The correct option (B) ѵ ∝ x(–1/2)
Explanation:
![](https://www.sarthaks.com/?qa=blob&qa_blobid=5748428590667507645)
For equilibrium,
T cos θ = mg (1)
T sin θ = Fe = [(k ∙ q ∙ q) / x2] coulomb's law
i.e. T sin θ = [(kq2)/x2] (2)
Dividing (2) by (1)
[(T sin θ)/(T cos θ)] = [(kq2) / (x2 ∙ mg)]
∴ tan θ = [(kq2) / (x2mg)] also tan θ = [(x/2)/ℓ] = (x/2ℓ)
∴ (x/2ℓ) = [(kq2)/(x2mg)]
∴ q2 = [(x3mg)/(2kℓ)] (2)
differential (2) wrt t.
∴ 2q(dq/dt) = [(mg)/(2kℓ)] × 3x2 ∙ (dx/dt)
∴ 2q(dq/dt) = 3x2(dx/dt) × (q2/x3) from (2)
∴ 2(dq/dt) = (q/x)3 ∙ (dx/dt)
∴ (2/3) (dq/dt) = (ѵ/x)[x(3/2){(mg)/(2kℓ)(1/2)}]
∴ [{(2/3) (dq/dt)}/{(mg)/(2kℓ)(1/2)}] = ѵx(1/2)
∴ ѵx(1/2) = constant
hence ѵ ∝ x(–1/2)