NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

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 __________.

Your Input ________

Answer

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.

Joint Entrance Examination

JEE Main JEE Advanced WB JEE

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

Medical

NEET

CBSE

Class 12