1

### JEE Main 2019 (Online) 9th January Evening Slot

Two point charges q1$\left( {\sqrt {10} \mu C} \right)$ and q2($-$ 25 $\mu$C) are placed on the x-axis at x = 1 m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is,
[take ${1 \over {4\pi { \in _0}}}$ = 9 $\times$ 109 Nm2C$-$2]
A
$\left( {63\widehat i - 27\widehat j} \right) \times {10^2}$
B
$\left( { - 63\widehat i + 27\widehat j} \right) \times {10^2}$
C
$\left( {81\widehat i - 81\widehat j} \right) \times {10^2}$
D
$\left( { - 81\widehat i + 81\widehat j} \right) \times {10^2}$

## Explanation

Electric field due to $\sqrt {10} \,\mu C$ charge :

$\overrightarrow {{E_1}} = -$ E1 sin$\theta$1 $\widehat i$ + E1 cos$\theta$1 $\widehat j$

Where,

E1 $= {1 \over {4\pi {\varepsilon _0}}} \times {{\left| {{q_1}} \right|} \over {{r_1}^2}}$

$= 9 \times {10^9} \times {{\sqrt {10} \times {{10}^{ - 6}}} \over {{{\left( {\sqrt {{1^2}} + {3^2}} \right)}^2}}}$

$= {{9 \times {{10}^3}} \over {\sqrt {10} }}\,v/m$

sin $\theta$1 $=$ ${1 \over {\sqrt {10} }}$

and cos$\theta$1 = ${3 \over {\sqrt {10} }}$

$\therefore$   $\overrightarrow {{E_1}} = {{9 \times {{10}^3}} \over {\sqrt {10} }}\,\left( { - {1 \over {10}}\widehat i + {3 \over {\sqrt {10} }}\widehat j} \right)$

$= 9 \times {10^2}\left( { - \widehat i + 3\widehat j} \right)$

Electric field due to $-$ 25 $\mu$C charge,

$\overrightarrow {{E_2}} =$ E2 sin$\theta$2$\widehat i$ $-$ E2 cos$\theta$2 $\widehat j$

where

E2 $= {1 \over {4\pi {\varepsilon _0}}} \times {{\left| {{9_2}} \right|} \over {r_2^2}}$

$= 9 \times {10^9} \times {{25 \times {{10}^{ - 6}}} \over {{{\left( {\sqrt {{4^2} + {3^2}} } \right)}^2}}}$

$= 9 \times {10^3}$ V/m

sin$\theta$2 = ${4 \over 5}$

and cos$\theta$2 = ${3 \over 5}$

$\therefore$   $\overrightarrow {{E_2}} = 9 \times {10^3}\,\,\left( {{4 \over 5}\widehat i - {3 \over 5}\widehat j} \right)$

$= 18 \times {10^2}\left( {4\widehat i - 3\widehat j} \right)$

$\therefore$   Net electric field,

$\overrightarrow E$ = ${\overrightarrow E _1}$ + ${\overrightarrow E _2}$

$= \left( {63\widehat i - 27\widehat j} \right) \times {10^2}\,\,V/m$
2

### JEE Main 2019 (Online) 9th January Evening Slot

Charge is distributed within a sphere of radius R with a volume charge density $\rho \left( r \right) = {A \over {{r^2}}}{e^{ - {{2r} \over s}}},$ where A and a are constants. If Q is the total charge of this charge distribution, the radius R is :
A
a log $\left( {1 - {Q \over {2\pi aA}}} \right)$
B
${a \over 2}$ log $\left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)$
C
a log $\left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)$
D
${a \over 2}$ log $\left( {1 - {Q \over {2\pi aA}}} \right)$

## Explanation

Volume of this spherical layer,

dv = (4$\pi$r2)dr

charge present in this layer,

dq = $\rho$ (4$\pi$r2 dr)

