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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A vehicle , with a horn of frequency `n` is moving with a velocity of `30 m//s` in a direction prependicular to the straight line joining the observer and the vehicle . The observer perceives the sound to have a grequency `(n + n_(1))` . If the sound velocity in air is `330 m//s` , thenA. `n_1=10n`B. `n_1=-n`C. `n_1=0.1n`D. `n_1=0` |
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Answer» Correct Answer - D No doppler effect, because velocity is perpendicular to line joining vehicle and observer. |
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| 2. |
A vibrating tuning fork is first held in the hand and then its end is broght in contact with a table. Which of the following statement (s) is are correct in respect of this situation?A. The sound is louder when the tuning fork is held in handB. The sound is louder when the tuning fork is in contact with table.C. The sound dies away sooner when tuning fork is brought in contact with the table.D. The sound remains for a longer duration when turning fork is held in hand. |
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Answer» Correct Answer - B::C::D When the vibrating tuning fork is brought in contact with the table, the vobration of the tuning fork are being transmitted to the surface of table whose surface area is very large as compared to the surface area of tuning fork and hence sound becomes louder and due to the energy transmitted over the table, the sound dies sooner. |
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| 3. |
A vibrating tuning fork tied to the end of a string 1.988 m long is whirled round a circle. If it makes two revolutions in a second, calculate the ratio of the frequencies of the highest and the lowest notes heard by an observer situated in the plane of the tuning fork. Valocity of sound is 350 m/s. |
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Answer» The number of revolutions per second is 2. Radius of the circle is `1.988m`. The linear velocity of the tuning fork is `v=2xx2pir=4xx(22)/(7)xx1.988=25(m)/(s)` i. Apparent frequency when the tuning fork is approaching the listener is `n_1=(vn)/(v-v_s)=(350n)/(350-25)=(14)/(13)n` ii. Apparent frequency when the tuning fork is moving away from the listener is `n_2=(vn)/(v+v_s)=(350N)/((350-25))=(14)/(15)n` The ratio of highest note to the lowest note is given by `(n_1)/(n_2)=(14n)/(13)xx(15)/(14n)=(15)/(13)=1.154` |
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| 4. |
Explain: if an observer places his ear to one end of a long iron pipe line, he can distinctly hear two sounds when a man hammers the other end of the pipeline. |
| Answer» Sound travels faster in iron that in air. So there is a difference between the times taken by sound to travel through iron and sir. This is why the observer hears two sounds. | |
| 5. |
As shown if Fig. a vibrating tuning fork of frequency 512 Hz is moving towards the wall with a speed `2(m)/(s)`. Take speed of sound as `v=340(m)/(s)` and answer the following questions. Q. If the listener, along with the source, is moving towards the wall with the same speed i.e., `2(m)/(s)`, such that the source remains between the listerner and the wall, number of beats heard by the listerner per second will beA. 4B. 8C. 0D. 6 |
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Answer» Correct Answer - B As no relative motion is there between observer and listener, hence frequency heard by observer is 512 Hz. He will observe frequency reflected from wall, `f_1=515Hz`. Hence, the wave reflected from wall will act as another source of frequency 515 Hz. Therefore, the frequency received by the observer from wall is `f_2=(v+v_0)/(v)f_1` `=(342)/(340)xx515=518Hz` Hence, beats observed is `f_2f=518-512=6`. |
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| 6. |
As shown if Fig. a vibrating tuning fork of frequency 512 Hz is moving towards the wall with a speed `2(m)/(s)`. Take speed of sound as `v=340(m)/(s)` and answer the following questions. Q. If the listerner along with the source is moving towards the wall with the same speed i.e., `2(m)/(s)`, such that he (listener) remains between the source and the wall, number of beats heard by him will beA. 2B. 6C. 8D. 4 |
