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Physics NEET MCQ
Oscillations And Wave Mcq
Quiz 7
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Q.1
When a sound wave of frequency 300 HZ passes through a medium the maximum displacement of a particle of the medium is 0.1 cm. The maximum velocity of the particle is equal to .. [ MNR 1992]
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a) 60π cm/sec
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b) 30π cm/sec
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c) 30 cm/sec
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d) 60 cm/sec
Explanation
Vmax=aω=2πfa=2π×300×0.1 Vmax=60π cm/sec Answer: (a)
Q.2
The Laplace's correction in the expression for the velocity of sound given by Newton is needed because sound wave ..
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a)Are longitudinal
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b) Propagate isothermally
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c)Propagate adiabatically
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d)Are of long wavelengths
Explanation
Answer: (c)
Q.3
Wave equations of two particles are y1=a sin(ωt-kx) and y2=a sin(kx+ωt)then the particles are .. [ BHU 1995]
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a) Moving in opposite directions
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b) Phase between is 90°
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c)Phase between is 180°
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d)Phase between is 0°
Explanation
One wave is moving along positive x axis while other is moving along negative x axisAnswer: (a)
Q.4
A sound wave has frequency 500Hz and velocity 350 m/sec. What is the distance between the two particles having phase difference of 60° [ CPMT 1990]
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a) 0.7 cm
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b) 120cm
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c)70 cm
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d)120.0 cm
Explanation
wave length λ=velocity /frequency λ=350/500 m Now phase difference 2π=λ distance between two particles Phase difference π/3= Answer:(b)
Q.5
Velocity of sound in a gas is proportional to .. [ CPMT 1985]
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a) Square root of adiabatic elasticity
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b) Adiabatic elasticity
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c) Square root of isothermal elasticity
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d) isothermal elasticity
Explanation
Answer: (a)
Q.6
A mass m=100 gm is attached at the end of a light spring which oscillates on a frictionless horizontal table with an amplitude equal to 0.16 m and time period equal to 2 sec. Initially the mass is released from rest at t=0 and displacement of the mass at any time (t) is .. [ PMT 1995]
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a)x=0.16 cos(πt)
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b) x=-0.16 cos(πt)
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c)x=0.16 cos(πt+π)
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d)x=-0.16cos(πt+π)
Explanation
The particle starts from the left extreme position. Therefore its initial phase is π. Its time period is T=2sec ∴ ω=2π/T=πSo the equation of motion is x=0.16 cos( πt + π)Answer: (c)
Q.7
A traveling wave passes a point of observation. At this point the time interval between successive crests is 0.2 sec ... [ MPPMT 1990]
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a) Wavelength is 5 metre
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b) Frequency is 5Hz
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c)Velocity of propagation is 5 metre/sec
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d)Wavelength is 0.2 metre
Explanation
Time between two successive crest=periodic time ( 0.2sec frequency=1 /T=1/02=5 HzAnswer: (b)
Q.8
A thin plane membrane separates hydrogen at 27° from hydrogen at 127°C, both being at the same pressure. A plane sound wave enters from the colder to the hotter side. If the angle of incidence on the membrane is 30°, then the angle of refraction is ..
