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Physics - NLM Question with Solution | TestHub

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The diagram below shows two different masses hung from identical Hooke's law springs. The Hooke's law constant k for the springs is equal to

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Solution:

$K (ℓ - ℓ_{0}) = mg K (ℓ^{'} - ℓ_{0}) = m^{'} g \frac{ℓ ^{'} - ℓ _{0}}{ℓ - ℓ _{0}} = \frac{m ^{'}}{m} = \frac{3}{2} \Rightarrow ℓ_{0} = 4 cm K = \frac{mg}{( 6 - 4 ) \times 1 0 ^{- 2}} \Rightarrow K = \frac{10}{2} = 5 N / cm$

Stream:JEESubject:PhysicsTopic:NLM

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