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Answer» Correct Answer - D Frequency received from source directly by observer will remain same. Hence frequency received by observer is `512Hz` let `f_1` be the frequency reflected by wall Then, `f_1=(v)/(v-v_S)xxf=(340)/(338)xx512=515Hz` The frequency received by observer (reflected from wall) is `f_2=((v_v_0)/(v))f=(342)/(340)xx515=518Hz` Hence beats heard is `f_2-f_1=518-512=6`. |
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| 7. |
As shown if Fig. a vibrating tuning fork of frequency 512 Hz is moving towards the wall with a speed `2(m)/(s)`. Take speed of sound as `v=340(m)/(s)` and answer the following questions. Q. If the listerner along with the source is moving towards the wall with the same speed i.e., `2(m)/(s)`, such that he (listener) remains between the source and the wall, number of beats heard by him will beA. 2B. 6C. 8D. 4 |
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Answer» Correct Answer - D Frequency received from source directly by observer will remain same. Hence frequency received by observer is `512Hz` let `f_1` be the frequency reflected by wall Then, `f_1=(v)/(v-v_S)xxf=(340)/(338)xx512=515Hz` The frequency received by observer (reflected from wall) is `f_2=((v_v_0)/(v))f=(342)/(340)xx515=518Hz` Hence beats heard is `f_2-f_1=518-512=6`. |
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| 8. |
As shown if Fig. a vibrating tuning fork of frequency 512 Hz is moving towards the wall with a speed `2(m)/(s)`. Take speed of sound as `v=340(m)/(s)` and answer the following questions. Q. Suppose that a listener is located at rest between the tuning fork and the wall. Number of beats heard by the listener per second will beA. 4B. 3C. 0D. 1 |
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Answer» Correct Answer - B The frequency heard directly from source is given by `f_1=((v)/(v-v_S))f` Here `v=340(m)/(s)`,`v_S=2(m)/(s)`,`f=512Hz` `f_1=(340)/(338)xx512=515Hz` the frequency of the wave reflected from wall will be same (no relative motion between wall and listener, so no change in frequency). Hence no beats are observed. |
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| 9. |
A small source of sound vibrating frequency 500 Hz is rotated in a circle of radius `(100)/(pi)` cm at a constant angular speed of `5.0` revolutions per second. The speed of sound in air is `330(m)/(s)`. Q. For an observer who is at rest at a great distance from the centre of the circle but nearly in the same plane, the minimum `f_(min)` and the maximum `f_(max)` of the range of values of the apparent frequency heard by him will beA. `f_(min)=455Hz`,`f_(max)=535Hz`B. `f_(min)=484Hz`,`f_(max)=515Hz`C. `f_(min)=484Hz`,`f_(max)=500Hz`D. `f_(min)=500Hz`,`f_(max)=515Hz` |
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Answer» Correct Answer - B Since the observer is in the same plane as the circle, at one instant of time, the source will move directly towards the observer when the apparent frequency will be maximum given by `f_(max)=((V)/(V-u))500=((330)/(330-u))500` Now, `u=romega=((100)/(pi))5xx2pixx10^-2(m)/(s)` `=10(m)/(s)` `f_(max)=((330)/(330-10))500=(330)/(320)xx500=515Hz` Similarly, at another instant, the source will move directly away from the observer with velocity `u=10(m)/(s)`. At this instant, the apparent frequency will be minimum given by `f_(min)=((330)/(330+10))500=(33)/(34)xx500=485Hz` |
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| 10. |
A source `S` of acoustic wave of the frequency `v_0=1700Hz` and a receiver `R` are located at the same point. At the instant `t=0`, the source start from rest to move away from the receiver with a constant acceleration `omega`. The velocity of sound in air is `v=340(m)/(s)`. If `omega=10(m)/(s^2)` for 10s and then `omega=0` for `tgt10s`, the apparent frequency recorded by the receiver at `t=15s`A. `1700Hz`B. `1313Hz`C. `850Hz`D. `1.23Hz` |
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Answer» Correct Answer - B Since `omega=0` for `tgt10s` the source will move at constant speed of `u=10xx10=100(m)/(s)` after `t=10s`. At `t=10s`, its distace r from the observer is `r=((1)/(2))omegat^2=((1)/(2))xx10xx100=500m`. Since the wave will take a time `r=10+(500)/(340)` which is less than 15 s, the wave that is received by the observer at `r=15s` would have left the source after 10 s, i.e., when its speed was constant at `100(m)/(s)`. Hence apparent frequency is `n_a=((340)/(340+100))1700` `=((17)/(22)xx1700)=1313Hz` |
