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NEET Exam  >  NEET Tests  >  Physics Class 11  >  Test: Waves - NEET MCQ

Test: Waves - NEET MCQ


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9 Questions MCQ Test Physics Class 11 - Test: Waves

Test: Waves for NEET 2024 is part of Physics Class 11 preparation. The Test: Waves questions and answers have been prepared according to the NEET exam syllabus.The Test: Waves MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Waves below.
Solutions of Test: Waves questions in English are available as part of our Physics Class 11 for NEET & Test: Waves solutions in Hindi for Physics Class 11 course. Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free. Attempt Test: Waves | 9 questions in 60 minutes | Mock test for NEET preparation | Free important questions MCQ to study Physics Class 11 for NEET Exam | Download free PDF with solutions
Test: Waves - Question 1
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The path difference between two wavefronts emitted by coherent sources of wavelength 5460 Å is 2.1 micron . The phase difference between the wavefronts at that point is _

Detailed Solution for Test: Waves - Question 1

Test: Waves - Question 2
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 Monochromatic light is that light in which

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Test: Waves - Question 3
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The path difference between two waves 
y1= A1 sin wt and y2= A2 cos (wt + f) will be

Detailed Solution for Test: Waves - Question 3

y1= A1 sin wt
y2= A2 cos (wt + f)
Now, y2 = A2 sin (wt + f + pi/2)
So, phase difference = f + pi/2
Path Difference = (λ/2pi)(f + pi/2)
Hence B.

Test: Waves - Question 4
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The resultant amplitude in interference with two coherent source depends upon _
Detailed Solution for Test: Waves - Question 4

Two sources are said to be coherent if there always exists a constant phase difference between the waves emitted by these sources. But when the sources are coherent, then the resultant intensity of light at a point will remain constant and so interference fringes will remain stationary.

Test: Waves - Question 5
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The necessary condition for phenomenon of interference to occur is
Detailed Solution for Test: Waves - Question 5

The necessary condition for phenomenon of interference to occur are:
1. There should be two coherent sources.
2. The frequency and amplitude of both the waves should be same.
3. The propagation of waves should be simultaneously and in same direction.
These are the conditions, no explanation.

*Answer can only contain numeric values
Test: Waves - Question 6
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A source (S) of sound has frequency 240 Hz. When the observer (O) and the source move towards each other at a speed v with respect to the ground (as shown in Case 1 in the figure), the observer measures the frequency of the sound to be 288 Hz. However, when the observer and the source move away from each other at the same speed v with respect to the ground (as shown in Case 2 in the figure), the observer measures the frequency of sound to be n Hz. The value of n is _______.

Detailed Solution for Test: Waves - Question 6

Calculation:

For Case 1 :

For Case 2 :

From (i) and (ii)

288 fapp = 240 × 240

fapp = 200 Hz

Test: Waves - Question 7
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An earthquake generates both transverse (S) and longitudinal (P) sound waves in the earth. The speed of S waves is about 4.5 km/s and that of P waves is about 8.0 km/s. A seismograph records P and S waves from an earthquake. The first P wave arrives 4.0 min before the first S wave. The epicenter of the earthquake is located at a distance about: 
Detailed Solution for Test: Waves - Question 7

Let the distance to the epicenter of the earthquake be d km. The time taken by the P waves to travel this distance is tP​, and the time taken by the S waves to travel this distance is tS​.

We know that:

tS ​− tP​ = 4minutes = 240seconds.

The relationship between speed, distance, and time is:

t = d/v​,

where vvv is the velocity and ttt is the time.

For the P waves:

tP ​= d/8.0​.

For the S waves:

tS ​= d​/4.5

From the time difference:

tS − tP = 240

Substitute the expressions for tS​ and tP​:

Find a common denominator and solve:

Simplify:

Thus, the distance to the epicenter is approximately 2500 km.

Test: Waves - Question 8
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An earthquake generates both transverse (S) and longitudinal (P) sound waves in the earth. The speed of S waves is about 4.5 km/s and that of P waves is about 8.0 km/s. A seismograph records P and S waves from an earthquake. The first P wave arrives 4.0 min before the first S wave. The epicenter of the earthquake is located at a distance about: 
Detailed Solution for Test: Waves - Question 8

Let the distance to the epicenter of the earthquake be d km. The time taken by the P waves to travel this distance is tP​, and the time taken by the S waves to travel this distance is tS​.

We know that:

tS ​− tP​ = 4minutes = 240seconds.

The relationship between speed, distance, and time is:

t = d/v​,

where vvv is the velocity and ttt is the time.

For the P waves:

tP ​= d/8.0​.

For the S waves:

tS ​= d​/4.5

From the time difference:

tS − tP = 240

Substitute the expressions for tS​ and tP​:

Find a common denominator and solve:

Simplify:

Thus, the distance to the epicenter is approximately 2500 km.

Test: Waves - Question 9
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A source of the sound of frequency 256 Hz is moving rapidly towards a wall with a velocity of 5 m/s. The speed of sound is 330 m/s. If the observer is between the wall and the source, then beats per second heard will be.
Detailed Solution for Test: Waves - Question 9

The observer will hear two sounds, one directly from the source and the other reflected image of sound. Hence, the number of beats heard per second is

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