1

### JEE Main 2017 (Online) 9th April Morning Slot

The electric field component of a monochromatic radiation is given by

$\overrightarrow E$ = 2 E0 $\widehat i$ cos kz cos $\omega$t

Its magnetic field $\overrightarrow B$ is then given by :
A
${{2{E_0}} \over c}$ $\widehat j$ sin kz cos $\omega$t
B
$-$ ${{2{E_0}} \over c}$ $\widehat j$ sin kz sin $\omega$t
C
${{2{E_0}} \over c}$ $\widehat j$ sin kz sin $\omega$t
D
${{2{E_0}} \over c}$ $\widehat j$ cos kz cos $\omega$t
2

### JEE Main 2018 (Offline)

In an a.c. circuit, the instantaneous e.m.f. and current are given by
e = 100 sin 30 t
i = 20 sin $\left( {30t - {\pi \over 4}} \right)$
In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively
A
50, 0
B
50, 10
C
${{1000} \over {\sqrt 2 }},10$
D
${{50} \over {\sqrt 2 }}$

## Explanation

Wattless current,

here  $\phi$  is the angle between i and e.

Average power,

Pav = Vrms Irms cos$\phi$

= ${{100} \over {\sqrt 2 }} \times {{20} \over {\sqrt 2 }}$ cos${\pi \over 4}$

= ${{1000} \over {\sqrt 2 }}$ watt.
3

### JEE Main 2018 (Offline)

For an RLC circuit driven with voltage of amplitude vm and frequency ${\omega _0}$ = ${1 \over {\sqrt {LC} }}$ the current exhibits resonance. The quality factor, Q is given by :
A
${{CR} \over {{\omega _0}}}$
B
${{{\omega _0}L} \over R}$
C
${{{\omega _0}R} \over L}$
D
${R \over {\left( {{\omega _0}C} \right)}}$

## Explanation

Quality factor (Q) = ${{Angular\,\,{\mathop{\rm Re}\nolimits} sonance} \over {Bandwith}}$

= ${{{1 \over {\sqrt {LC} }}} \over {{R \over L}}}$

= ${{{\omega _0}} \over {{R \over L}}}$

= ${{{\omega _0}L} \over R}$
4

### JEE Main 2018 (Offline)

An EM wave from air enters a medium. The electric fields are
$\overrightarrow {{E_1}}$ = ${E_{01}}\widehat x\cos \left[ {2\pi v\left( {{z \over c} - t} \right)} \right]$ in air and $\overrightarrow {{E_2}}$ = ${E_{02}}\widehat x\cos \left[ {k\left( {2z - ct} \right)} \right]$ in medium, where the wave number k and frequency $\nu$ refer to their values in air. The medium is non-magnetic. If ${\varepsilon _{{r_1}}}$ and ${\varepsilon _{{r_2}}}$ refer to relative permittivities of air and medium respectively, which of the following options is correct ?
A
${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = 4$
B
${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = 2$
C
${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = {1 \over 4}$
D
${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = {1 \over 2}$

## Explanation

Electric field in air,

$\overrightarrow {{E_1}}$  =  E01 $\widehat x$ cos ( ${{2\pi vz} \over c}$ $-$ 2$\pi$vt )

$\therefore\,\,\,$ Velocity in air = ${{2\pi v} \over {{{2\pi v} \over c}}}$ = c

Also,    c = ${1 \over {\sqrt {\mu \varepsilon {r_1}{\varepsilon _0}} }}$ . . . . . . (1)

$\overrightarrow {{E_2}}$  =  E02 $\widehat x$ cos(2kz $-$ kct)

$\therefore\,\,\,$ Velocity in medium = ${{kc} \over {2k}}$ = ${c \over 2}$

Also, ${c \over 2}$ = ${1 \over {\sqrt {\mu {\varepsilon _{r2}}\,{\varepsilon _0}} }}$ . . . . . (2)

As, medium is non magnetic,

So,   $\mu$medium = $\mu$air = $\mu$

Dividing (1) by (2), we get

2 = $\sqrt {{{{\varepsilon _{r2}}} \over {{\varepsilon _{r1}}}}}$

$\Rightarrow$$\,\,\,$ ${{{{\varepsilon _{r1}}} \over {{\varepsilon _{r2}}}}}$ = ${1 \over 4}$