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| 11. |
A small source of sound vibrating frequency 500 Hz is rotated in a circle of radius `(100)/(pi)` cm at a constant angular speed of `5.0` revolutions per second. The speed of sound in air is `330(m)/(s)`. Q. For an observer situated at a great distance on a straight line perpendicular to the plane of the circle, through its centre, the apparent frequency of the source will beA. greater that 500 HxB. smaller than 500 HzC. always remain 500 HzD. greater for half the circle and smaller during the other half |
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Answer» Correct Answer - C Since the source is moving on a small circle in a plane perpendicular to the direction of the wave moving towards the observer located on the axis of the circle, there would be no change in the observed frequency which will be the the same as the real frequency, i.e., 500 Hz. |
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| 12. |
Spherical sound waves are emitted uniformly in all direction from a point source. The variation in sound level SL. As a function of distance r from the source can be written as where a and b are positive constants.A. `SL=-blogr^a`B. `SL=a-b(logr)^2`C. `SL=a-blogr`D. `SL=(a-b)/(r^2)` |
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Answer» Correct Answer - C `SL=10log((k)/(I_0))` `=10log((k)/(I_0r^2))` `=10logK-10log(I_0r^2)` `=10logk-10logI_0-20logr` `=a-blogr` |
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| 13. |
Does the velocity of sound in a solid increase significantly on heating the solid? |
| Answer» No, the velocity of sond in a solid does not increase on heating the solid because neither the density `rho` nor the modulus of elasticity E change appreciably with change in temperature. | |
| 14. |
Two sound waves from two different sources interfere at a point to yield a sound of varying intensity. The intensity level between the maximum and minimum is 20 dB. What is the ratio of the intensities of the individual waves? |
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Answer» If `I_(max)` and `I_(min)` are the maximum and minimum intensities of the resultant waves, then the intensity level between them will be `L=10log((I_(max))/(I_(min)))dB` or, `20dB=10log((I_(max))/(I_(min)))dB` or, `(I_(max))/(I_(min))=(100)/(1)` If `A_1` and `A_2` are the amplitudes of the individual waves, intereferring, then `(I_(max))/(I_(min))=((A_1+A_2)^2)/((A_1-A_2)^2)` `((A_1_A_2)/(A_1-A_2))^2=(100)/(1)` `implies(A_1+A_2)/(A_1-A_2)=(10)/(1)` or `(A_1)/(A_2)=(11)/(9)` Therefore the intensity ratio between the individual waves will be `(I_1)/(I_2)=((A_1)/(A_2))^2=(121)/(81)` |
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| 15. |
Figure. Represents the displacement `y` versus distance `x` along the direction of propagation of a longitudinal wave. The pressure is maximum at position markedA. PB. QC. RD. S |
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Answer» Correct Answer - C `P=-B(dy)/(dx)` At R, `(dy)/(dx)` is most negative. So pressure is maximum. |
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| 16. |
Do displacement, particle velocity and pressure variation in a longitudinal wave vary with the same phase? |
| Answer» No the particle velcotu is `(pi)/(2)` out of phase with the displacement, and the pressure variation is out of phase by `pi` with the displacement. | |
| 17. |
Consider a souce of sound `S`, and an observer/detector `D`. The source emits a sound wave of frequency `f_0`. The frequency observed by `D` is found to be (i) `f_1`, if `D` approaches `S` and `S` is stationary (ii) `f_2`, if `S` approaches `S` and `S` is stationary (iii) `f_3`,if both `S` and `D` and `D` is stationary Speed In all three cases, relative velocity of `S` wrt `D` is the same. For this situation which is incorrect?A. `f_1nef_2nef_3`B. `f_1ltf_2`C. `f_3ltf_0`D. `f_1ltf_3ltf_2` |
