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JEE Main Physics question

PhysicsWork, Energy and PowerJEE Main

A particle free to move along the x-axis has potential energy given by U(x) = k[1 − exp(−x²)] for −∞ ≤ x ≤ ∞, where k is a positive constant of appropriate dimensions. Then [IIT JEE 1999, UPSEAT 2003]

Answer: (D)

Solution

For small x, expanding the exponential: U(x) ≈ k[1 − (1 − x² + x⁴/2! − ...)] ≈ kx², so F = −dU/dx ≈ −2kx, i.e., F ∝ −x, the signature of simple

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Question ID VK-154126