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Physics - NLM Question with Solution | TestHub

PhysicsNLMMiscellaneousEasy2 minQB

PhysicsEasysingle choice

A mass M is suspended by two springs A and B of force constants $k_{1}$ and $k_{2}$ respectively as shown in the diagram. The total stretch of springs in equilibrium is :

Options:

Answer:

Solution:

Both springs carry the same tension $M g$ (series). Stretch of A $= \frac{M g}{k _{1}}$ , stretch of B $= \frac{M g}{k _{2}}$ .

Total stretch $= M g (\frac{1}{k _{1}} + \frac{1}{k _{2}}) = \frac{M g ( k _{1} + k _{2} )}{k _{1} k _{2}}$ .

Stream:JEESubject:PhysicsTopic:NLMSubtopic:Miscellaneous

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