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JEE Main Physics question

PhysicsWork, Energy and PowerJEE Main

If a long spring is stretched by 0.02 m, its potential energy is U. If the spring is stretched by 0.1 m, the potential energy will be

Answer: (D)

Solution

Since U ∝ x², U₂/U₁ = (x₂/x₁)² = (0.1/0.02)² = 5² = 25, so U₂ = 25U.

Question ID VK-153891