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Physics NEET MCQ
Oscillations And Wave Mcq
Quiz 8
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Q.1
A stretched string of length l fixed at both ends can sustain stationary waves of wave length λ given by : [ CPMT 1998,] where n is a whole number
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a)λ=n2 / 2l
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b) λ=l2 / 2n
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c)λ=2l /n
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d)λ=2ln
Explanation
let n be the number of loops in the string thenn ( λ/2)=l or λ=2l/n Answer: (c)
Q.2
If oil of density higher than that of water is used in place of water in a resonance tube, its frequency will... [ MNR 1986]
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a) Increase
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b) Decrease
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c)Remain the same
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d)Depend upon the density of the material of tube
Explanation
Answer: (c)
Q.3
The fundamental frequency of a string stretched with weight of 4kg is 256Hz. The weight required to produce its octave is .. [ MPPMT 1988]
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a)4 kg wt
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b) 12 kg wt
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c)16 kg wt
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d)24 kg wt
Explanation
Answer:(c)
Q.4
An open pipe of length 90cm emitting its second overtone is in unison with an open pipe emitting its third overtone. The length of the open pipe will be .. [ EAMCET 1994]
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a) 1.6 m
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b) 1.2 m
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c) 3.2 m
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d) 2.5 m
Explanation
Answer: (b)
Q.5
The frequency of fundamental note emitted by a string of length l, clamped at both ends is ( v=velocity of sound in string) ..[ CPMT 1993]
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a)v/l
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b) 2v/l
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c)v/2l
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d)v/4l
Explanation
Answer: (c)
Q.6
Two organ pipes both closed at one end, have length l and l+Δl. Neglect end-correction. if the velocity of sound in air is V. ten the number of beats per second is ..[ BHU 1995]
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a) V/ 4l
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b) V/2l
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c)(V/4l2) Δl
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d)(V/2l2)Δl
Explanation
Answer: (c)
Q.7
The amplitude of vibration of any particle in a standing wave, produced along a stretched string depends on [ EAMCET 1991]
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a) Frequency of incident wave
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b) Time period of reflected wave
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c)Location of particle
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d)Time
Explanation
Answer:(c)
Q.8
Which of the following statement is wrong:
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a) In an open pipe the fundamental frequency is v/2L
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b) In a closed pipe the closed end is displacement node
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c) In an open pipe only the odd harmonics of fundamental frequency are present
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d) In a closed pipe the fundamental frequency is v/4L
Explanation
Answer: (c)
Q.9
Air is blown at the mouth of tube ( length=25cm and diameter=3cm) closed at one end. Velocity of sound is 330 m/sec. The sound which is produced will correspond to the frequencies:.. [ CPMT 1986]
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a)330 Hz
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b) Combination of frequencies 330 Hz, 662Hz, 990 Hz, 1320Hz,...
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c)Combination of frequencies 330 Hz, 990Hz, 1650 Hz, 2310Hz,...
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d)Combination of frequencies 330 Hz, 990Hz, 1500 Hz, 1200Hz,...
