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This mock test of Waves NAT Level - 1 for IIT JAM helps you for every IIT JAM entrance exam.
This contains 10 Multiple Choice Questions for IIT JAM Waves NAT Level - 1 (mcq) to study with solutions a complete question bank.
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*Answer can only contain numeric values

QUESTION: 1

Plane harmonic waves of frequency 500 Hz are produced in air with displacement amplitude 1 x 10^{-3} cm. Deduce energy flux in J/m^{2}) in the wave (density of air =1.29 gm/lit, speed of sound in air = 340 m/s).

Solution:

The correct answer is: 0.22

*Answer can only contain numeric values

QUESTION: 2

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

Solution:

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/m^{3}

Pressure amplitude

= 2 x 3.14 x 10^{-5} x 500 x 340 x 1.29

= 13.8 N/m^{2}

The correct answer is: 13.8

*Answer can only contain numeric values

QUESTION: 3

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?

Solution:

∴ 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

*Answer can only contain numeric values

QUESTION: 4

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 ?

Solution:

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

*Answer can only contain numeric values

QUESTION: 5

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?

Solution:

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** = 0.2**m**

The correct answer is: 0.2

*Answer can only contain numeric values

QUESTION: 6

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 :

Solution:

sin kx_{1}, or sin kx_{2} is not zero.

Therefore, neither of x_{1} or x_{2} is a node

The correct answer is: 0.857

*Answer can only contain numeric values

QUESTION: 7

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 ).

Solution:

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

*Answer can only contain numeric values

QUESTION: 8

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 :

Solution:

Wave number

or

The correct answer is: 15.7

*Answer can only contain numeric values

QUESTION: 9

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 ?

Solution:

*Answer can only contain numeric values

QUESTION: 10

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 :

Solution:

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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