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Answer» Correct Answer - B Let relative velocity of `v` and the speed of sound be `v_0`. Then, `f_1=(v_0-(-v))/(v_0)xxf_0=(v_0+v)/(v_0)f_0` `f_2=(v_0)/(v_0-v)xxf_0` `f_3=(v_0+(v)/(2))/(v_0-(v)/(2))xxf_0` It is clear from above that `f_1nef_2nef_3`,`f_3gtf_0` and we can prove that `f_2gtf_3gtf_1`. |
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| 18. |
Consider a souce of sound `S`, and an observer/detector `D`. The source emits a sound wave of frequency `f_0`. The frequency observed by `D` is found to be (i) `f_1`, if `D` approaches `S` and `S` is stationary (ii) `f_2`, if `S` approaches `S` and `S` is stationary (iii) `f_3`,if both `S` and `D` and `D` is stationary Speed In all three cases, relative velocity of `S` wrt `D` is the same. For this situation which is incorrect?A. `n_1=n_2=n_3`B. `n_1ltn_2`C. `n_3gtn_0`D. `n_3` lies between `n_1` and `n_2` |
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Answer» Correct Answer - B::C::D When observer `P` approaches the statinary source at speed `v`. `n_1=(V+n)/(V)xxn_0` .(i) (V is speed of sound) When source `S` approaches the stationary observer `P` at speed `v` `n_2=(V)/(V-v)xxn_0` ..(ii) Thus, `n_2gtn_1`, i.e., choice (b) is correct when both `S` and `P` approach each other with speed `(v)/(2)` `n_3=(V+((v)/(2)))/(V-((v)/(2)))n_0` .(iii) Hence `n_3gtn_0` and `n_3` lies between `n_1` and `n_2`. |
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| 19. |
The resultant loudness at a point P is `n` dB higher than the loudness of `S_1` which is one of the two identical sound sources `S_1` and `S_2` reaching at that point in phase. Find the value of n. |
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Answer» Loudness due to `S_1=I_1=ka^2` where a is the amplitude and loudness due to `S_1` and `S_2` both `I_2=k(2a)^2=4I_1` `n=10log_(10)((4I_1)/(I_1))=10log_(10)(4)=10(0.6)=6` |
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| 20. |
A siren emitting a sound of frequency 2000 Hz moves away from you towards a cliff at a speed of 8 m/s. (a) What is the frequency of the sound you hear coming directly from the siren. (b) What is the frequency of sound you hear reflected off the cliff. Speed of sound in air is `330(m)/(s)`. |
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Answer» (a) The frequency of sound heard directly. `f_1=f_0((v)/(v+v_S))` `v_S=8(m)/(s)` `f_1=((330)/(330+8))xx2000` `f_1=(330)/(338)xx2000=1953Hz` (b) the frequency of the reflected sound is given by `f_2=f_0((v)/(v-v_S))` `f_2=((330)/(330-8))xx2000` `f_2=(330)/(322)xx2000=2050Hz` |
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| 21. |
A source of sound and a detector are placed at the same place on ground At `t=0`, the source `S` is projected towards reflector with velocity `v_0` in vertical upward directon and reflector starts moving down with constant velocity `v_0` At `t-0`, the vertical separation between the reflector and source is `H((gtv_0^2)/(2g))`. The speed of sound in air is `v(gtgtv_0)`, Take `f_0` as the frequency of sound waves emitted by source. Based on above information answer the following questions. Q. Frequency of sound waves emitted by source at `t=(v_0)/(2g)` isA. `f_0`B. `f_0[(v)/(v)+(v_0)/(2)]`C. `f_0[((v-v_0)/(2))^(2)/(v)]`D. `f_0[(((v-v_0)/(2))/(v+v_0))/(2)]` |
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Answer» Correct Answer - B Actual frequency emitted by source does not depend upon the velocity of source but frequency heard may change due to relative motion between the observer and the source. |
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| 22. |
A source of sound and a detector are placed at the same place on ground At `t=0`, the source `S` is projected towards reflector with velocity `v_0` in vertical upward directon and reflector starts moving down with constant velocity `v_0` At `t-0`, the vertical separation between the reflector and source is `H((gtv_0^2)/(2g))`. The speed of sound in air is `v(gtgtv_0)`, Take `f_0` as the frequency of sound waves emitted by source. Based on above information answer the following questions. Q. Frequency of sound waves emitted by source at `t=(v_0)/(2g)` isA. `f_0`B. `f_0[(v)/(v)+(v_0)/(2)]`C. `f_0[((v-v_0)/(2))^(2)/(v)]`D. `f_0[(((v-v_0)/(2))/(v+v_0))/(2)]` |
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Answer» Correct Answer - B Actual frequency emitted by source does not depend upon the velocity of source but frequency heard may change due to relative motion between the observer and the source. |
