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Physics NEET MCQ
Oscillations And Wave Mcq
Quiz 9
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Q.1
A string fixed at both the ends is vibrating in two segments. The wavelength of the corresponding wave is [ UPSC 1994]
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a) l/4
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b) l/2
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c) l
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d) 2l
Explanation
Answer: (c)
Q.2
A wire of tension 225 N produces 6 beats per second when it is tune with fork, when the tension changes to 256 N, it is again tuned with same tuning fork, the number of beats remain unchanged, the frequency of tuning fork will be [ Raj.PET 1996]
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a)256
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b) 225
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c)280
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d)186
Explanation
Frequency of string n ∝√T n=k√T=k√225Let n be the frequency of tuning fork. According to question, initially frequency of wire is less than that of fork, so n-k√225=60n-6=k√225=15kNet time frequency of wire is more, thus 6+n=k√256=16k∴ n-6 / n+6=15/16 or n=186 HzAnswer: (d)
Q.3
The fundamental frequency of open pipe is 30 C/sec, the fundamental frequency of closed pipe of same length will be [ Raj.PET 1996]
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a) 10 C/sec
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b) 20 C/sec
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c)30 C/sec
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d)15 C/sec
Explanation
Formula for open end tube fundamental frequency=v/2l formula for one end closed tube fundamental frequency=v/4lAnswer: (d)
Q.4
A cylindrical tube, open at both ends, has a fundamental frequency n. If one of the ends is closed, the frequency becomes/ remains [ UGET 1995]
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a)0.5n
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b) n
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c)2n
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d)4n
Explanation
Answer:(a)
Q.5
An air column in a pipe closed at one end will be inn resonance with a vibrating tuning fork of frequency 680Hz if the length of the column in metre be [ ISM Dhanbad 1994]
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a) 0.25
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b) 0.105
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c) 0.375
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d) 0.5
Explanation
for closed end pipe n=v/4l 680=340 / 4l l=340 / ( 680×4)=1/8=0.125m Next resonance occurs at 0.125×3=0.375 m Answer: (c)
Q.6
An observer is moving away from source of sound of frequency 100Hz. His speed is 33 m/s, speed of sound is 330m/s, observed frequency is ..[ EAMCET 1995]
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a)90 Hz
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b) 100 Hz
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c)91 Hz
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d)110 Hz
Explanation
Source is stationary and observer is moving away formula for apparent frequency is Answer: (a)
Q.7
If a rocket moving with velocity v towards the moon, sends signals of frequency n to moon surface and after reflection of these signals receives the reflected signals, the change in frequency will be [ v<
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a)
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b)
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c)
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d)
Explanation
First source is moving ( rocket) to stationary object (Moon) apparent frequency Second Source is stationary (Moon) and observer is moving ( Rocket) to wards the sourceChange in frequency=n"-n Δ n=(1 +2v/c)n - nΔn=2vn/cAnswer: (c)
Q.8
A source of sound is moving with constant velocity of 20m/s emitting a note of frequency 1000 Hz. The ratio of frequencies observed by a stationary observer while the source is approaching him and after it crosses him will be ( speed of sound V=340 m/s) ..[ MPPMT 1994]
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a)9:8
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b)8:9
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c)1:1
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d)9:10
Explanation
Source moving towards stationary observer Source is moving away from stationary observerBy taking ratio n' / n" we get Answer:(a)
Q.9
Two cars are moving on two perpendicular roads a crossing with speeds of 72km/hr and 36km/he. If first car blows horn of frequency 280Hz, then the frequency of horn heard by the driver of second car when line joining the cars makes 45° angle with the roads, will be [ Raj.PMT 1997]
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a) 321 Hz
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b) 298Hz
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c) 289 Hz
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d) 280 Hz
Explanation
Velocity of first car v1=20m/s and second car v2=10 m/s The components of velocities of the two cars along the direction of propagation of sound=20cos45°=14.14 m/sec and 10cos45=7.07 m/s Since source and observer approach each other apparent frequency Answer: (b)
Q.10
A star is moving away from the earth surface with speed of 106 m/sec. If the wave-length of spectral lines received from the star is 5700 Å . The doppler's shift will be [ Raj.PMT 1996]
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a)200Å
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b) 19Å
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c)20Å
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d)0.2Å
Explanation
Star moving away from observer red shift Δλ=(v/c)λ Δλ=106 / 3×108) 5700=19ÅAnswer: (b)
Q.11
The frequency of radar is 780Mhz. The frequency of the reflected wave from aeroplane is increased by 2.6kHz. The velocity of the aeroplane is ..[ CPMT 1991]
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a) 2 km/sec
