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 [2016]

Question

Smart Achievers · Accepted Answer

(1)

From figure, $T \cos θ = m g$ ...(i)

$v$ ...(ii)

From eqns. (i) and (ii), $\tan θ = \frac{k q^{2}}{x^{2} m g}$

Since $θ$ is small, $∴$ $\tan θ \approx \sin θ = \frac{x}{2 l}$

$∴$ $\frac{x}{2 l} = \frac{k q^{2}}{x^{2} m g} \Rightarrow q^{2} = x^{3} \frac{m g}{2 l k} or q \propto x^{3 / 2}$

$\Rightarrow \frac{d q}{d t} \propto \frac{3}{2} \sqrt{x} \frac{d x}{d t} = \frac{3}{2} \sqrt{x} v$

Since, $\frac{d q}{d t} = constant$ ; $∴$ $v \propto \frac{1}{\sqrt{x}}$

Solution

1 $v \propto x^{- 1 / 2}$

2 $v \propto x^{- 1}$

3 $v \propto x^{1 / 2}$

4 $v \propto x$

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Solution

1 v∝x-1/2

2 v∝x-1

3 v∝x1/2

4 v∝x

Ans. (1) From figure, Tcosθ=mg ...(i) Tsinθ=kq2x2 ...(ii) From eqns. (i) and (ii), tanθ=kq2x2mg Since θ is small, ∴ tanθ≈sinθ=x2l ∴ x2l=kq2x2mg ⇒ q2=x3mg2lk or q∝x3/2 ⇒dqdt∝32xdxdt=32x v Since, dqdt=constant ; ∴ v∝1x

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1 $v \propto x^{- 1 / 2}$

2 $v \propto x^{- 1}$

3 $v \propto x^{1 / 2}$

4 $v \propto x$