Practice Start Test

Electromagnetic Waves

NEET Physics · 97 questions · Page 2 of 10 · Click an option or "Show Solution" to reveal answer

Q11
Which colour of the light has the longest wavelength?
A blue
B violet
C red
D green
Correct Answer
Option C
Solution

Since, Red colour has least frequency. So, red has the longest wavelength among the given colour. As,

ν=vλ\nu = v\lambda
Q12
An EM wave is propagating in a medium with a velocity v→=vi^\overrightarrow v = v\widehat i. The instantaneous oscillating electric field of this em wave is along +y axis. Then the direction of oscillating magnetic field of the em wave will be along
A z direction
B +z direction
C –y direction
D –x direction
Correct Answer
Option B
Solution

As we know,

E→×B→=V→\overrightarrow E \times \overrightarrow B = \overrightarrow V

⇒\Rightarrow

Ej^×B→=vi^E\widehat j \times \overrightarrow B = v\widehat i

[ As given Electric field vector is along +y axis and

v→=vi^\overrightarrow v = v\widehat i

]

i^=j^×k^\widehat i = \widehat j \times \widehat k

so

B→=Bk^\overrightarrow B = B\widehat k

. ∴\therefore Direction of magnetic field vector is along +z direction.

Q13
In an electromagnetic wave in free space that root mean square value of the electric field is E rms = 6 V m −-1 . The peak value of the magnetic field is
A 2.83 × \times 10 −-8 T
B 0.70 × \times 10 −-8 T
C 4.23 × \times 10 −-8 T
D 1.41 × \times 10 −-8 T
Correct Answer
Option A
Solution

The rms value of magnetic field B rms =

Ermsc{{{E_{rms}}} \over c}

=

6(3×108){6 \over {\left( {3 \times {{10}^8}} \right)}}

= 2 × 10 – 8 T Now peak value of magnetic field B peak =

2Brms\sqrt 2 {B_{rms}}

= 2.83 ×\times 10 −-8 T

Q14
Out of the following options which one can be used to produce a propagating electromagnetic wave ?
A A chargeless particle
B An accelerating charge
C A charge moving at constant velocity
D A stationary charge
Correct Answer
Option B
Solution

For producing electromagnetic waves, accelerating charged particle is required.

Q15
The energy of the em waves is of the order of 15 keV. To which part of the spectrum does it belong ?
A Ultraviolet rays
B γ−rays\gamma - rays
C X-rays
D Infra-red rays
Correct Answer
Option C
Solution

Energy of x-ray is (100 eV to 100 keV). Hence energy of the order of 15 keV belongs to X-rays.

Q16
A radiation of energy 'E' falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (C = Velocity of light)
A 2EC2{{2E} \over {{C^2}}}
B EC2{E \over {{C^2}}}
C EC{E \over C}
D 2EC{{2E} \over C}
Correct Answer
Option D
Solution

Momentum transferred to the surface = change in momentum = P f - P i =

+EC−(−EC)+ {E \over C} - \left( { - {E \over C}} \right)

=

2EC{{2E} \over C}
Q17
Light with an energy flux of 25 × \times 10 4 W m −-2 falls on a perfectly reflecting surface at normal incidence. If the surface area is 15 cm 2 , the average force exerted on the surface is
A 1.25 × \times 10 −-6 N
B 2.50 × \times 10 −-6 N
C 1.20 × \times 10 −-6 N
D 3.0 × \times 10 −-6 N
Correct Answer
Option B
Solution

For a perfectly reflecting surface, the average force exerted on the surface is F =

2IAc{{2IA} \over c}

=

2×25×104×15×10−43×108{{2 \times 25 \times {{10}^4} \times 15 \times {{10}^{ - 4}}} \over {3 \times {{10}^8}}}

= 2.50 ×\times 10 −-6 N

Q18
An electromagnetic wave of frequency υ=3.0\upsilon = 3.0 MHz passes from vaccum into a dielectric medium with relative permittivity εr{{\varepsilon _r}} = 4.0. Then
A Wavelength is doubled and frequency becomes half.
B Wavelength is halved and frequency remains unchanged.
C Wavelength and frequency both remain unchanged.
D Wavelength is doubled and frequency unchanged.
Correct Answer
Option B
Solution

Velocity of electromagnetic wave in vacuum c =

1μ0ε0{1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}

=

νλ\nu \lambda

vacuum .....(1) Velocity of electromagnetic wave in the medium v medium =

1μ0μrε0εr{1 \over {\sqrt {{\mu _0}{\mu _r}{\varepsilon _0}{\varepsilon _r}} }}

=

cμrεr{c \over {\sqrt {{\mu _r}{\varepsilon _r}} }}

For dielectric medium, μ\mu r = 1 ∴\therefore v medium =

cεr{c \over {\sqrt {{\varepsilon _r}} }}

=

c4=c2{c \over {\sqrt 4 }} = {c \over 2}

........(2) Wavelength of the wave in medium λ\lambda medium =

vmediumν{{{v_{medium}}} \over \nu }

=

c2ν{c \over {2\nu }}

=

λvaccum2{{{\lambda _{vaccum}}} \over 2}
Q19
The condition under which a microwave oven heats up a food item containing water molecules most efficiently is
A Microwaves are heat waves, so always produce heating.
B Infra-red waves produce heating in a microwave oven.
C The frequency of the microwaves must match the resonant frequency of the water molecules.
D The frequency of the microwaves has no relation with natural frequency of water molecules.
Correct Answer
Option C
Solution

The natural frequency of vibration of the water molecules in the food item should be same as that of the frequency of microwaves.

Q20
The ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagatting in vacuum is equal to
A the speed of light in vacuum
B reciprocal of speed of light in vacuum
C the ratio of magnetic permeability to the electric susceptibility of vacuum
D unity
Correct Answer
Option B
Solution

If the speed of light c then E 0 = B 0 c ⇒\Rightarrow

B0E0=1c{{{B_0}} \over {{E_0}}} = {1 \over c}

Hence ratio of amplitude of magnetic field to amplitude of electric field for electromagnetic wave in a vacuum is reciprocal of speed of light.

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