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| 23. |
A source of sound and a detector are placed at the same place on ground At `t=0`, the source `S` is projected towards reflector with velocity `v_0` in vertical upward directon and reflector starts moving down with constant velocity `v_0` At `t-0`, the vertical separation between the reflector and source is `H((gtv_0^2)/(2g))`. The speed of sound in air is `v(gtgtv_0)`, Take `f_0` as the frequency of sound waves emitted by source. Based on above information answer the following questions. Q. Frequency of sound waves emitted by source at `t=(v_0)/(2g)` isA. `f_0`B. `f_0[(v)/(v)+(v_0)/(2)]`C. `f_0[((v-v_0)/(2))^(2)/(v)]`D. `f_0[(((v-v_0)/(2))/(v+v_0))/(2)]` |
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Answer» Correct Answer - B Actual frequency emitted by source does not depend upon the velocity of source but frequency heard may change due to relative motion between the observer and the source. |
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| 24. |
Speed of sound wave is v. If a reflector moves towards a stationary source emitting waves of frequency f with speed u, the wavelength of reflected waves will beA. `(v-u)/(v+u)f`B. `(v+u)/(v)f`C. `(v+u)/(v-u)f`D. `(v-u)/(v)f` |
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Answer» Correct Answer - C Apparent frequency for reflector (which will act here as an observer) would be `f_1=((v+u)/(v))f` Where f is the actual frequency of source. The reflector will now behave as a source. The apparent frequency will now become `f_2=((v)/(v-u))f_1` Substituting the value of `f_1` we get `f_2=((v+u)/(v-u))f` |
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| 25. |
Two sound sources are moving away from a stationary observer in opposite direction with velocities `V_1` and `V_2(V_1gtV_2)`. The frequency of both the sources is 900 Hz. `V_1` and `V_2` are both quite less than speed of sound, `V=300(m)/(s)`. Find the value of `(V_1-V_2)` so that beat frequency observed by observer is 9 Hz. (in m/s). |
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Answer» `f_1=900((300)/(300+V_1))` or `f_1=900[1+(V_1)/(300)^-1]=900-3V_1` Likewise, `f_2=900-3V_2` given `f_2-f_1=9impliesV_1-V_2=3(m)/(s)` |
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| 26. |
Two sound sources are moving in opposite directions with velocities `v_1` and `v_2(v_1gtv_2)`. Both are moving away from a stationary observer. The frequency of both the sources is 900 Hz. What is the value of `v_1-v_2` so that the beat frequency observed by the observer is 6 Hz? Speed of sound `v=300(m)/(s)`. Given that `v_1` and `v_2ltltltv`.A. `1(m)/(s)`B. `2(m)/(s)`C. `3(m)/(s)`D. `4(m)/(s)` |
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Answer» Correct Answer - B `f_1=900((300)/(300+v_1))` `cong900(1+(v_1)/(300))^-1` `=900-3v_1` similarly, `f_2=900((300)/(300+v_2))=900-3v_s` `f_2-f_1=6` `3(v_1-v_2)=6` `3(v_1-v_2)=6` or `v_1-v_2=2(m)/(s)` |
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| 27. |
The frequency changes by `10%` as the source approaches a stationary observer with constant speed `v_S`, What sould be the percentage change in ferquency as the source recedes from the observer with the same speed? Given that `v_Sltltv` (v speed of sound in air).A. `14.3%`B. `20%`C. `16.7%`D. `10%` |
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Answer» Correct Answer - D When the source approaches the observer `f_1=f((v)/(v-v_s))=f(1-(v_s)/(v))^(-)=f(1+(v_s)/(v))` or `((f_1-f)/(f))xx100=(v_s)/(v)xx100=10` In the second case, when the source recedes from the observer `f_2=f((v)/(v+v_s))=f(1+(v_s)/(v))^(-)=f(1-(v_s)/(v))` `((f_2-f)/(f))xx100=-(v_s)/(v)xx100=-10` In the first case, observed frequency increases by `10%` while in the second case, observed frequency decreases by `10%`. |
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| 28. |
Sound is more clearly heard with the wind How? |
| Answer» If the wind blows from the source to the listener, the velocity of the wind is added to that of the sound. The upper layer of air possesses a greater velocity of sound than that in the lower layers. The result is that the wavefront turn downwards and the sound rays curl down. A listener (O) on the ground thus has a better chance of hearing | |
