1

### JEE Main 2018 (Online) 16th April Morning Slot

In the following circuit, the switch S is closed at t = 0. The charge on the capacitor C1 as a function of time will be given by $\left( {{C_{eq}} = {{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right)$

A
${C_1}E\left[ {1 - \exp \left( { - tR/{C_1}} \right)} \right]$
B
${C_2}E\left[ {1 - \exp \left( { - t/R{C_2}} \right)} \right]$
C
${C_{eq}}E\left[ {1 - \exp \left( { - t/R{C_{eq}}} \right)} \right]$
D
${C_{eq}}E\,\,\exp \left( { - t/R{C_{eq}}} \right)$

## Explanation

Charge on capacitor at time t,

q = q0 [ 1 $-$ e$-$t/RCeq]

= Ceq E [ 1 $-$ e$-$t/RCeq]

[ as $\,\,\,\,\,\,$ q0 = Ceq E]

this charge q will be on both capacitor C1 and C2, as both are in series.
2

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

Three charges + Q, q, + Q are placed respectively, at distance, 0, d/2 and d from the origin, on the x-axis. If the net force experienced by + Q, placed at x = 0, is zero, then value of q is :
A
$-$ ${Q \over 4}$
B
+ ${Q \over 2}$
C
+ ${Q \over 4}$
D
$-$ ${Q \over 2}$

## Explanation

Force on + Q charge at x = 0 due to q charge, F1 = ${{KQq} \over {{{\left( {{d \over 2}} \right)}^2}}}$

Force on +Q charge at x = 0 due to + Q charge at x = d is,

F2 = ${{KQQ} \over {{d^2}}}$

According to the question,

F1 + F2 = 0

$\Rightarrow$   F1 = $-$ F2

$\Rightarrow$   ${{KQq} \over {{{\left( {{d \over 2}} \right)}^2}}}$ = $-$ ${{KQQ} \over {{d^2}}}$

$\Rightarrow$   4q = $-$ Q

$\Rightarrow$   q = $-$ ${Q \over 4}$
3

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

A resistance is shown in the figure. Its value and tolerance are given respectively by :

A
270 $\Omega$, 10 %
B
27 k$\Omega$, 10 %
C
27 k$\Omega$, 20 %
D
270 $\Omega$, 5 %

## Explanation

From color code table :

For Red value is 2

For Violet value is 7

For Orange multiplier is 103

For Silver tolarence is 10%

$\therefore$ Resistance and tolerance is

= 27 $\times$ 103 $\pm$ 10%

= 27 k$\Omega$ $\pm$ 10%
4

### 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,

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

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