Waves Test

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Waves Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

A water wave in a shallow tank passes through a gap in a barrier. What happens to the speed and what happens to the wavelength of the wave as it passes through the gap?

A
speed - decreases ; wavelength - decreases

B
speed - decreases ; wavelength - remains constant

C
speed - remains constant ; wavelength - decreases

D
speed - remains constant ; wavelength - remains constant

Solution

Assuming that the width of the gap is comparable to the wavelength of the wave. this is a case of diffraction in which speed and wavelength of the wave will remain unaltered
Hence option D is correct
Question 2
1 / -0

Wavelength of light of frequency $$100\ Hz$$

A
$$2 \times 10^6\ m$$

B
$$3 \times 10^6\ m$$

C
$$4 \times 10^6\ m$$

D
$$5 \times 10^6\ m$$

Solution

$$\lambda=\dfrac{c}{v}$$
$$=\dfrac{3 \times 10^8}{100}=3 \times 10^6\ m$$
Question 3
1 / -0

In how many parts radio wave frequency is divided

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

Radio waves al have the same basic structure with wavelength and amplitude being the only things that change when modulated. The main two parts of a radio wave are crest and trough. (Option A)
Question 4
1 / -0

A source of sound of frequency 600 Hz is placed inside water. The speed in water is 1500 m/s and in air it is 300 m/s. The frequency of sound recorded by an observer standing in air is

A
200 Hz

B
3000 Hz

C
120 Hz

D
600 Hz

Solution

The frequency is a characteristic of source. It is independent of the medium . Hence, the correct option is (d)
Question 5
1 / -0

The principle of superposition of waves can not be used to explain wave phenomena like

A
Interference

B
Diffraction

C
Stationary Waves and beats

D
Dispersion

Solution

The principle of superposition of waves is used to find the resultant displacement, which is used in interference and diffraction to find the change at edges.
Stationary waves are obtained by reflection of a wave at a solid surface. The resultant amplitude can be found out using superposition principle.
Dispersion of waves is a phenomenon which depends on nature of matter and hence superposition can't be used to find resultant displacement.
Question 6
1 / -0

Transverse waves are produced in a long string by attaching its free end to a vibrating tuning fork. Figure shows the shape of a part of the string. The points in phase are

A
A and D

B
B and E

C
C and F

D
A and G

Solution

Points in phase will have a phase difference of $$0 \ or \ 2\pi $$
$$\Rightarrow $$ A and G are the points in same phase.
Question 7
1 / -0

Human audible frequency range is 20Hz to 20KHz. If velocity of sound in air is 340m/s, the minimum wavelength of audible sound wave is

A
0.17mm

B
1.77mm

C
17mm

D
170mm

Solution

Using the well known relation between wavelength frequency and velocity of a wave,

V $$= \nu \lambda $$

$$ \dfrac{340}{20} $$ > $$ \lambda > \dfrac{340}{20000} $$

Thus,

$$ \lambda = [0.017,17] $$ in meters
Question 8
1 / -0

Two light waves are represented by $$y_{1} =3\sin\omega t$$ and $$y_{2}=4\sin( \omega t+\dfrac{\pi }{3} )$$. The resultant amplitude due to interference will be

A
$$\sqrt{21}$$

B
$$\sqrt{26}$$

C
$$\sqrt{37}$$

D
$$\sqrt{41}$$

Solution

The resultant wave $$y=y_{1}+y_{2}$$
$$y=3\sin \omega t+4\sin( \omega t+\pi /3)$$
$$=3\sin \omega t+2\sin \omega t+2\sqrt{3}\cos \omega t$$
$$=5\sin \omega t+2\sqrt{3}\cos \omega t$$

Amplitude of the resultant wave is given by $$=\sqrt{(5)^{2}+(2\sqrt{3})^{2}}$$
$$=\sqrt{25+12}$$
$$=\sqrt{37}$$
Question 9
1 / -0

During propagation of longitudinal plane wave in a medium, two particles separated by a distance equivalent to one wavelength at an instant will be/have

A
In phase, same displacement

B
In phase, different displacement

C
Different phase, same displacement

D
Different phase, different displacement

Solution

The wave pattern replicates after every wavelength seperation so the particle seperated by the distances of a wavelength has same displacement and are in same phase.
Question 10
1 / -0

The frequency of vibration of a rod is $$200Hz$$. If the velocity of sound in air is $$340m/s$$, the wave length of the sound produced is

A
$$1.7m$$

B
$$6.8 m$$

C
$$3.4 m$$

D
$$13.6 m$$

Solution

$$V=f\lambda$$ where $$V =$$ velocity, $$\lambda$$ = wavelength and $$f =$$ frequency

$$\lambda=\dfrac{V}{f}$$$$=\dfrac{340}{200}$$

$$\lambda=1.7 m$$

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Re-Attempt Weekly Quiz Competition

Waves Test - 21

Result

Waves Test - 21

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SHARING IS CARING

Question 1

speed - decreases ; wavelength - decreases

speed - decreases ; wavelength - remains constant

speed - remains constant ; wavelength - decreases

speed - remains constant ; wavelength - remains constant

Solution

Question 2

$$2 \times 10^6\ m$$

$$3 \times 10^6\ m$$

$$4 \times 10^6\ m$$

$$5 \times 10^6\ m$$

Solution

Question 3

$$2$$

$$3$$

$$4$$

$$5$$

Solution

Question 4

200 Hz

3000 Hz

120 Hz

600 Hz

Solution

Question 5

Interference

Diffraction

Stationary Waves and beats

Dispersion

Solution

Question 6

A and D

B and E

C and F

A and G

Solution

Question 7

0.17mm

1.77mm

17mm

170mm

Solution

Question 8

$$\sqrt{21}$$

$$\sqrt{26}$$

$$\sqrt{37}$$

$$\sqrt{41}$$

Solution

Question 9

In phase, same displacement

In phase, different displacement

Different phase, same displacement

Different phase, different displacement

Solution

Question 10

$$1.7m$$

$$6.8 m$$

$$3.4 m$$

$$13.6 m$$

Solution

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