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JEE Advanced Physics Electrostatics 2020 JEE Advanced 2020 (Paper 1)

JEE Advanced Physics Question (2020) — Solution

Question

One end of a spring of negligible unstretched length and spring constant k is fixed at the origin ( 0 , 0 ) . A point particle of mass m carrying a positive charge q is attached at its other end. The entire system is kept on a smooth horizontal surface. When a point dipole p → pointing towards the charge q is fixed at the origin, the spring gets stretched to a length l and attains a new equilibrium position (see figure below). If the point mass is now displaced slightly by Δ l ≪ l from its equilibrium position and released, it is found to oscillate at frequency 1 δ k m , The value of δ is _______.

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

Original frequency f = 1 2 π k m If dipole appears, At equilibrium 2 K p ℓ + x 0 3 · q = k x 0 . . . i Now disturbed 2 K p ℓ + x 0 + x 3 · q - k x 0 + x = m a ⇒ m a = 2 K pq ℓ + x 0 3 1 + x ℓ + x 0 - 3 - k x 0 + x ⇒ m a = 2 K p q ℓ + x 0 3 1 - 3 x ℓ + x 0 - 3 - k x 0 + x = - 6 K p q x ℓ + x 0 4 - k x = - 3 x ℓ + x 0 k x 0 - k x = - k x 3 x 0 ℓ + x 0 + 1 As ℓ is negative, m a = - 4 k x ; ⇒ a = 4 k m x New frequency : f ' = 1 2 π 4 k m = 2 f = 1 δ k m ∴ δ = π = 3 . 14

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