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IIT JAM Exam  >  IIT JAM Tests  >  Waves NAT Level - 1 - IIT JAM MCQ

Waves NAT Level - 1 - IIT JAM MCQ


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10 Questions MCQ Test - Waves NAT Level - 1

Waves NAT Level - 1 for IIT JAM 2024 is part of IIT JAM preparation. The Waves NAT Level - 1 questions and answers have been prepared according to the IIT JAM exam syllabus.The Waves NAT Level - 1 MCQs are made for IIT JAM 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Waves NAT Level - 1 below.
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Waves NAT Level - 1 - Question 1
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Plane harmonic waves of frequency 500 Hz are produced in air with displacement amplitude 1 x 10-3 cm. Deduce energy flux in J/m2) in the wave (density of air =1.29 gm/lit, speed of sound in air = 340 m/s).


Detailed Solution for Waves NAT Level - 1 - Question 1


The correct answer is: 0.22

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Waves NAT Level - 1 - Question 2
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Plane harmonic waves of frequency 500 Hz are produced in air with displacement amplitude 1 x 10-3 cm.  Deduce pressure amplitude (in N/m2).  (density of air = 1.29 gm/lit, speed of sound in air = 340m/s)


Detailed Solution for Waves NAT Level - 1 - Question 2

Given A = 1 x 10-3 cm = 10-5 m, v = 500 Hz, v = 340 m/s, p = 1.29 g/lit = 1.29 kg/m3
Pressure amplitude

= 2 x 3.14 x 10-5 x 500 x 340 x 1.29
= 13.8 N/m2 
The correct answer is: 13.8

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Waves NAT Level - 1 - Question 3
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If the frequency of a tuning fork is 400 Hz and the velocity of sound in air is 320 m/s, find how far does the sound travel (in m) while the fork completes 30 vibrations?


Detailed Solution for Waves NAT Level - 1 - Question 3


∴ Distance travelled by the wave when the fork completes 1 vibration = 0.8 m.
So distance travelled by the wave when the fork completes 30 vibrations.
= 0.8 x 30
= 24m
The correct answer is: 24

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Waves NAT Level - 1 - Question 4
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The vibrations of a string of length 60 cm fixed at both ends are represented by the equation

where x and y are in cm and t in seconds. What is the maximum displacement of point x = 5cm ?

Detailed Solution for Waves NAT Level - 1 - Question 4

The given equation is :

This can be written as :

This shows that a = 2 cm, λ =30 cm and ω = 1440 cm/s
The maximum displacement is given by :

For x = 5 cm

The correct answer is: 3.464

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Waves NAT Level - 1 - Question 5
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A transverse harmonic wave of amplitude 0.02 m is generated at one end (x = 0) along horizontal string by a tuning fork of frequency 500 Hz. At a given instant of time, the displacement of the particle at x = 0.1 m is 0.005 m and that of the particle at x = 0.2 m is 0.005 m. Calculate the wavelength of the wave?

Detailed Solution for Waves NAT Level - 1 - Question 5

The general equation of this wave is given by

Here A = 0.02 m
Again, when x = 0.1m, y = -0.005 m

or


Again x = 0.2m, y = 0.005m





or λ = 0.2m
The correct answer is: 0.2

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Waves NAT Level - 1 - Question 6
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Equations of a stationary and a travelling waves are as follows :

The phase difference between two points  respectively for the two waves. The ratio  is :

Detailed Solution for Waves NAT Level - 1 - Question 6


sin kx1, or sin kx2 is not zero.
Therefore, neither of x1 or x2 is a node


The correct answer is: 0.857

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Waves NAT Level - 1 - Question 7
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A simple harmonic wave travelling x-axis is given by y = 5 sin 2π ( 0.2t - 0.5x) [x is in m and t in s]. Calculate the amplitude (in m ).

Detailed Solution for Waves NAT Level - 1 - Question 7

Here
But the standard progressive wave equation is :

Comparing equation (i) and (ii), we have

A = 5cm

The correct answer is: 5/100 

= 0.05m

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Waves NAT Level - 1 - Question 8
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The equation for the vibration of a string fixed at both ends vibrating in its third harmonic is given by :
 The length of the string is :

Detailed Solution for Waves NAT Level - 1 - Question 8


Wave number


or

The correct answer is: 15.7

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Waves NAT Level - 1 - Question 9
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A wave of frequency 400 Hz is travelling with a velocity 800 m/s. How far are two points situated whose displacement differs in phase by  ?

Detailed Solution for Waves NAT Level - 1 - Question 9

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Waves NAT Level - 1 - Question 10
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The amplitude of wave disturbance propagating in positive x-axis is given by  at t = 0 andat t= 2s, where x and y are in meters. The shape of the disturbance does not change during the propagation. The velocity (in m/s) of the wave is :

Detailed Solution for Waves NAT Level - 1 - Question 10

The given pulse is of the form

where v is the wave velocity 
Given equation is

Comparing eq. (1) and (2), we get

The correct answer is: 0.5

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