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### IIT-JEE 2011 Paper 1 Offline

Numerical

Four point charges, each of +q, are rigidly fixed at the four corners of a square planar soap film of side a. The surface tension of the soap film is $$\gamma$$. The system of charges and planar film are in equilibrium, and $$a = k{\left[ {{{{q^2}} \over \gamma }} \right]^{1/N}}$$, where k is a constant. Then N is __________.

Correct Answer is 3

## Explanation

The net force on one of the charges due to other charges is

$$F = {{2k{q^2}} \over {{a^2}}} + {{k{q^2}} \over {2{a^2}}} = {5 \over 2}\left( {{{k{q^2}} \over {{a^2}}}} \right)$$

where $$k = {1 \over {4\pi \varepsilon }}$$. Here, as shown in the figure, line AB divided the soap film into two equal parts. The free-body diagram of half part is also depicted in the figure here. At equilibrium, the surface tension balances the force.

Therefore,

$${F_{surface}} = 2\sqrt 2 a\gamma$$

That is, $$2\sqrt 2 a\gamma = {5 \over 2}\left( {{{k{q^2}} \over {{a^2}}}} \right)$$

$$\Rightarrow {a^3} = {5 \over {4\sqrt 2 }}\left( {{{{q^2}} \over \gamma }} \right)$$

Therefore,

a = Any constant $$\times {\left( {{{{q^2}} \over \gamma }} \right)^{1/3}}$$

Hence, N = 3.

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