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b) 1 km/sec
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c)0.5km/sec
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d)0.25 km/sec
Explanation
Change in frequency=Δn=2vn/c Answer: (c)
Q.12
A source of frequency 150Hz is moving in the direction of person with velocity of 110m/sec. The frequency heard by the person will be ( speed of sound in the medium=330 m/sec) ..[ CPMT 1989]
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a) 225 Hz
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b) 200 Hz
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c)150 Hz
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d)100 Hz
Explanation
Source is moving towards stationary observer Answer:(a)
Q.13
A source of sound S is moving with velocity 50 m/s towards a stationary observer. The observer measures the frequency of the source as 1000Hz. What will be the apparent frequency of the source when it is moving away from the observer after crossing him? The velocity of sound in the medium is 30m/s [ MPPMT 1994]
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a) 750 Hz
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b) 857 Hz
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c) 1143 Hz
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d) 1333 Hz
Explanation
When source approaches observer frequency When source go away from observer frequency Answer: (a)
Q.14
The apparent wavelength of light from a star moving away from the earth is 0.4%more than its real wavelength. The velocity of the star is [ MPPMT 1997]
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a)150 km/sec
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b) 300 km/sec
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c)600 Km/sec
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d)1200 km/sec
Explanation
Star is moving away red shift Change in wave length Δλ=(v/c)λ Answer: (d)
Q.15
The difference between the apparent frequency of a source of sound as perceived by an observer during its approach and recession is 2% of the natural frequency of the source. If the velocity of the sound is ..[ CPMT 1998]
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a) 6 m/sec
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b) 3 m/sec
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c)1.5 m/sec
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d)12 m/sec
Explanation
Frequency while approach to wards stationary source Frequency while recession from stationary sourceClearly n' > n" Thus Answer: (b)
Q.16
An observer standing at station observes frequency 212 when a train approaches and 189 when train goes away from him. If velocity of sound in air is 340 m/s, the velocity of train and actual frequency of whistle will be [ RajPET 1997]
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a) 15.5 m/s, 200 Hz
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b) 19.5 m/s, 205 Hz
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c)19.5 m/s, 200 Hz
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d)32.5 m/s, 205 Hz
Explanation
Source moving towards stationary observer Source moving away from stationary observer By taking ratio of above two equationsby substituting value in first equation we get Answer:(c)
Q.17
THe Doppler's shift is independent of ..[ MPPMT 1993]
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a) The speed of the source
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b) The frequency of the observer
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c) The frequency of the emitted waves
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d) The distance between the source and the observer
Explanation
Answer: (d)
Q.18
A source of frequency 480 Hz is at rest. An observer also at rest observes the frequency of the source. If wind is blowing from source towards the observer at a speed of 30m/s, the frequency of the source as heard by the observer is ( speed of sound = 330 m/s)
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a) 480 Hz
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b) 467 Hz
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c) 523 Hz
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d) -480Hz
Explanation
there no relative motion between source and observer Answer: (a)
Q.19
If a source emitting frequency f moves towards an observer with a velocity v/3 and the observer moves away from the source with velocity v/4, the apparent frequency as heard by the observer will be ..[( v= velocity of sound)
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a) (9/8)f
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b) (8/9)f
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c) (3/4)f
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d) (4/3)f
Explanation
Observer and source are moving in same direction Answer:(a)
Q.20
Two whistles A and B have frequencies 660Hz and 590Hz respectively. An observer is standing in the middle of the line joining the two sources. Source B and observer are moving towards right with velocity 30m/sec and A is standing at left side. If the velocity of sound in air is 300 m/sec, the number of beats listened by the observer are [ Raj.PET 1996]
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a) 2
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b) 4
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c) 6
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d) 8
Explanation
There is no relative velocity between B and observer thus frequency heard by observer is 59Hz observer is moving towards stationary source thus ∴ beats frequency = (594-590) = 4 per/sec Answer: (b)
Q.21
Two trains, one coming towards and another going away from an observer both of 4m/s produced whistle simultaneously of frequency 30Hz. Find the number of beats produced .. [ BHU 1998]
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a) 5
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b) 6
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c)7
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d)12
Explanation
