1

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

For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance h from its center. Then value of h is :
A
${R \over {\sqrt 5 }}$
B
${R \over {\sqrt 2 }}$
C
R
D
R$\sqrt 2$

## Explanation

Electric field on the axis of the ring,

$E = {{KQh} \over {{{\left( {{R^2} + {h^2}} \right)}^{{3 \over 2}}}}}$

For maximum electric field,

${{dE} \over {dh}} = 0$

$\Rightarrow$   $h = {R \over {\sqrt 2 }}$
2

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

Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross section 5 mm2, is $\upsilon$. If the electron density in copper is 9 $\times$ 1028/m3 the value of $\upsilon$. in mm/s is close to (Take charge of electron to be = 1.6 $\times$ 10$-$19C)
A
0.02
B
3
C
2
D
0.2

## Explanation

We know,

I = neAVd

$\therefore$   Vd = ${{\rm I} \over {neA}}$

=   ${{1.5} \over {9 \times {{10}^{28}} \times 1.6{ \times ^{ - 19}} \times 5 \times {{10}^{ - 6}}}}$

=   0.02 m/s
3

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

A parallel plate capacitor is made of two square plates of side 'a', separated by a distance d (d < < a). The lower triangular portion is filled with a dielectric of dielectric constant K, as shown in the figure. Capacitance of this capacitor is :

A
${{K{ \in _0}{a^2}} \over {2d(K + 1)}}$
B
${{K{ \in _0}{a^2}} \over {d(K - 1)}}\ln K$
C
${{K{ \in _0}{a^2}} \over d}\ln K$
D
${1 \over 2}{{K{ \in _0}{a^2}} \over d}$

## Explanation

Let the capacitance of the upper part of the strip of length dx is dC1 and lower part of the strip is dC2

$\therefore$   dC1 = ${{{\varepsilon _0}\,adx} \over {d - y}}$

and dC2 = ${{K{\varepsilon _0}\,adx} \over y}$

Here dC1 and dC2 are in series.

So, equivalent capacitance,

${1 \over {dC}}$ = ${1 \over {d{C_1}}} + {1 \over {d{C_2}}}$

$\Rightarrow$    ${1 \over {dC}}$ = ${{d - y} \over {{\varepsilon _0}a\,dx}} + {y \over {{\varepsilon _0}K\,adx}}$

$\Rightarrow$   ${1 \over {dC}}$ = ${1 \over {{\varepsilon _0}\,adx}}$ ( d $-$ y + ${y \over K}$)

$\Rightarrow$   dC = ${{{\varepsilon _0}\,adx} \over {\left( {d - y} \right) + {y \over K}}}$

We can divide entire parallel plate capacitor into similar part like the strip of length dx and all the strips will have common end A and B. So they are in parallel.

So equivalent capacitance is the sum of all the strips capacitance.

$\therefore$   $\int {dC}$ = $\int\limits_0^a {{{{\varepsilon _0}\,adx} \over {\left( {d - y} \right) + {y \over K}}}}$

$\Rightarrow$   C = $\int\limits_0^a {{{{\varepsilon _0}\,a\,dx} \over {\left( {d - y} \right) + {y \over K}}}}$

From above diagram you can find this $\to$

tan$\theta$ = ${y \over x}$

Also tan$\theta$ = ${d \over a}$

$\therefore$   ${y \over x}$ = ${d \over a}$

$\Rightarrow$   y = ${d \over a}$ x

By putting this value of y in the integration we get,

C = $\int\limits_0^a {{{{\varepsilon _0}\,a\,dx} \over {d - {d \over a}x + {d \over {Ka}}x}}}$

= $\int\limits_0^a {{{{\varepsilon _0}\,a\,dx} \over {d + \left( {{1 \over K} - 1} \right){d \over a}x}}}$

= $\int\limits_0^a {{{{\varepsilon _0}\,{a^2}\,dx} \over {da + \left( {{{1 - K} \over K}} \right)xd}}}$

= $\int\limits_0^a {{{K{\varepsilon _0}\,{a^2}\,dx} \over {Kda + \left( {1 - K} \right)xd}}}$

= ${{K{\varepsilon _0}{a^2}} \over {d\left( {1 - k} \right)}}\left[ {\ln \left( {\left( {1 - K} \right)x + Ka} \right)} \right]_0^a$

= ${{K{\varepsilon _0}{a^2}} \over {d\left( {1 - k} \right)}}\left[ {\ln \left( a \right) - \ln \left( {Ka} \right)} \right]$

= ${{K{\varepsilon _0}\,{a^2}} \over {d\left( {1 - k} \right)}}\ln \left( {{a \over {Ka}}} \right)$

= ${{K{\varepsilon _0}\,{a^2}} \over {d\left( {1 - k} \right)}}\ln \left( {{1 \over K}} \right)$

= ${{K{\varepsilon _0}\,{a^2}} \over {d\left( {1 - k} \right)}}\ln \left( K \right)$
4

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

A series AC circuit containing an inductor (20 mH), a capacitor (120 $\mu$F) and a resistor (60 $\Omega$) is driven by an AC source of 24V/50 Hz. The energy dissipated in the circuit in 60 s is :
A
5.65 $\times$ 102J
B
2.26 $\times$ 103J
C
5.17 $\times$ 102 J
D
3.39 $\times$ 103 J

## Explanation

Energy dissipated in 60 Sec

= (Pavg) $\times$ 60

= Vrms $\times$ Irms $\times$ cos$\phi$ $\times$ 60

= Vrms $\times$ ${{{V_{rms}}} \over Z} \times$ cos$\phi$ $\times$ 60

XL = $\omega$L = 2$\pi$FL

= 2$\pi$(50)$\times$ 20 $\times$ 10$-$3

= 2$\pi$ $\Omega$

XC = ${1 \over {\omega C}}$

= ${1 \over {2\pi fC}}$

= ${1 \over {2\pi \left( {50} \right) \times 120 \times {{10}^{ - 6}}}}$

= 26.52 $\Omega$

$\therefore$   XC $-$ XL = 20.24 $\simeq$ 20

$\therefore$   Z = $\sqrt {{{\left( {{X_C} - {X_L}} \right)}^2} + {R^2}}$

= $\sqrt {{{\left( {20} \right)}^2} + {{60}^2}}$

= 20 $\sqrt {10}$ $\Omega$

Also cos$\phi$ = ${R \over Z}$ = ${{60} \over {20\sqrt {10} }} = {3 \over {\sqrt {10} }}$

$\therefore$   Energy dissipated in 60 sec

= ${{{{\left( {24} \right)}^2}} \over {20\sqrt {10} }} \times {3 \over {\sqrt {10} }} \times 60$

= 5.17 $\times$ 102 J