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 2 of 8 pages

Question 11. Two waves of equal amplitude A, and equal frequency travel in the same direction in a medium. The amplitude of the resultant wave is

0
A
2A
between 0 and 2A

Discuss Question

Answer: Option D. -> between 0 and 2A
:
D
When two waves of amplitude A interfere the maximum amplitude is 2A and minimum is zero. The resultant amplitude in any general case is between 0 and 2A depending upon the phase difference.

Question 12. Two pulses of identical shape overlap such that the displacement of the rope is momentarily zero at all points, what happens to the energy at this time.

Energy becomes zero
The total energy is in form of potential energy
The total energy is in form of kinetic energy
None of these

Discuss Question

Answer: Option C. -> The total energy is in form of kinetic energy
:
C
When the string is such that the net displacement because of both the pulses is not zero. The total energy is distributed in kinetic and potential energy. When string has no displacement, there is no potential energy. Potential energy basically arises due to stretching of string which would be zero in such case and as we know all energy is conserved. The total energy is only kinetic!

Question 13. Two wave pulses travel in opposite directions on a string and approach each other. The shape of one pulse is inverted with respect to the other.

The pulses will collide with each other and vanish after collision.
The pulses will reflect from each other, i.e., the pulse going towards right will finally move towards left and vice versa.
The pulses will pass through each other but their shapes will be modified.
The pulses will pass through each other without any change in their shapes after the overlap.

Discuss Question

Answer: Option D. -> The pulses will pass through each other without any change in their shapes after the overlap.
:
D
The shapes of wave pulses after the overlap are identical to their original shapes.
After overlap each pulse travels just as it did before.

Question 14. Both the strings, shown in figure, are made of same material and have same cross-section. The pulleys are light. The wave speed of a transverse wave in the string

A B

v_{1}

and in

C D

it is

v_{2}

. Then

\frac{v_{1}}{v_{2}}

Both The Strings, Shown In Figure, Are Made Of Same Material...

1
2
√2
1√2

Discuss Question

Answer: Option D. -> 1√2
:
D

Tension in string

A B

T

Tension in string

C D

2 T

V

\propto

\sqrt{T}

\Rightarrow

\frac{V_{1}}{V_{2}}

\sqrt{\frac{T}{2 T}}

\frac{1}{\sqrt{2}}

Question 15. Doing the Mexican wave' in a sports stadium is when people stand and sit in a continuuous order. So as to make a wave [as looked from outside] people don't leave their respective seat and just move up and down. The wave, transverse in nature appears to be moving. Is this example of a real physical wave?

True
False
1.50 v
0.75 v

Discuss Question

Answer: Option B. -> False
:
B
A wave or a 'real' physics wave transports both momentum and energy which of course this wave doesn't do. [we are not talking about sports enthusiasm] This is not a real physics wave!

Question 16. Two strings

A

and

B

, made of same material, are stretched by same tension. The radius of string

A

is double the radius of

B

. A transverse wave travels on

A

with speed

v_{A}

and on

B

with speed

v_{B}

. The ratio

(\frac{v_{A}}{v_{B}})

Discuss Question

Answer: Option A. -> 12
:
A
Two strings

A

and

B

, made of same material, are stretched by same tension. The radius of string

A

is double of the radius of

B

.A transverse wave travels on

A

with speed

v_{A}

and on

B

with speed

v_{B}

.The ratio

(\frac{v_{A}}{v_{B}})

v

\sqrt{\frac{T}{μ}}

; as

T

is constant,

v

\propto

\frac{1}{\sqrt{μ}}

But

μ

\frac{m a s s}{l e n g t h}

π r^{2} ρ

\to

μ

\propto

r^{2}

\Rightarrow

v

\propto

\frac{1}{r}

\to

\frac{v_{A}}{v_{B}}

\frac{r_{B}}{r_{A}}

\frac{r}{2 r}

\frac{1}{2}

Question 17. The equation

y = A s i n^{2} (k x - ω t)

represents a wave motion with

amplitude A, frequency ω2π
amplitude A2, frequency ωπ
amplitude 2A, frequencyω4π
does not represent a wave motion

Discuss Question

Answer: Option B. -> amplitude A2, frequency ωπ
:
B

y = A s i n^{2} (k x - ω t)

This is not ideal equation of wave we don't have

s i n^{2} θ

in ideal equation so let's remove the square.
Use trigonometry

c o s 2 θ = c o s^{2} θ - s i n^{2} θ

= 1 - 2 s i n^{2} θ

s i n^{2} θ = \frac{1 - c o s 2 θ}{2}

y = A \frac{{1 - c o s 2 (k x - w t)^{2}}}{2}

y = \frac{A}{2} - \frac{A}{2} c o s (2 k x - 2 ω t)

y^{'} = (y - \frac{A}{2}) = - \frac{A}{2} c o s (2 k x - 2 ω t)

So amplitude =

- \frac{A}{2}

angular frequency

(ω^{'}) = 2 ω

frequency(f)=

\frac{ω}{2 π} = \frac{2 ω}{2 π} = \frac{ω}{π}

Question 18. Two waves, each having a frequency of 100 H

_{z}

and a wavelength of 2.0 cm, are travelling in the same direction on a string. What is the phase difference between the waves (a) if the second wave was produced 0.015 s later than the first one at the same place, what would be the resultant amplitude? [Given individual amplitudes = 2.0 mm]

4 mm
2 mm
2√2mm
zero

Discuss Question

Answer: Option D. -> zero
:
D
Phase diagram method can be used here but for that we need to calculate the phase difference.

Δ ϕ = 2 π \frac{Δ t}{T}

Here

Δ t = 0.015 s

T_{1}

which is the time period =

\frac{1}{f}

T = \frac{1}{100}

Δ ϕ = 2 π \frac{0.015}{\frac{1}{100}} = 3 π = 2 π + π

2 π

is the period, the effective phase difference is

π

now let's draw
the phase diagram

Two Waves, Each Having A Frequency Of 100 Hz And A Wavelengt...

So we can see that net amplitude would be zero.

Question 19. What is the equation of a sinusoidal wave moving in the positive x direction with wave length 6 m and time period 0.5s. The amplitude of the wave is 5 m.

y=6sin(2π5x−4πt)
y=5sin(π3x−4πt)
y=5sin(x−t)
None of these

Discuss Question

Answer: Option B. -> y=5sin(π3x−4πt)
:
B
here

λ = 6 m

∴ k = \frac{2 π}{6} = \frac{π}{3}

ω = \frac{2 π}{T} = \frac{2 π}{0.5} = 4 π

given,amplitude =5m

y = 5 s i n (\frac{π}{3} x - 4 π t)

[the standard equation is y=Asin(kx-

ω

t)]

Question 20. A laser beam can be focussed on an area equal to the square of its wavelength. A He-Ne laser radiates energy at the rate of 1mW and its wavelength is 632.8 nm. The intensity of focussed beam will be