Three small identical bubbles of water having same charge on each coalesce to form a bigger bubble. Then the ratio of the potentials on one initial bubble and that on the resultant bigger bubble is :

Three small identical bubbles of water having same charge on each coalesce to form a bigger bubble

Q. Three small identical bubbles of water having same charge on each coalesce to form a bigger bubble. Then the ratio of the potentials on one initial bubble and that on the resultant bigger bubble is :

(A) $1 : 3^{2/3}$

(B) $3^{2/3} : 1$

(D) $1 : 2^{2/3}$

Correct Answer: $1 : 3^{2/3}$

Explanation

Let the radius of each small bubble be $r$ and charge on each bubble be $q$.

When three identical bubbles coalesce, volume is conserved. Therefore,

$$ \frac{4}{3}\pi R^3 = 3 \times \frac{4}{3}\pi r^3 $$

$$ R^3 = 3r^3 \Rightarrow R = 3^{1/3} r $$

Total charge on the bigger bubble,

$$ Q = 3q $$

Electric potential of a spherical bubble is given by

$$ V = \frac{kQ}{R} $$

Potential of one initial bubble,

$$ V_1 = \frac{kq}{r} $$

Potential of the resultant bigger bubble,

$$ V_2 = \frac{k(3q)}{3^{1/3}r} = 3^{2/3}\frac{kq}{r} $$

Thus,

$$ \frac{V_1}{V_2} = \frac{1}{3^{2/3}} $$

Hence, the required ratio is

$1 : 3^{2/3}$

Explanation

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