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### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2007

MCQ (Single Correct Answer)
Charges are placed on the vertices of a square as shown. Let $\overrightarrow E$ be the electric field and $V$ the potential at the center. If the charges on $A$ and $B$ are interchanged with those on $D$ and $C$ respectively, then
A
$\overrightarrow E$ changes, $V$ remains unchanged
B
$\overrightarrow E$ remains unchanged, $V$ changes
C
both $\overrightarrow E$ and $V$ change
D
$\overrightarrow E$ and $V$ remain unchanged

## Explanation

As shown in the figure, the resultant electric fields before and after interchanging the charges will have the same magnitude, but opposite directions.

Also, the potential will be same in both cases as it is a scalar quantity.

2

### AIEEE 2007

MCQ (Single Correct Answer)
An electric charge ${10^{ - 3}}\,\,\mu \,C$ is placed at the origin $(0,0)$ of $X-Y$ co-ordinate system. Two points $A$ and $B$ are situated at $\left( {\sqrt 2 ,\sqrt 2 } \right)$ and $\left( {2,0} \right)$ respectively. The potential difference between the points $A$ and $B$ will be
A
$4.5$ volts
B
$9$ volts
C
zero
D
$2$ volts

## Explanation

The distance of point $A\left( {\sqrt 2 ,\sqrt 2 } \right)$ from the origin,

$OA = \left| {\overrightarrow {{r_1}} } \right|$

$= \sqrt {{{\left( {\sqrt 2 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}}$

$= \sqrt 4 = 2$

The distance of point $B(2,0)$ from the origin,

$OB = \left| {\overrightarrow {{r_2}} } \right| = \sqrt {{{\left( 2 \right)}^2} + {{\left( 0 \right)}^2}} = 2$ units.

Now, potential at $A,$ ${V_A} = {1 \over {4\pi { \in _0}}}.{Q \over {\left( {OA} \right)}}$

Potential at $B,$ ${V_B} = {1 \over {4\pi { \in _0}}}.{Q \over {\left( {Ob} \right)}}$

$\therefore$ Potential difference between the points $A$ and $B$ is zero.
3

### AIEEE 2006

MCQ (Single Correct Answer)
Two insulating plates are both uniformly charged in such a way that the potential difference between them is ${V_2} - {V_1} = 20\,V.$ (i.e., plate $2$ is at a higher potential). The plates are separated by $d=0.1$ $m$ and can be treated as infinitely large. An electron is released from rest on the inner surface of plate $1.$ What is its speed when it hits plate $2$?
$\left( {e = 1.6 \times {{10}^{ - 19}}\,C,\,\,{m_e} = 9.11 \times {{10}^{ - 31}}\,kg} \right)$
A
$2.65 \times {10^6}\,m/s$
B
$7.02 \times {10^{12}}\,m/s$
C
$1.87 \times {10^6}\,m/s$
D
$32 \times {10^{ - 19}}\,m/s$

## Explanation

$eV = {1 \over 2}m{v^2}$

$\Rightarrow v = \sqrt {{{2ev} \over m}} = \sqrt {{{2 \times 1.6 \times {{10}^{ - 19}} \times 20} \over {9.31 \times {{10}^{ - 31}}}}}$

$= 2.65 \times {10^6}\,m/s$
4

### AIEEE 2006

MCQ (Single Correct Answer)
Two spherical conductors $A$ and $B$ of radii $1$ $mm$ and $2$ $mm$ are separated by a distance of $5$ $cm$ and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres $A$ and $B$ is
A
$4:1$
B
$1:2$
C
$2:1$
D
$1:4$

## Explanation

After connection, ${V_1} = {V_2}$

$\Rightarrow K{{{Q_1}} \over {{r_1}}} = K{{{Q_2}} \over {{r^2}}}$

$\Rightarrow {{Q{}_1} \over {{r_1}}} = {{{Q_2}} \over {{r_2}}}$

The ratio of electric fields

${{{E_1}} \over {{E_2}}} = {{K{{{Q_1}} \over {r_1^2}}} \over {K{{{Q_2}} \over {r_2^2}}}} = {{{Q_1}} \over {r_1^2}} \times {{r_2^2} \over {{Q_2}}}$

$\Rightarrow {{{E_1}} \over {{E_2}}} = {{{r_1} \times r_2^2} \over {r_1^2 \times {r_2}}} \Rightarrow {{{E_1}} \over {{E_2}}} = {{{r_2}} \over {{r_1}}} = {2 \over 1}$

Since the distance between the spheres is large as compared to their diameters, the induced effects may be ignored.

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