Two identical charged spheres suspended from a common point by two massless strings of length l are initially a distance d ( d l ) apart because of th - Physics MCQ FIITJEE AIEEE 2011 (Set Q) 2026 | ExamTest.live

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Two identical charged spheres suspended from a common point by two massless strings of length l are initially a distance d (d ≪ l) apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the charges approach each other with a velocity v . Then as a function of distance x between them:

Solution & Step-by-step Explanation

For small θ, equilibrium gives tan θ \approx sin θ = \frac{x}{2 l} . F_{e} = m g tan θ \approx m g \frac{x}{2 l} \frac{k q ^{2}}{x ^{2}} = \frac{m gx}{2 l} ⟹ q^{2} \propto x^{3} ⟹ q \propto x^{3/2} Differentiating with respect to time: \frac{d q}{d t} \propto \frac{3}{2} x^{1/2} \frac{d x}{d t} Since \frac{d q}{d t} (leakage rate) is constant: constant \propto x^{1/2} v ⟹ v \propto x^{- 1/2}

Try it yourself before checking the explanation above.

Two identical charged spheres suspended from a common point by two massless strings of length l are initially a distance d (d ≪ l) apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the charges approach each other with a velocity v . Then as a function of distance x between them:

v \propto x^{- 1}

v \propto x^{1/2}

v \propto x

v \propto x^{- 1/2}

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