Explanation
Fundamental frequency of pipe=v/4l=330 /(4×0.25)=330HzOther frequencies will be=3v/4l=990 Hz5v/4l=1650 Hz7v/4l=2310 Hz Answer: (c)
Q.10
The end correction for the vibrations of air column in a tube of circular cross-section will be more if the tube is ..[ CPMT 1985]
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a) Reduced in length
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b) Increased in length
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c)made thinner
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d)Widened
Explanation
Answer: (d)
Q.11
An air-column in a pipe, which is closed at one end, will be in resonance with a vibrating tuning fork of frequency 256Hz, if the length of the column is cm is ( velocity of sound in air is 340 m/s) [ CBSE 1997]
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a) 21.25
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b) 125
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c) 62.5
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d) 33.2
Explanation
fundamental frequency n=v/4l l=v/ 4n l=340 / (4×256)=33.2 cm Answer: (d)
Q.12
A closed organ pipe and an open organ pipe have their first overtones identical in frequency. Their length are in the ratio..[ MPPMT 1987]
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a)1:2
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b) 2:3
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c)3:4
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d)4:5
Explanation
first overtone of closed pipe=3v/4lfirst overtone of open pipe=v/2l 3v/4l1=2v/2l2 l1 / l2=3/4 Answer: (c)
Q.13
The radius, density and tension of a string A are twice the radius, density and tension of another string B. If the length of the strings are equal, the ratio nA / nB of their frequencies of vibration, will be ..[ MPPMT 1990]
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a) 1
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b) 2
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c)1/2
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d)1/4
Explanation
Given density ρA=2ρBradius rA=2 rB Tension TA=2TB Thus Answer: (c)
Q.14
The sonometer wire is vibrating in the second overtone. We may say that there are
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a) Two nodes and two antinodes
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b) One node and two antinodes
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c)Four nodes and three antinodes
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d)Three nodes and three antinodes
Explanation
Answer:(c)
Q.15
A sonometer wire vibrates with frequency n. It is replaced by another wire of three times the diameter, if tension and other parameters of the wire will be: [ CPMT 1985]
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a) 9n
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b) 3n
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c) n/3
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d) n/9
Explanation
Since density and length and tension is same frequency n ∝ 1/ √m Thus n ∝ / r Answer: (c)
Q.16
When a tuning fork vibrates, the waves produced in the fork are
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a)longitudinal in stem and transverse in prong
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b) longitudinal in prong and transverse in stem
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c)longitudinal in both prong and in stem
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d)transverse in both prong and in stem
Explanation
Answer: (a)
Q.17
Which one of the following is a simple harmonic motion [ CBSE 1994]
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a) Ball bouncing between two rigid vertical walls
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b) Particle moving in a circle with uniform speed
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c)Wave moving through a string fixed at both ends
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d)Earth spinning about its own axis
Explanation
Answer: (c)
Q.18
Energy is not carried by [ M.N.R 1990]
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a)transverse progressive waves
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b) longitudinal progressive waves
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c)stationary waves
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d)electromagnetic waves
Explanation
Answer:(c)
Q.19
A tuning fork vibrating with sonometer having 20 cm wire produces 5 beats per second. the beat frequency does not change if the length of the wire is changed to 21 cm the frequency of the tuning fork ( in Hertz ) must be [ MNR 1991]
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a) 200
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b) 210
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c) 205
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d) 215
Explanation
Let the frequency of tuning fork be n Then, the possible frequencies of the sonometer wire are n±5 i.e. (n+5) and (n-5) Formula for frequency of stretched wire For frequency n+5 For frequency n-5 by taking ratioAnswer: (c)
Q.20
String of wires of same material of length l and 2l vibrate with frequencies 100 and 150 respectively. The ratio of their tension is [ MPPMT 1990]
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a)2:3
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b) 3:2
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c)1:9
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d)1:3
Explanation
From the formula for frequency of stretched wire since radius and density is same mass per unit length is same Thus n ∝ √T / l Answer: (c)
Q.21
Four wires of identical lengths, diameters and materials are stretched on a sonometer box. The ratio of their tensions is 1:4:9:The ratio of their fundamental frequencies is .. [ CPMT 1989]
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a) 16:9:4:1