| 29. |
The equation of a sound wave in air is given by `trianglep=(0.02)sin[(3000)t-(9.0)x]`, where all variables are is SI units. (a) find the frequency, wavelength and the speed of sound wave in air. (b) If the equilibrium pressure of air is `1.01xx10^5(N)/(m^2)`, What are the maximum and minimum pressure at a point as the wave passes through that point? |
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Answer» a. comparing with the standard form of a travelling wave, `trianglep=trianglep_(max)sin[omegat-(x)/(v)]` we see that `omega=3000s^-1.` The frequency is `f=(omega)/(2pi)=(3000)/(2pi)Hz` Also from the same comparison, `(omega)/(v)=9.0m^-1` or `v=(omega)/(9.0m^-1)=(3000s^-1)/(9.0m^-1)=(1000)/(3)(m)/(s)` The wavelength is `lamda=(v)/(f)=((1000)/(3(m)/(s)))((3000)/(2piHz))=(2pi)/(9)m` b. The pressure amplitude is `trianglep_(max)=0.02(N)/(m^2)`. Hence the maximum and minimum pressures at a point in the wave motion will be `1.01xx10^5+-0.02(N)/(m^2)` |
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| 30. |
The sound from a very high burst of fireworks takes 5 s to arrive at the observer. The burst occurs 1662 m above the observer and travels vertically through two stratifier layers of air, the top one of thickness `d_1` at `0^@C` and the bottom one of thickness `d_2` at `20^@C`. Then (assume velocity of sound at `0^@C` is `330(m)/(s)`)A. `d_1=342m`B. `d_2=1320m`C. `d_1=1485m`D. `d_2=342m` |
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Answer» Correct Answer - A Time taken is given by `T=t_1+t_2=(d_1)/(v_1)+(d_2)/(v_2)` `v_1=v_(0^@C)=330(m)/(s)` `v_2=(330+0.6t)=342(m)/(s)` `d=1662m` `T=(d_1)/(330)+((d-d_1))/(342)=5s` `(d_1(342-330))/(330xx342)+(d)/(342)=5s` `12d_1=5(342xx330)-330xx1662` `d_1=1320m` `d_2=342m` |
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| 31. |
A siren placed at a railway platform is emitting sound of frequency `5 kHz`. A passenger sitting in a moving train `A` records a frequency of `5.5kHz` while the train approaches the siren. During his return journey in a different train `B` he records a frequency of `6.0 kHz` while approaching the same siren. the ratio the velocity of train `B` to that of train `A` isA. `(242)/(252)`B. 2C. `(5)/(6)`D. `(11)/(6)` |
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Answer» Correct Answer - D `f_(approach)_(A)=5.5kHz=((v+v_A)/(v))5` ..(i) `f_(approach)_(B)=6kHz=((v+v_B)/(v))5` ..(ii) Where `v` is the velocity of sound. Now, `5.5=(1+(v_A)/(v))5` `implies(v_A)/(v)=0.1` .(iii) similarly, `6=(1+(v_B)/(v))5` `implies(v_B)/(v)=0.2` .(iv) `implies(v_B)/(v_A)=2` |
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| 32. |
A source of sound and detector are moving as shown in Fig. at `t=0`. Take velocity of sound wave to be `v`. For this situation mark out the correct statement(s).A. The frequency received by the detector is always greater than `f_0`B. Initially, frequency received by the detector is greater than `f_0`, becomes equal to `f_0`, and then decreases with the time.C. Frequency received by the detector is equal to `f_0` at `t=(dcottheta_0)/((2v_0))`.D. Frequency received by the detector can never be equal to `f_0` |
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Answer» Correct Answer - B::C As time increases, the source and detector are relatively approaching each other up to `t=t_0` where `t_0` is the instant when the source and detector are located perpendicular to direction of motion. `v_0xxt_0=(dcottheta_0)/(2)` `t_0=(dcottheta_0)/(2v_0)` For `tltt_0` `f_(ap)gtf_0` For `tgtf_0`, `f_(ap)ltf_0` |
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| 33. |
The average power transmitted across a cross section by two sound waves moving in the same direction are equal. The wavelength of two sound waves are in the ratio of `1:2`, then find the ratio of their pressure amplitudes. |
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Answer» Intensity is given bt `I=(p_0^2)/(2rhov)`. Here v and `rho` are same for both. And also given that `I` is same for both. So pressure amplitude is also same for both |
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| 34. |