Train approaching stationary observer frequency Train going away from stationary observer frequency ∴ Beats heard=243-237=5 Answer: (a)
Q.22
A whistle whirled in circle of radius one metre traverses the circular path twice per second. An observer is situated outside the circle but in same plane. If the velocity of sound is 332m/s, then the interval between the highest and lowest observed pith is [ JIPMER1998]
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a) 2:
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b) 1.08:1
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c)332:1
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d)332:4π
Explanation
Liner velocity of whistle v=2πr/T=2π/0.5=4π=12.56 m/s When whistles is moving towards the observer frequency is maximum given by When whistles is moving away the observer frequency is minimum given by By taking maximum and minimum frequency Answer: (b)
Q.23
A train has just completed a U-curve in a track which is semicircle. The engine is at forward end of the semicircular part of the track while the last carriage is at the rear end of the semi-circular track. The driver blows a whistle of frequency 200Hz. Velocity of sound is 340 m/sec. Then the apparent frequency as observed by a passenger in the middle of the train, when the speed of the train is 30m/sec is
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a) 2/9 Hz
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b) 288 Hz
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c)200 Hz
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d)181 Hz Hz
Explanation
Velocity component of the source in the direction of motion of sound=30cos45 Along BAVelocity component of observer in the direction BA=30cos45 ∴ There is no relative motion between the source and the observer, hence no change in real frequency is observed. Answer:(c)
Q.24
The frequency of note emitted by source changes by 20% as it approaches observer. As it recedes away from him, the apparent frequency will be different from the actual frequency by
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a) 20%
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b) 17.4%
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c) 16.67%
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d) 14.3%
Explanation
Case I source is moving towards observer Apart frequency n'=1.2n thus CAse II source is moving away from observer Apparent frequency n" Answer: (d)
Q.25
A wave travelling in the +ve x-direction havingdisplacement along y-direction as 1 m, wavelength2π m and frequency of 1/πHz is represented by
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a) y = sin(x – 2t)
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b) y = sin(2πx – 2πt)
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c) y = sin(10πx – 20πt)
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d) y = sin(2πx + 2πt)
Explanation
Standard equation for wave Answer:(a)
Q.26
If we study the vibration of a pipe open at both ends, then the following statement is not true
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a) Open end will be anti-node
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b) Odd harmonics of the fundamental frequencywill be generated
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c) All harmonics of the fundamental frequencywill be generated
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d) Pressure change will be maximum at both ends
Explanation
Antinode is formed at open end. At open ends pressure change will be zero. Answer:(d)
Q.27
A source of unknown frequency gives 4 beats/s,when sounded with a source of known frequency250 Hz. The second harmonic of the source ofunknown frequency gives five beats per second,when sounded with a source of frequency 513 Hz.The unknown frequency is [ NEET -2013]
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a) 254 Hz
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b) 246 Hz
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c) 240 Hz
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d) 260 Hz
Explanation
Unknown frequency using 250 Hz first harmonics = 250±4 or 254 or 246 Now second harmonic using 250 Hz first harmonics = 2×first harmonics = 2×254 = 508 or 493 Now this second harmonics gives give five beets with source frequency of 513 Possible frequency = 513±5 = 508 or 518 From above second harmonic is 508 Therefore unknown frequency = 254 Answer:(a)
Q.28
The oscillation of a body on a smooth horizontal surface is represented by the equation, X =Acos (ωt)where X= displacement at time t, ω=frequency of oscillation. Which one of the following graphs shows correctly the variation 'a' with 't'?
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a)
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b)
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c)
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d)
Explanation
Formula for acceleration = -Aω2 cos(ωt) Time t= 0, a = A Time t= T/2 Answer:(a)
Q.29
) If n1, n2 and n3 are the fundamental frequencies of three segments in to which a string is divided, then the original fundamental frequency n of the string is given by …[ AIPMT 2014]
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a)
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b) n = n1 + n2 + n3
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c)
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d)
Explanation
Fundamental frequency is given by Now string is divided in three parts so L = L1 + L2 + L3 …(i) Similarly Substituting value of L, L1, L2 and L3 in (i) Answer:(c)
Q.30
The number of possible natural oscillations of air column in a pipe closed at one end of length85 cm whose frequencies lie below 1250Hz are: (velocity of sound = 340ms–1)
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a) 7
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b) 6
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c) 4
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d) 5
Explanation
f3 = 3f1 ; f5 = 5f1; f7 = 7f1; f9=9f1 ; f11=11f1 Thus fundamental frequency and five overtone, total six possible natural oscillations Answer:(b)
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