The vibrations of a string of length 60 cm fixed at both the ends

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 Multiple Choice QuestionsMultiple Choice Questions

21.

A wave of frequency 500 Hz travels with a speed of 360 ms-1. The distance between two nearest points which are 60° out of phase is 

  • 12 cm

  • 18 cm

  • 50 cm

  • 24 cm


22.

The apparent frequency observed by a moving observer away from a stationary source is 20% less than the actual frequency. If the velocity of sound in air is 330 ms-1, then the velocity of the observer is

  • 660 ms-1

  • 330 ms-1

  • 66 ms-1

  • 33 ms-1


23.

A string under tension of 129.6 N produces 10 beats/second, when it vibrates along with a tuning fork. When the tension in the string is increased to 160 N, it vibtrates in unison with the tuning fork. Then, frequency of the tuning fork is

  • 100 Hz

  • 110 Hz

  • 90 Hz

  • 220 Hz


24.

The ionospheric layer acts as a reflector for the frequency range

  • 1 kHz to 10 kHz

  • 3 to 30 MHz

  • 3 to 30 kHz

  • 100 kHz to 1 MHz


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25.

Two travelling waves, y1 = A sin [k(x + ct)] and y2 = A sin [k(x- ct)] are superposed on a string. The distance between adjacent antinodes is

  • ctπ

  • ct2π

  • πk

  • kπ


26.

If a stretched wire is vibrating in the second overtone, then the number of nodes and antinodes between the ends of the string are respectively

  • 2 and 2

  • 1 and 2

  • 3 and 4

  • 2 and 3


27.

Pick out the correct statement in the following with reference to stationary wave pattern

  • In a tube closed at one end, all the harmonics are present

  • In a tube open at one end, only even harmonics are present

  • The distance between successive nodes is equal to the wavelength

  • In a stretched string, the first overtone is the same as the second harmonic


28.

The bulk modulus of a liquid of density 8000 kgm-3 is 2 x 109 Nm-2. The speed of sound in that liquid is (in ms-1)

  • 200

  • 250

  • 500

  • 350


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29.

The vibrations of a string of length 60 cm fixed at both the ends are represented by the equation y = 2 sin 4πx15 cos 96πt, where x and y are in cm. The maximum number of loops that can be formed in it, is

  • 6

  • 16

  • 5

  • 15


B.

16

Given, y = 2 sin 4πx15 cos 96πt

and      l = 60 cm

We know that, f = P2l × v

where, P = number of loops

96π2π = P2 × 60 × ωK96π2π = P2 × 60 × 96π4π15  P = 16


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30.

The pressure variations in the propagation of sound waves are

  • isobaric

  • isochoric

  • isobaric and isochoric

  • adiabatic


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