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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 $\upsilon$ . Then as a function of distance $x$ between them

Option 1)
$\upsilon\: \alpha\: x^{-1/2}$

Option 2)
$\upsilon\: \alpha\: x^{-1}$

Option 3)
$\upsilon\: \alpha\: x^{1/2}$

Option 4)
$\upsilon\: \alpha\: x$

Answers (3)

best_answer

As we learnt in

Coulombic force -

$F\propto Q_{1}Q_{2}=F\propto \frac{Q_{1}Q_{2}}{r^{2}}=F=\frac{KQ_{1}Q_{2}}{r^{2}}$

- wherein

K - proportionality Constant

Q₁ and Q₂ are two Point charge

$T\cos \Theta =mg$

$T\sin \Theta =f_{e}=\frac{1}{4\pi\varepsilon _{o}}\:\frac{q^{2}}{d^{2}}$

$tan \Theta =\frac{1}{4\pi\varepsilon _{o}}\:\frac{q^{2}}{d^{2}mg}$

$\therefore tan \Theta=\frac{x}{2l}$

$\frac{x}{2l}=\frac{q^{2}}{4\pi\varepsilon _{o}x^{2}mg}$

$\Rightarrow \frac{x}{2l}\propto \frac{q^{2}}{x^{2}}$

$\Rightarrow q^{2}\propto x^{3}$

$\Rightarrow q\propto x^{3/2}$

$\Rightarrow\frac{dq}{dt}\propto\frac{3}{2}x^{1/2}\frac{dx}{dt} \therefore \left (\frac{dq}{dt}= constant \right )$

$\therefore \frac{dx}{dt}=v\propto x^{-1/2}$

Option 1)

$\upsilon\: \alpha\: x^{-1/2}$

This is correct option

Option 2)

$\upsilon\: \alpha\: x^{-1}$

This is incorrect option

Option 3)

$\upsilon\: \alpha\: x^{1/2}$

This is incorrect option

Option 4)

$\upsilon\: \alpha\: x$

This is incorrect option

Posted by

prateek

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Study 40% syllabus and score up to 100% marks in JEE

Option 1 ans

Posted by

Sujal

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