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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2004

MCQ (Single Correct Answer)
Four charges equal to -$$Q$$ are placed at the four corners of a square and a charge $$q$$ is at its center. If the system is in equilibrium the value of $$q$$ is
A
$$ - {Q \over 2}\left( {1 + 2\sqrt 2 } \right)$$
B
$${Q \over 4}\left( {1 + 2\sqrt 2 } \right)$$
C
$$ - {Q \over 4}\left( {1 + 2\sqrt 2 } \right)$$
D
$${Q \over 2}\left( {1 + 2\sqrt 2 } \right)$$

Explanation


Net field at A should be zero

$$\sqrt 2 \,{E_1} + {E_2} = {E_3}$$

$$\therefore$$ $${{kQ \times \sqrt 2 } \over {{a^2}}} + {{kQ} \over {\left( {\sqrt 2 a} \right)}} = {{kq} \over {{{\left( {{a \over {\sqrt 2 }}} \right)}^2}}}$$

$$ \Rightarrow {{Q\sqrt 2 } \over 1} + {Q \over 2} = 2q$$

$$ \Rightarrow q = {Q \over 4}\left( {2\sqrt 2 + 1} \right).$$
2

AIEEE 2004

MCQ (Single Correct Answer)
A charge particle $$'q'$$ is shot towards another charged particle $$'Q'$$ which is fixed, with a speed $$'v'$$. It approaches $$'Q'$$ upto a closest distance $$r$$ and then returns. If $$q$$ were given a speed of $$'2v'$$ the closest distances of approaches would be
A
$$r/2$$
B
$$2r$$
C
$$r$$
D
$$r/4$$

Explanation

$${1 \over 2}m{v^2} = {{kQq} \over r}$$

$$ \Rightarrow {1 \over 2}m{\left( {2v} \right)^2} = {{kqQ} \over {r'}}$$

$$ \Rightarrow r' = {r \over 4}$$
3

AIEEE 2004

MCQ (Single Correct Answer)
Two spherical conductors $$B$$ and $$C$$ having equal radii and carrying equal charges on them repel each other with a force $$F$$ when kept apart at some distance. A third spherical conductor having same radius as that $$B$$ but uncharged is brought in contact with $$B,$$ then brought in correct with $$C$$ and finally removed away from both. The new force of repulsion between $$B$$ and $$C$$ is
A
$$F/8$$
B
$$3$$ $$F/4$$
C
$$F/4$$
D
$$3$$ $$F/8$$

Explanation

$$F \propto {{{Q_A}{Q_C}} \over {{x^2}}}$$

$$x$$ is distance between the spheres. After first operation charge on $$B$$ is halved i.e $${Q \over 2}.$$

and charge on third sphere becomes $${Q \over 2}.$$ Now it is touched to $$C$$, charge then equally

distributes them selves to make potential same, hence charge on $$C$$ becomes

$$\left( {Q + {Q \over 2}} \right){1 \over 2} = {{3Q} \over 4}.$$

$$\therefore$$ $${F_{new}} \propto {{{Q_C}{Q_B}} \over {{x^2}}}$$

$$ = {{\left( {{{3Q} \over 4}} \right)\left( {{Q \over 2}} \right)} \over {{x^2}}}$$

$$ = {3 \over 8}{{{Q^2}} \over {{x^2}}}$$

or, $${F_{new}} = {3 \over 8}F$$
4

AIEEE 2003

MCQ (Single Correct Answer)
The length of a given cylindrical wire is increased by $$100\% $$. Due to the consequent decrease in diameter the change in the resistance of the wire will be
A
$$200\% $$
B
$$100\% $$
C
$$50\% $$
D
$$300\% $$

Explanation

$${R_f} = {n^2}{R_1}$$

Here $$n=2$$ (length becomes twice)

$$\therefore$$ $${R_f} = 4{R_i}$$

New resistance $$=400$$ of $${R_i}$$

$$\therefore$$ Increase $$ = 300\% $$

Questions Asked from Electrostatics

On those following papers in MCQ (Single Correct Answer)
Number in Brackets after Paper Indicates No. of Questions
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