Solved: PHYSICS Question for NEET

NEET PHYSICSHard

Two identical charged spheres suspended from a common point by two massless strings of lengths 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:

v \propto x

v \propto x⁻¹/²

v \propto x⁻¹

v \propto x¹/²

Step-by-Step Solution

At equilibrium, the electrostatic force ( $F_e$ ) is balanced by the horizontal component of the tension, while gravity balances the vertical component. For small angles (since $x \ll l$ ), $\tan \theta \approx \sin \theta = \frac{x}{2l}$ .

Equilibrium Condition: $F_e = mg \tan \theta \Rightarrow \frac{kq^2}{x^2} = mg \frac{x}{2l}$ .
Relation: This implies $q^2 \propto x^3$ , or $q \propto x^{3/2}$ .
Differentiation: Since charge leaks at a constant rate ( $\frac{dq}{dt} = C$ ), differentiating with respect to time gives $\frac{dq}{dt} \propto \frac{d}{dt}(x^{3/2}) \Rightarrow C \propto x^{1/2} \frac{dx}{dt}$ .
Velocity: Substituting $v = \frac{dx}{dt}$ , we get $C \propto x^{1/2} v$ , which leads to $v \propto x^{-1/2}$ .

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