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Physics - NLM Question with Solution | TestHub

PhysicsNLMFBD & EquilibriumMedium2 minQB

PhysicsMediummultiple choice

A mass 2 m suspended from a given spring causes it to stretched relative to its relaxed length. The mass and the spring are then each cut into two identical pieces and connected as shown in figure.

Options:(select one or more)

Answer:

A, D

Solution:

F.B.D. of mass 2m

Now $kx = 2 mg$

$x = \frac{2 mg}{k}$

Distance of bottom of mass $= ℓ + \frac{2 mg}{k}$ ,

where $ℓ$ is the natural length of spring

Now spring is cut into two equal parts of $ℓ /2$ long each, so from hook's law : $k ℓ =$ constant

$\Rightarrow k \times ℓ = k^{'} \times \frac{ℓ}{2} \Rightarrow k^{'} = 2 k$

Now in second case : Let elongation in upper spring be $x_{1}$ \& lower spring be $x_{2}$ .

Stream:JEESubject:PhysicsTopic:NLMSubtopic:FBD & Equilibrium

⏱ 2mℹ️ Source: QB

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