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 6 of 8 pages

Question 51. Which of these will travel in a wave along with it?

particles
energy
both
none of these

Discuss Question

Answer: Option B. -> energy
:
B
In a wave, energy travels without any physical transfer of particles.

Question 52. Velocity of sound in air is

332 m s^{- 1}

. Its velocity in vacuum will be

> 332ms−1
=332ms−1
None of these

Discuss Question

Answer: Option D. -> None of these
:
D
Sound waves are mechanical waves i.e., they need material medium to propagate. So they will not travel in vacuum at all! Huh!

Question 53. Two sine waves travelling in the same direction through the same region, have equal frequencies, wavelengths and amplitudes. If the amplitude of each wave is 4 mm and the phase difference between the waves is 90

^{\circ}

, what is the resultant amplitude?

8 mm
4√3mm
2√2mm
4√2mm

Discuss Question

Answer: Option D. -> 4√2mm
:
D
Giventhe frequency and wavelength of the wave is same, the phase method of calculating resultant amplitude can be used. Here given is that phase difference

ϕ

\frac{π}{2}

Resultant amplitude

A = \sqrt{a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} c o s ϕ}

Also

a_{1} = a_{2} = 4 m m

A = 4 \sqrt{2} m m

Question 54. The extension in a string, obeying Hook's law, is

x

. The speed of sound in the stretched string is

v

. If the extension in the string is increased to

1.5 x

, the speed of sound will be :

1.22 v
0.61 v
1.50 v
0.75 v

Discuss Question

Answer: Option A. -> 1.22 v
:
A
From Hooke's law
Tension in a string (

T

)

\propto

extension (

x

)
and speed of sound in string

v

\propto

\sqrt{T}

Therefore,

v \propto \sqrt{x}

X

is increased to 1.5 times i.e., speed will increase by

\sqrt{1.5}

times or 1.22 times. Therefore, speed of Sound in new position will be 1.22 v.

Question 55. Two waves passing through a region are represented by

y = 1.0 c m s i n [(3.14 c m^{- 1}) x - (157 s^{- 1}) t]

and

y = 1.5 c m s i n [(1.57 c m^{- 1}) x - (314 s^{- 1}) t]

Find the displacement of the particle at x = 4.5 cm at time t = 5.0 ms.

0.35 cm
(1√2)cm
(1.5√2)cm
- 0.35 cm

Discuss Question

Answer: Option D. -> - 0.35 cm
:
D
According to the principle of superposition, each wave produces its disturbance independent of the other and the resultant disturbance is equal to the vector sum of the individual disturbances. The displacements of the particle at x = 4.5 cm at time t = 5.0 ms due to the two waves are,

y_{1} = (1.0 c m) s i n (3.14 c m^{- 1}) (4.5 c m) - (157 s^{- 1}) (5.0 \times 10^{- 3} s)

= (1.0 c m) s i n [4.5 π - \frac{π}{4}]

= (1.0 c m) s i n [4.5 π - \frac{π}{4}] = \frac{1.0 c m}{\sqrt{2}}

and

y_{2} = (1.5 c m) s i n [(1.57 c m^{- 1}) (4.5 c m) - (314 s^{- 1}) (5.0 \times 10^{- 3} s)]

(1.5 c m) s i n [2.25 π - \frac{π}{2}]

(1.5 c m) s i n [2 π - \frac{π}{4}]

- (1.5 c m) s i n \frac{π}{4} = - \frac{1.5 c m}{\sqrt{2}}

The net displacment is

y = y_{1} + y_{2} = - \frac{0.5 c m}{\sqrt{2}} = - 0.35 c m

Question 56. On December 2006, a great earthquake occurred off the coast of Sumatra and triggered immense waves (Tsunami) that killed some 200,000 people. Satellites observing these waves from space measured 800 km from one wave crest to the next and a period between waves of 1 hr. what was the speed of the wave?

400 km/hr
1600 km/hr
800 km/hr
Data insufficient

Discuss Question

Answer: Option C. -> 800 km/hr
:
C
crest to crest distance is wavelength λ = 800 km
Time period:Time taken for wave to travel a distance of one wavelength = 1 hr
Wave velocity =

\frac{D i s t a n c e}{T i m e} = \frac{1 w a v e l e n g t h (λ)}{T i m e p e r i o d (T)}

\frac{800 k m}{1} = 800 k m / h r

Question 57. A sinusoidal wave propagates along a string in figure (a) and (b). 'y' represents displacement of particle from the mean position. 'x' & 't' have usual meanings. Find:
1. Wavelength, frequency and speed of the wave
2. Maximum velocity and maximum acceleration of the particles

2m, 0.25 Hz, 2 ms−1, 1.5 mms−1, 0.75 mms−2
4m, 0.25 Hz, 10 ms−1, 1.5 πmms−1, 0.75 mms−2
4m, 0.25 Hz, 1 ms−1, 1.5 πmms−1, 0.75 π2mms−2
None of these

Discuss Question

Answer: Option C. -> 4m, 0.25 Hz, 1 ms−1, 1.5 πmms−1, 0.75 π2mms−2
:
C
From the y-t graph we can see the wave repeats itself on interval of 4s. So the time period is 4s. from the y-x graph we see that the distance between the points after which the wave repeats itself is 4m. Hence the wave length is 4m.

f = \frac{1}{T} = \frac{1}{4} S^{- 1} = 0.25 s^{- 1}

v = λ f = 4 \frac{1}{4} = 1 m s^{- 1}

v_{p} m a x = A ω

= A \times (2 π f

= 3 \times \times 2 π \times 0.25

[A=3 from y vst graph]

a_{p} m a x = ω^{2} A

= (2 π \times 0.25)^{2} \times 3

\frac{3}{4} π^{2} = 0.075 π^{2} m m s^{- 2}

Question 58. A wave propagates on a string in the positive x-direction at a velocity v. The shape of the string at

t = t_{0}

is given by

g (x, t_{0})

= A sin

(\frac{x}{a})

Write the wave equation for a general time t.

A sin[xa−v(t−t0)]
A sin[x−v(t−t0)a]
A sin[x+v(t−t0)a]
None of these

Discuss Question

Answer: Option B. -> A sin[x−v(t−t0)a]
:
B
Here given is

g (x, t_{0}) = A s i n (\frac{x}{a})

t = t_{0}

Comparing with the general equation g(x,t)=

A s i n (k x - ω t + ϕ)

We get at

t = t_{0}

k = \frac{1}{a}

also

- ω t_{0} + ϕ = 0

\Rightarrow ϕ = ω t_{0}

We also known that

k = \frac{ω}{v}

\Rightarrow ω = \frac{v}{a}

so we get

g (x, t) = A s i n (\frac{x}{a} - \frac{v t}{a} + \frac{v t_{0}}{a})

g (x, t) = A s i n [\frac{x}{a} - \frac{v t}{a} + \frac{v t_{0}}{a}]

= A s i n (\frac{x - v (t - t_{0})}{a})

Question 59. A string of length 0.4 m and mass

10^{- 2}

kg is tightly clampled at its ends. The tension in the string is 1.6 N. identical wave pulses are produced at one end at equal intervals of time ∆t. The minimum value of ∆t, which allows constructive interference between succesive pulses, is :