Electromagnetic Waves(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

ELECTROMAGNETIC WAVES MCQs

Total Questions : 30 | Page 1 of 3 pages

Question 1.

According to Maxwell’s hypothesis, a changing electric field gives rise to

An e.m.f.
Electric current
Magnetic field
Pressure radiant

Discuss Question

Answer: Option C. -> Magnetic field
:
C

According to the Maxwell’s EM theory, the EM waves propagation contains electric and magnetic field vibration in mutually perpendicular direction. Thus the changing of electric field give rise to magnetic field.

Question 2.

In an electromagnetic wave, the electric and magnetising fields are 100V $m^{- 1}$ and 0.265 $A m^{- 1}$ . The maximum energy flow is

26.5 $W / m^{2}$
36.5 $W / m^{2}$
46.7 $W / m^{2}$
765 $W / m^{2}$

Discuss Question

Answer: Option A. -> 26.5

W / m^{2}

:
A

Here $E_{0} = 100 V / m, B_{0} = 0.265$ A/m.
$∴$ Maximum rate of energy flow $S = E_{0} \times B_{0}$
$= 100 \times .265 = 26.5 \frac{W}{m^{2}}$

Question 3.

The 21 cm radio wave emitted by hydrogen in interstellar space is due to the interaction called the hyperfine interaction is atomic hydrogen. The energy of the emitted wave is nearly

$10^{- 17}$ Joule
1 Joule
$7 \times 10^{- 8}$ Joule
$10^{- 24}$ Joule

Discuss Question

Answer: Option D. ->

10^{- 24}

Joule
:
D

$E = \frac{h c}{λ} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{21 \times 10^{- 2}} = 0.94 \times 10^{- 24} \approx 10^{- 24}$ J

Question 4.

An electromagnetic wave going through vacuum is described by E = $E_{o} s i n (k x - ω t); B = B_{o} s i n (k x - ω t);$ . Which of the following equation is true

$E_{o} k = B_{o} ω$
$E_{o} ω = B_{o} k$
$E_{o} B_{o} = ω k$
None of these

Discuss Question

Answer: Option A. ->

E_{o} k = B_{o} ω

:
A

$\frac{E_{0}}{B_{0}} = C .$ also $k = \frac{2 π}{λ}$ and $w = 2 π v$
These relation gives $E_{0} K = B_{0} w$

Question 5.

TV waves have a wavelength range of 1-10 meter. Their frequency range in MHz is

30-300
3-30
300-3000
3-3000

Discuss Question

Answer: Option A. -> 30-300
:
A

$v = \frac{C}{λ} \Rightarrow v_{1} = \frac{3 \times 10^{8}}{1} = 3 \times 10^{8}$ Hz=300 MHz
and $v_{2} = \frac{3 \times 10^{8}}{10} = 3 \times 10^{7}$ Hz=30 MHz

Question 6.

An LC resonant circuit contains a 400 pF capacitor and a 100 mH inductor. It is set into oscillation coupled to an antenna. The wavelength of the radiated electromagnetic waves is

377 mm
377 metre
377 cm
3.77 cm

Discuss Question

Answer: Option B. -> 377 metre
:
B

$v = \frac{1}{2 π \sqrt{L C}}$ and $λ = \frac{C}{V}$

Question 7.

A radio receiver antenna that is 2 m long is oriented along the direction of the electromagnetic wave and receives a signal of intensity $5 \times 10^{- 16} W / m^{2}$ . The maximum instantaneous potential difference across the two ends of the antenna is

1.23 mV
1.23 mV
1.23 V
12.3 mV

Discuss Question

Answer: Option A. -> 1.23 mV
:
A

$I = \frac{1}{2} ϵ_{0} C E_{0}^{2}$
$\Rightarrow E_{0} = \sqrt{\frac{2 I}{ϵ_{0} C}} = \sqrt{\frac{2 \times 5 \times 10^{- 16}}{8.85}} = 0.61 \times 10^{- 6} \frac{V}{m}$
Alsop $E_{0} = \frac{V_{0}}{d} \Rightarrow V_{0} = E_{0} d = 0.61 \times 10^{- 6} \times 2 = 1.23 μ V$

Question 8.

Television signals broadcast from the moon can be received on the earth while the TV broadcast from Delhi cannot be received at places about 100 km distant from Delhi. This is because

There is no atmosphere around the moon
Of strong gravity effect on TV signals
TV signals travel straight and cannot follow the curvature of the earth
There is atmosphere around the earth

Discuss Question

Answer: Option C. -> TV signals travel straight and cannot follow the curvature of the earth
:
C

Over a long distance, earths’ surface will be curved and since TV signals travel in straight line television broad cast signals cannot be received.

Question 9.

A TV tower has a height of 100 m. The average population density around the tower is 1000 per $k m^{2}$ . The radius of the earth is $6.4 \times 10^{6}$ m. The population covered by the tower is

$2 \times 10^{6}$
$3 \times 10^{6}$
$4 \times 10^{6}$
$6 \times 10^{6}$

Discuss Question

Answer: Option C. ->

4 \times 10^{6}

:
C

Population covered = $2 π h R \times$ Population density
$= 2 π \times 100 \times 6.4 \times 10^{6} \times \frac{1000}{(10^{3})^{2}} = 4 \times 10^{6}$

Question 10.

In an electromagnetic wave, the amplitude of electric field is 1 V/m. the frequency of wave is $5 \times 10^{14}$ Hz. The wave is propagating along z-axis. The average energy density of electric field, in $J o u l e m^{3}$ , will be

$1.1 \times 10^{- 11}$
$2.2 \times 10^{- 12}$
$3.3 \times 10^{- 13}$
$4.4 \times 10^{- 14}$

Discuss Question

Answer: Option B. ->

2.2 \times 10^{- 12}

:
B

Average energy density of electric field is given by
$u_{e} = \frac{1}{2} ϵ_{0} E^{2} = \frac{1}{2} ϵ_{0} {(\frac{E_{0}}{\sqrt{2}})}^{2} = \frac{1}{4} ϵ_{0} E_{0}^{2}$
$= \frac{1}{4} \times 8.85 \times 10^{- 12} (1)^{2} = 2.2 \times 10^{- 12} J / m^{3} .$