Which of the following statements are correct?A. Changes inair temperature have no effect on the speed of sound.B. Changes in air pressure have no effect on the speed of sound.C. The speed of sound in water is higher that in air.D. The speed of sound in water is lower than in air. |
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Answer» Correct Answer - B::C `v=sqrt((gammaRT)/(M))` Change in temperature affects the velocity of sound is air but as long as temperature remains same change in pressure has no effect. `v=sqrt((B)/(P))` Bulk modulus of water is very high, so velocity of sound in water is higher than that in air. |
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| 35. |
An observer `A` is moving directly towards a stationary sound source while another observer `B` is moving away from the source with the same velocity. Which of the following statements are correct?A. Average of freqeuncies recorded by `A` and `B` is equal to natural frequency of the sourceB. Wavelength of wave received by `A` is less than that of waves received by `B`.C. Wavelength of waves received by two observers will be same.D. Both the observers will observe the wave travelling with same speed. |
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Answer» Correct Answer - A::C Let velocity of each observer be `u` as shown in the figure. `_Ararr^u` `_Brarr^u` Then freqeuency received by `A` will be `n_1=n_0((v+u)/(v))` where `n_0` is natural frequency of the source and `v` is sound propagation velocity. The frequency received by `B` will be `n_2=n_0((v-u)/(v))` Since `((n_1+n_2))/(2)=n_0`, therefore, option (a) is correct. |
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| 36. |
A boy is walking away from a well at a speed of `1.0 m//s` in a direction at right angles to the wall. As he walks, he below a whistle steadily. An observer towards whom the boy is walking hears `4.0` beats per second. If the speed of sound is `340 m//s` , what is the frequency of the whistle?A. 480 HzB. 680 HzC. 840 HzD. 1000 Hz |
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Answer» Correct Answer - B The frequency of direct sound of whistle heard by observer is `n_1=(v)/(v-v_S)n=(340)/(340-1)xxn=(340)/(339)n` ..(i) Frequency of sound of whistle reflected by wall is `n_2=(v)/(v+v_S)n=(340)/(341)n` .(ii) Given, `n_1-n_2=4` Therefore, `(340)/(339)n-(340)/(341)n=4` `impliesn=680Hz` |
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| 37. |
A sound wave of frequency n travels horizontally to the right with speed with speed c. It is reflected from a broad wall moving to the left with speed v. The number of beats heard by a stationary observer to the left of the wall isA. zeroB. `(n(c+v))/(c-v)`C. `(nv)/(c-v)`D. `(2nv)/(c-v)` |
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Answer» Correct Answer - D Initially wall behaves as an approaching observer, so frequency of sound reaching the wall is `n_1=(c+v)/(c )n` while reflecting, the wall behaves as an approaching source, so frequency received by stationary observer is `n_2=(c)/(c-v)n_1=(c)/(c-v)xx(c+v)/(c)n=(c+v)/(c-v)n` Direct frequency received by observer in n. the number of beats is `x=n_2-n=(c+v)/(c-v)n-n=(2nv)/(c-v)` |
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| 38. |
A tuning fork of frequency 380 Hz is moving towards a wall with a velocity of `4(m)/(s)` Then the number of beats heard by a stationary listener between direct and reflected sound will be (velocity of sound in air is `340(m)/(s)`) |
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Answer» Correct Answer - A The frequency of direct and reflected sound is same. |
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| 39. |
Statement I: Intensity of sound wave changes when the listener moves towards or away from the stationary source. Statement II: The motion of listener causes the apparent change in wavelength.A. Statement I is true, Statement II is true: Statement II is a correct explanation for statement I.B. Statement I is true, Statement II is true, Statement II is NOT a correct explanation for Statement I.C. Statement I is true, Statement II is falseD. Statement I is falce: Statement II is true |
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Answer» Correct Answer - C Intensity of sound at any point is dependent upon frequency as well as amplitude. As due to Doppler effect the apparent frequency changes, so intensity as perceived by the listener also changes. When listener is moving wavelength remains the same. |