= ${A \over {{r^2}}}{e^{ - {{2r} \over a}}}\,\,\left( {4\pi {r^2}dr} \right)$

= $A\,{e^{ - {{2r} \over a}}}\left( {4\pi dr} \right)$

$\therefore$  Total charge in the sphere,

Q= $\int\limits_0^R {4\pi A{e^{ - {{2r} \over a}}}} \,dr$

= 4$\pi$A$\int\limits_0^R {{e^{ - {{2r} \over a}}}} \,dr$

= 4$\pi$A$\left[ {{{{e^{ - {{2r} \over a}}}} \over { - {2 \over a}}}} \right]_0^R$

= 4$\pi$A $\left( { - {a \over 2}} \right)\left( {{e^{ - {{2R} \over a}}} - 1} \right)$

$\therefore$  Q = 2$\pi$aA $\left( {1 - {e^{ - {{2R} \over a}}}} \right)$

$\Rightarrow$  ${1 - {e^{ - {{2R} \over a}}}}$ = ${Q \over {2\pi aA}}$

$\Rightarrow$  ${{e^{ - {{2R} \over a}}}}$ = 1 $-$ ${Q \over {2\pi aA}}$

$\Rightarrow$  ${e^{{{2R} \over a}}}$ = ${1 \over {1 - {Q \over {2\pi aA}}}}$

$\Rightarrow$  ${{2R} \over a} = \log \left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)$

$\Rightarrow$  R = ${a \over 2}$ log $\left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)$
3

### JEE Main 2019 (Online) 9th January Evening Slot

In the given circuit the the internal resistance of the 18 V cell is negligible. If R1 = 400 $\Omega$, R3 = 100 $\Omega$ and R4 = 500 $\Omega$ and the reading of an ideal voltmeter across R4 is 5V, then the value of R2 will be :

A
300 $\Omega$
B
450 $\Omega$
C
550 $\Omega$
D
230 $\Omega$

## Explanation

Voltage accross resistance R4 = 5 V

$\therefore$   IR4 = 5 V

$\Rightarrow$   500 $\times$ I = 5

$\Rightarrow$   I = ${1 \over {100}}$ A

$\therefore$   Voltage across resistor R3 = ${1 \over {100}}\left( {100} \right)$ = 1 A

$\therefore$   Total drop in resistance R3 and R4 = 5 + 1 = 6V

So, voltage accross R2 resistance is also 6V as R3, R4 and R2 are in parallel

$\therefore$   Voltage accross R1 resistor R1 resistor = 18 $-$ 6 = 12 V

$\therefore$   Current through R1 resistor = ${{12} \over {400}}$ = ${3 \over {100}}$ A

$\therefore$   Current through R2 resistor

= ${3 \over {100}} - {1 \over {100}}$

= ${2 \over {100}}$ A

$\therefore$   $\left( {{2 \over {100}}} \right)$ R2 = 6

$\Rightarrow$   R2 = 300 $\Omega$
4

### JEE Main 2019 (Online) 10th January Morning Slot

Two electric dipoles, A, B with respective dipole moments ${\overrightarrow d _A} = - 4qai$ and ${\overrightarrow d _B} = - 2qai$ are placed on the x-axis with a separation R, as shown in the figure. The distance from A at which both of them produce the same potential is -

A
${{\sqrt 2 R} \over {\sqrt 2 + 1}}$
B
${R \over {\sqrt 2 + 1}}$
C
${{\sqrt 2 R} \over {\sqrt 2 - 1}}$
D
${R \over {\sqrt 2 - 1}}$

## Explanation

V $= {{4qa} \over {\left( {R + x} \right)}} = {{2qa} \over {\left( {{x^2}} \right)}}$

$\sqrt 2 x = R + x$

$x = {R \over {\sqrt 2 - 1}}$

dist $= {R \over {\sqrt 2 - 1}} + R = {{\sqrt 2 R} \over {\sqrt 2 - 1}}$