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Electrostatics question

2016 · Q119
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Electrostatics question

2016 · Q119

NEETPhysicsElectrostaticsMCQ+4 / −1
Two identical charged spheres suspended from a common point by two massless strings of lengths lll, are initially at a distance d(d < < lll) 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
  1. A
    v ∝\propto∝ x−-−1/2
  2. B
    v ∝\propto∝ x−-−1
  3. C
    v ∝\propto∝ x1/2
  4. D
    v ∝\propto∝ x
View written solutionFree

Correct answer: A

NEET 2016 Phase 1 Physics - Electrostatics Question 68 English Explanation

From figure tan⁡θ=Femg≃θ\tan \theta = {{{F_e}} \over {mg}} \simeq \theta tanθ=mgFe​​≃θ

kq2x2mg=x2ℓ{{k{q^2}} \over {{x^2}mg}} = {x \over {2\ell }}x2mgkq2​=2ℓx​

⇒\Rightarrow⇒ x3∝q2{x^3} \propto {q^2}x3∝q2   ...(i)

⇒x3/2∝q \Rightarrow {x^{3/2}} \propto q⇒x3/2∝q

Differentiating eq. (1) w.r.t. time

3x2dxdt∝2qdqdt3{x^2}{{dx} \over {dt}} \propto 2q{{dq} \over {dt}}3x2dtdx​∝2qdtdq​ but dqdt{{dq} \over {dt}}dtdq​ is constant

so x2(v) ∝\propto∝ q Replace q from eq. (2)

x2(v) ∝\propto∝ x3/2 ⇒\Rightarrow⇒ v ∝\propto∝ x–1/2

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