JEE Main Physics question
Statement 1: Two springs of force constants k₁ and k₂ are stretched by the same force. If k₁ > k₂, then work done in stretching the first (W₁) is less than work done in stretching the second (W₂). Statement 2: F = k₁x₁ = k₂x₂ ⟹ x₁/x₂ = k₂/k₁ ⟹ W₁/W₂ = (k₁x₁²/2)/(k₂x₂²/2) = (k₁/k₂)(k₂/k₁)² = k₂/k₁, and since k₁ > k₂, W₁ < W₂.
Answer: (A)
Solution
For a given force, energy stored W = F²/(2k) ∝ 1/k, so the softer spring (smaller k₂) stores
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