NEET 2016 Phase 1 · PhysicsCoulomb forcePrevious Year Question
Two identical charged spheres suspended from a common point by two massless strings of length l are initially at a distance d (d << l) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity v. Then v varies as a function of the distance x between the spheres as
Answer: (C) v ∝ x^(-1/2). Answer: (c) Solution: For each sphere in equilibrium (small-angle, x << l): The Coulomb repulsion is balanced by the horizontal component of tension.
- A.v ∝ x^(1/2)
- B.v ∝ x
- C.v ∝ x^(-1/2)✓
- D.v ∝ x⁻¹
Correct Answer
(C) v ∝ x^(-1/2)
Solution & Explanation
Answer: (c) Solution: For each sphere in equilibrium (small-angle, x << l): The Coulomb repulsion is balanced by the horizontal component of tension. tan θ ≈ sin θ = (x/2)/l Force balance: F_coulomb = mg·tan θ (1/4πε₀)·q²/x² = mg·(x/2l) So q² ∝ x³ → q ∝ x^(3/2) Charge leaks at a constant rate: dq/dt = constant. Since q ∝ x^(3/2): dq/dt ∝ x^(1/2)·(dx/dt) = x^(1/2)·v = constant Therefore v ∝ x^(-1/2).
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