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a) sin-1 (1/√3)
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b) sin-1(2/3)
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c)sin-1(3/8)
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d)sin-1(2/8)
Explanation
We know that v1 /v2=√T1 / √T2 And refractive index sini/ sinr=v1 /v2∴sini/ sinr=√T1 / √T2 sin30/sinr=√300 / √400 Or sinr=sin30 × (√400 /√300) sinr=(1/2)× ( √4 / √3)sinr=1/√3r=sin-1 (1 / √3) Answer:(a)
Q.9
A wave equation is given by where y is in cm. Frequency of wave and maximum acceleration will be .. [ Raj.PET 1997]
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a) 100 HZ, 4.7×103 cm/s2
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b) 50 HZ, 7.5×103 cm/s2
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c) 25 HZ, 4.7×103 cm/s2
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d) 25 HZ, 7.5×103 cm/s2
Explanation
Comparing with standard equation of progressive wave Maximum Acceleration=ω2A a=(2πn)2A a=(50π)2 ×3=7.5×104 m/sec2 Answer: (d)
Q.10
The particles of medium vibrate about their mean positions whenever a wave travels through that medium. The phase difference between the vibrations of two such particles .. [ SCRA 1994]
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a)varies with the time
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b) varies with distance separating them
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c)varies with time as well as distance
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d)is always zero
Explanation
Answer: (c)
Q.11
An engine is moving on a circular path of radius 100 metres with speed of 20 metres per sec. What will be the frequency observed by an observer standing stationary at the centre of the circular path when the engine blows a whistle of frequency 500Hz
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a) More than 50 Hz
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b) Less then 500Hz
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c) 500 Hz
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d) No sound
Explanation
Answer: (c)
Q.12
At what temperature is the speed of sound in hydrogen will be the same as that in oxygen at 111°C? The ratio of the densities of oxygen and hydrogen is 16:1
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a) 249°C
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b) -249°C
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c)63°C
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d)-65°C
Explanation
From formula for velocityGiven Velocity of wave in Hydrogen and in Oxygen same thus Answer: (b)
Q.13
When a tuning fork is made to vibrate, then the vibrations of its prongs
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a) are in phase
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b) have a phase difference of π
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c)have a phase difference of π/2
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d)have a phase difference of π/4
Explanation
Answer:(b)
Q.14
student sees a jet plane flying from east to west when the jet is seen just above his head, the sound of the jet appears to reach him making angle of 60° with the horizontal from east. If the velocity of sound is v, then velocity of jet is
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a) 2v
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b) (√3 /2)v
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c) (2/√3) v
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d) v/2
Explanation
Let vp be the velocity of plane vs be the velocity of sound From figure Answer: (d)
Q.15
With increase in temperature, the velocity of sound in liquid, in general
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a)Increases slightly
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b) Decreases slightly
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c)Remains unchanged
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d)Changes erratically
Explanation
Answer: (b)
Q.16
The amplitude of a velocity of particle is given by Vm=Vo / (aω2 - bω+c ) where Vo, a , b, and c are positive. The condition for a single resonant frequency is ...[ CPMT 1998]
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d)b2=7ac
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a) b2 < 4ac
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b) b2 > 4ac
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c)b2 < 5ac
Explanation
Answer: (a)
Q.17
Speed of sound in air is 332 m/s at NTP. The speed of sound in hydrogen at NTP will be [ MNR 1995]
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a) 5312 m/s
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b) 2546 m/s
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c)1328 m/s
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d)664 m/s
Explanation
Answer:(c)
Q.18
Velocity of sound wave in air is 330 metre per sec. For a particular sound in air, a path difference of 40cm is equivalent to phase difference of 1.6π. The frequency of this wave is ...[ CBSE 1980]
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a) 165 hz
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b) 150 hz
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c) 660 hz
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d) 330 hz
Explanation
We know that ΔΦ=(2π/λ)Δx or 1.6π=(2π/λ) 40 λ=80 /1.6=50cm=0.5cm n=v/λ=3300 / 0.5=660 Hz Answer: (c)
Q.19
Two wires of different densities, are joined at x=An incident wave yi=ai (ωt-k1x) traveling from left to right is partly reflected and partly transmitted at x=If the amplitude of reflected and transmitted waves be ar and ai respectively, then ar / ai is
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a)
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b)
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c)
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d)
Explanation
Equation for reflected wave is yr=ar sin(ωt + k1x) Equation for transmitted wave is yr=at sin(ωt - k2x) Now yi=yr +yt Answer: (c)
Q.20
In a stretched string under tension and fixed at both the ends, the area of cross-section of its wire is halved and tension is doubled. The frequency becomes.
