \frac{4 Mmg}{K _{1} ( M + 2 m )}$

Stream:JEESubject:PhysicsTopic:NLM

⏱ 2mℹ️ Source: QB

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