1

JEE Main 2018 (Online) 16th April Morning Slot

Two identical conducting spheres A and B, carry equal charge. They are separated by a distance much larger than their diameters, and the force between theis F. A third identical conducting sphere, C, is uncharged. Sphere C is first touhed to A, then to B, and then removed. As a result, the force between A and B would be equal to :
A
F
B
${{3F} \over 4}$
C
${{3F} \over 8}$
D
${{F} \over 2}$

Explanation

Let, change of A and B = q

$\therefore\,\,\,$ Force between them, F = ${{k \times q \times q} \over {{r^2}}} = {{k{q^2}} \over {{r^2}}}$

When C touched with A then charge of A. Will fl;ow to C and divide into half parts.

$\therefore\,\,\,$ charge of A and C ,

qA = qB = ${q \over 2}$

Then C touched with B, then charge on B,

qB = ${{{q \over 2} + q} \over 2} = {{3q} \over 4}$

$\therefore\,\,\,$ Force between A and B,

F' = ${{k \times {q \over 2} \times {{3q} \over 4}} \over {{r^2}}}$

= ${{k \times 3{q^2}} \over {8{r^2}}}$

= ${3 \over 8} \times {{k{q^2}} \over {{r^2}}}$

= ${3 \over 8}\,F$
2

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.
3

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 = $-$ ${Q \over 4}$
4

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%