JEE Main Physics question
Two springs of spring constants 1500 N m⁻¹ and 3000 N m⁻¹ respectively are stretched with the same force. They will have potential energy in the ratio
Answer: (A)
Solution
For the same force, PE = F²/(2k) ∝ 1/k. Ratio U₁:U₂ = k₂:k₁ = 3000:1500 = 2:1, i.e., U₁/U₂ = 1500/3000 giving U₁:U₂ = 1:2.