JEE Main Physics question
Two springs have force constants k₁ and k₂. They are extended through the same distance x. If their elastic energies are E₁ and E₂, then E₁/E₂ is equal to
Answer: (A)
Solution
For the same extension x, E = ½kx² ∝ k, so E₁/E₂ = k₁/k₂.