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a) Twice
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b) Eight times
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c)Half
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d)For times
Explanation
frequency n ∝ √(T/A). since T is doubled and A is halved, frequency becomes 2 timesAnswer: (a)
Q.21
The end correction of a resonance column is 1.0 cm. If the shortest length resonating with a tuning fork is 15.0 cm, the next resonating length is [ MPPMT 1985]
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a) 31 cm
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b) 45 cm
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c)46 cm
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d)47 cm
Explanation
end correction e Answer:(d)
Q.22
The fundamental frequency of sonometer wire carrying a block of mass 1 kg and density 1.8 is 260Hz. When the block is completely immersed in a liquid of density 1.2, then what will be its new frequency? [ CPMT 1998]
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a) 300 Hz
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b) 150Hz
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c) 450Hz
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d) none
Explanation
Tension T in air=mg Given density of string ρ=1.8 and density of liquid ρo=1.2 Due to buoyant force gravitational acceleration will change given by g'=g( 1 -ρo / ρ) ∴ g'=g[ 1 - (1.2/1.8)]=g(1/3) Thus tension in string when in liquid=mg'=mg/3 According to formula for resonating frequency Answer: (b)
Q.23
With a closed organ pipe of length l, the fundamental ton has frequency. [ CPMT 1993]
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a) ( v/2l) and only even harmonics are present
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b) ( v/2l) and only odd harmonics are present
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c) (v/2l) and even as well as odd harmonics are present
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d) (v/4l) and only odd harmonics are present
Explanation
Answer: (d)
Q.24
In an open end organ pipe of length l, if the velocity of sound is v, then the fundamental frequency will be [ CPMT 1990]
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a) (v/2l) and all harmonics are present
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b) (v/4l) and all harmonics are present
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c)(v/2l) and even harmonic are absent
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d)(v/4l) and even harmonics are absent
Explanation
Answer: (a)
Q.25
The frequency of vibration of string can be increased by:
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a)Increasing the length of the string keeping the tension constant
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b) Decreasing the density of the string keeping the tension constant
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c)Increasing the thickness of the string keeping the length constant
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d)Decreasing the tension of the string keeping the length constant
Explanation
Answer:(b)
Q.26
A cylindrical tube, open at both ends has a fundamental frequency 'f' in air. The tube is vertically dipped in water so that half of it is in water, the fundamental frequency of air column is [ BHU 1995]
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a) f/2
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b) 3f/4
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c) f
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d) 2f
Explanation
For open tube f=v/2l. On dipping the tube in water it becomes a closed tube. For closed tube f'=v/4l'=v/ [4(l/2)=v/2l=f Answer: (c)
Q.27
The frequency of the fundamental mode of vibration of an organ pipe is 400Hz. When one end of the pipe is closed the fundamental frequency is ..[ EAMCET 1986]
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a)200 Hz
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b) 400 Hz
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c)600Hz
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d)800 Hz
Explanation
Fundamental frequency both open end n=v/2l Fundamental frequency for one end closed n'=v/4lThus n'=n/2 n'=400 /2=200Answer: (a)
Q.28
In order to increase the fundamental frequency of string from 100Hz to 400Hz, the tension must be increased by [ EAMCET 19987]
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a) 2 times
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b) 4 times
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c)8 times
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d)16 times
Explanation
frequency n ∝ √T 100 ∝ √T 400 ∝ √T' By taking the ratio T"=16T Answer: (d)
Q.29
With increase in temperature, the frequency of the sound from an organ pipe.. [ Raj. PET 1996]
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a)Decreases
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b) Increases
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c)Remain unchanged
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d)Changes erratically
Explanation
Answer:(b)
Q.30
The frequencies of the harmonic of the string are ..[ CPMT 1988]
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a) Unrelated
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b) Of the same pitch
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c) In the ratio 1:2:3
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d) In the ratio 1:3:5
Explanation
Answer: (c)
0 h : 0 m : 1 s
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