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b) 4:3:2:1
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c)1:2:3:4
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d)1:4:9:16
Explanation
Since frequency n∝ √T ∴ ratio of their fundamental frequencies is 1:2:3:4Answer: (c)
Q.22
A stretched string of one metre length, fixed at both ends, having a mass of 5×10-4 kg is under a tension of 20 newtons. It is plucked at a point situated at 25cm from one end. The stretched string would vibrate with a frequency of [ CPMT 1988]
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a) 400 Hz
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b) 100 Hz
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c)200 Hz
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d)256 Hz
Explanation
velocity of wave in string v=√ (T/m)=√ (20/(5×10-4)=200 m/secNow λ/4=25 λ=100 cm=1 m Now n=v/λ=200 /1=200 Hz Answer:(c)
Q.23
A hollow metallic tube of length L and closed at one end produces resonance with a tuning fork of frequency n. the entire tube is then heated carefully so that at equilibrium temperature its length changes by l. If the change in velocity V of the sound is v the resonance will now be produced by tuning fork whose frequency is [ CPMT 1988]
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a)
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b)
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c)
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d)
Explanation
Both the velocity and length of tube will increase Answer: (a)
Q.24
The velocity of sound in air is 330 metre/sec. the fundamental frequency of an organ pipe open at both ends and of length 0.3 metres will be [ MNR 1993]
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a)200 Hz
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b) 550 Hz
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c)300 Hz
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d)275 Hz
Explanation
fundamental frequency=v/2l n=330/(2×0.3)=550 HzAnswer: (b)
Q.25
A sonometer wire is in unison with tuning fork. Keeping the same tension, the length of the wire between the bridges is doubled. the tuning fork can be still in resonance with the wire, provided the wire now vibrates in [ MPPMT 1989]
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a)4 segments
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b) 6 segments
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c)3 segments
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d)2 segments
Explanation
Answer: (d)
Q.26
the length of sonometer wire AB is 110 cm. Where should the bridges be placed from A to divide fundamental frequencies in the ratio of 1:2:3 .. [ CBSE 1995]
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a) 30 cm and 90 cm
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b) 60 cm and 90 cm
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c)40 cm and 80 cm
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d)none of these
Explanation
let l1 , l2, l3 be the three segments. Then l1 + l2 + l3=110 cmFrom the law of vibrations of stretched strings n1l1=n2l2=n3l3given n1: n2 : n3=1 : 2 : 3 l2=l1 / 2 and l3=l1 /3 Thus l1 + l1 /2 + l1 /3=110 l1=60 , l2=30 cm and l3=20 cm Therefore two bridges should be placed at 60cm and ( 60+30=90) cm from end A Answer:(b)
Q.27
A stretched string 0.25 cm has a frequency of f=300Hz in the fundamental mode. The velocity of transverse wave in the string is ..[ EAMCET 1988]
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a) 75 m/s
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b) 150 m/s
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c) 330 m/s
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d) 1200 m/s
Explanation
n=v/2l v=n ×2l v=300×2×0.25 v=150 m/s Answer: (b)
Q.28
Air is blown at the mouth of a tube ( length=25cm, diameter=2cm) open at both the ends as shown in figure. Velocity of sound is 330m/sec. The sound emitted by the tube will have all the frequencies in the group..[ CPMT 1989]
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a)660, 1320, 1980 Hz
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b) 660, 1000, 3300Hz
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c)302, 664, 1320Hz
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d)330, 990, 1690Hz
Explanation
First harmonics=v/2l=330/(2×0.25)=660 Hz Second Harmonics=2v/2l=1320 Hz Third Harmonics=3v/2l=1980 HzAnswer: (a)
Q.29
An object of specific gravity ρ is hung from a thin steel wire. The fundamental frequency for transverse standing waves in the wire is 300Hz. The object is immersed in water so that one half of its volume is submerged. The new fundamental frequency, in Hz is ..[ IIT1995]
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a)
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b)
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c)
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d)
Explanation
Let V be the volume of the weight suspended from wire and ρ be the density of the material of weight then Tension T=weight of object suspended=Vρg From the formula for frequency When the weight is half submerged in water, an upthrust=Vg/2 acts on it. Thus the net tension in the wire=Vρg - Vg/2 By taking ratioAnswer: (a)
Q.30
A long glass tube is held vertically in water. A tuning fork is struck and held over the tube. Strong resonance are observed at two successive length 0.5m and 0.84m above the surface of water. If the velocity of sound is 340m/s, then the frequency of the tuning fork is .. [ SCRA 1994]
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a) 128 Hz
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b) 256 Hz
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c)284 Hz
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d)500 Hz
Explanation
From the formula for resonance column v=2n(l2 - l1) Answer:(d)
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