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

12th Grade > Physics

ELECTROMAGNETIC WAVES AND INDUCTION MCQs

Electromagnetic Waves, Electromagnetic Induction, Waves On A String

Total Questions : 73 | Page 7 of 8 pages

Question 61. A tuning fork of frequency 480 Hz is used to vibrate a sono-meter wire having natural frequency 240 Hz. The wire will vibrate with a frequency of

240 Hz
480 Hz
720 Hz
will not vibrate

Discuss Question

Answer: Option B. -> 480 Hz
:
B
This is not the case of free oscillations. This is forced oscillation where natural frequency is 240 Hz and driver frequency is 480 Hz. It vibrates with driver frequency i.e., 480 Hz, which corresponds to the second mode of vibration of the wire

Question 62. Consider two waves passing through the same string. Principle of superposition for displacement says that the net displacement of a particle on the string is sum of the displacements produced by the two waves individually. Suppose we state similar principles for the net velocity of the particle and the net kinetic energy of the particle. Such a principle will be valid for

Both the velocity and the kinetic energy
The velocity but not for the kinetic energy
The kinetic energy but not for the velocity
Neither the velocity nor the kinetic energy.

Discuss Question

Answer: Option B. -> The velocity but not for the kinetic energy
:
B
According to principle of super position

{⃗ y}_{n e t} =_{1} +_{2}

Here

y_{1} & y_{2}

are particle displacement vectors
differentiating with respect to time.

\frac{d {⃗ y}_{n e t}}{d t} = \frac{d_{1}}{d t} + \frac{d_{2}}{d t}

{⃗ v}_{n e t} = {⃗ v}_{1} + {⃗ v}_{2}

Here velocities are vectors.
So yes the superposition law is applicable to velocities.
Now let us square both sides

(v_{n e t})^{2} = ({⃗ v}_{1} + {⃗ v}_{2})^{2}

v_{n e t}^{2} = {v_{1}}^{2} + {v_{2}}^{2} + 2 v_{1} v_{2}

(all are magnitudes)
So clearly the law of super position is not applicable for net kinetic energy.

Question 63. A wave pulse is travelling on a string with a speed v towards the positive X-axis. The shape of the string at t = 0 is given by g(x) =

A s i n (\frac{x}{a})

where A and a are constants. Write the equation of the wave for a general time t, if the wave speed is v.

g(x)=A sin(x+vta)
g(x)=A sin(xa)
g(x)=A sin(x−vta)
g(x)=A sin(xt+va)

Discuss Question

Answer: Option C. -> g(x)=A sin(x−vta)
:
C
Given shape of the wave at t=0

g (x) = A s i n (\frac{x}{a})

let's say the wave looks like

A Wave Pulse Is Travelling On A String With A Speed V Toward...

It's wave speed is v. After time t, the wave would have moved vt.
Now if at t = 0
The equation is

g (x) = A s i n (\frac{x}{a})

This wave was at x = x, t= t will be same as the wave at x= x-vt at t = 0
So at t = 0 g(x) = A sin

\frac{x - v t}{a}

Alternate Solution:- Given g(x) = A sin

(\frac{x}{a})

General equation A sin (kx -

ω

Question 64. A steel wire of mass 4.0 g and length 80 cm is fixed at the two ends. The tension in the wire is 50N. The frequency of the fourth harmonic of the fundamental is

0.25 Hz
25 Hz
250 Hz
2500 Hz

Discuss Question

Answer: Option C. -> 250 Hz
:
C
Given mass

= 4 g = 4 \times 10^{- 3} k g

length = 80 cm = 0.8 m

μ = \frac{m a s s}{l e n g t h} = \frac{4 \times 10^{- 3}}{0.8} = 0.5 \times 10^{- 2}

given T = 50 N
Frequency of nth harmonic is given by

f_{n} = \frac{n v}{2 L}

v = \sqrt{\frac{T}{μ}} = \sqrt{\frac{50}{0.5 \times 10^{- 2}}}

= \sqrt{10^{4}} = 10^{2} m s^{- 1}

f_{4} = \frac{4 \times 10^{2}}{2 \times 0.8}

= 0.25 \times 10^{3} = 250 H z

Question 65. The displacement of a particle of a string carrying a travelling wave is given by y= (3.0 cm) sin 6.28(0.50x - 50 t),
where x is in centimeter and t in second. Find
(i) the amplitude (p) 50
(ii) the wavelength (q) 3
(iii) the frequency(r) 2
(iv) the speed (s) 100

(i) - (r); (ii) - (p); (iii) - (s); (iv) - (q)
(i) - (q); (ii) - (r); (iii) - (p); (iv) - (s)
(i) - (p); (ii) - (q); (iii) - (r); (iv) - (s)
(i) - (s); (ii) - (r); (iii) - (q); (iv) - (p)

Discuss Question

Answer: Option B. -> (i) - (q); (ii) - (r); (iii) - (p); (iv) - (s)
:
B
The standard wave equation for a wave is

y = A s i n (k x - ω t)

Given equation

y = 3 s i n 6.28 (0.5 x - 50 t)

\Rightarrow y = 3 s i n (6.28 \times 0.5 x - 6.28 \times 50 t)

Comparing we get
amplitude A = 3 cm
Right here, we can see that B is the only possible option, but we may as well calculate the other values too.
Wave number

k = \frac{2 π}{λ} = 6.28 \times 0.5

\Rightarrow λ = \frac{2 π}{k} = 2 c m

Angular frequency

ω = 6.28 \times 50

Frequency

f = \frac{1}{T} = \frac{ω}{2 π} = \frac{6.28 \times 50}{2 \times 3.14}

f = 50 H z

wave speed =

\frac{λ}{T} = 2 \times 50 = 100 \frac{c m}{s}

Question 66. 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 67. 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/m2
36.5 W/m2
46.7 W/m2
765 W/m2

Discuss Question

Answer: Option A. -> 26.5 W/m2
:
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 68. A wave is propagating in a medium of electric dielectric constant 2 and relative magnetic permeability 50. The wave impedance of such a medium is

5 Ω
376.6 Ω
1883 Ω
3776 Ω

Discuss Question

Answer: Option C. -> 1883 Ω
:
C
Wave impedance

Z = \sqrt{\frac{μ_{r}}{ϵ_{r}}} \times \sqrt{\frac{μ_{0}}{ϵ_{0}}}

= \sqrt{\frac{50}{2}} \times 376.6 = 1883 Ω

Question 69. A lamp emits monochromatic green light uniformly in all directions. The lamp is 3% efficient in converting electrical power to electromagnetic waves and consumes 100W of power. The amplitude of the electric field associated with the electromagnetic radiation at a distance of 10m from the lamp will be

1.34 V/m
2.68 V/m
5.36 V/m
9.37 V/m

Discuss Question

Answer: Option A. -> 1.34 V/m
:
A

S_{a v} = \frac{1}{2} ϵ_{0} c E_{0}^{2} = \frac{P}{4 π R^{2}}

\Rightarrow E_{0} = \sqrt{\frac{P}{2 π R^{2} ϵ_{0} C}}

= \sqrt{\frac{3}{2 \times 3.14 \times 100 \times 8.85 \times 10^{- 12} \times 3 \times 10^{8}}}

=1.34 V/m

Question 70. 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