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| 40. |
Statement I: If two people talk simulaneously and each creates an intensity of 60 dB at a point `P`, then total intensity level at the point P is 120 dB Statement II: sound level is defined on a non-linear scale.A. Statement I is true, Statement II is true: Statement II is a correct explanation for statement I.B. Statement I is true, Statement II is true, Statement II is NOT a correct explanation for Statement I.C. Statement I is true, Statement II is falseD. Statement I is falce: Statement II is true |
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Answer» Correct Answer - D `SL_f=10log((2I)/(I_0))` Where `I` is the intensity at point `P` due to one person. `60dB=SL=10log((I)/(I_0))` `impliesSL_f=10log2+10log((I)/(I_0))=63.01dB` |
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| 41. |
`Statement I:` A tuning fork is considered as a source of an acoustic wave of a single frequency as marked on its body. `Statement II:` The tuning fork cannot produce any of its harmonics due to its special nature of construction.A. Statement I is true, Statement II is true: Statement II is a correct explanation for statement I.B. Statement I is true, Statement II is true, Statement II is NOT a correct explanation for Statement I.C. Statement I is true, Statement II is falseD. Statement I is falce: Statement II is true |
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Answer» Correct Answer - A The tuning fork does produce harmonics, but the intensities of the harmonics are too weak to be effective, due to its special ditribution of mass and the use of prongs. |
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| 42. |
Statement I: The apparent freqeuncy which is the frequency as noted by an observer or an observing detection device of the acoustic wave that moves from the source to the observer propagating in a medium may be different from its true frequency. Statement II: A source in motion relative to an observer sends out less or more number of waves per metre distance in the medium and an observer in motion collects less or more number of waves per second that when both of them remain at rest relatively.A. Statement I is true, Statement II is true: Statement II is a correct explanation for statement I.B. Statement I is true, Statement II is true, Statement II is NOT a correct explanation for Statement I.C. Statement I is true, Statement II is falseD. Statement I is falce: Statement II is true |
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Answer» Correct Answer - C Both statements are true. The wavelength of a wave from a source moving towards or away from an observer changes due to the motion of the source. Similarly, for an observer moving towards or away from a source, it is the freqeuncy (number of waves passing him per second) that is affected by his motion. |
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| 43. |
A source of sonic oscillations with frequency `v_(0)=100Hz` moves at right angles to the wall with a velocity `u=0.17 m//s`. Two stationary receivers `R_(1)` and `R_(2)` are located on a straight line, coinciding with the trajectory of the source , in the following succession :` R_(1)-` source `-R_(2)-` wall. Which receiver registers the beatings and what is the beat frequency ? The velocity of sound is equal to `upsilon= 340 m//s`. |
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Answer» `R_2` is approached by the source and the image source at the same speed so `R_2` receives waves of the same frequency and hence, it does not register beats. `R_1` is approached by the image source while the source itself recedes from it Hence it it receives waves of different wavelength so it registers beats. `N_1`(apparent frequency of the source wave) `=(340-0)/(340-(0.17))xx1000=1000.5Hz` `N_2`(apparent frequency of the reflected wave) `=(340-0)/(340+0.17)xx1000=999.5Hz` Therefore beats registered`=1000.5-999.5=1.0Hz` |
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| 44. |
A whistle emitting a sound of frequency `440 H`z is tied to string of `1.5 m` length and rotated with an angular velocity of `20 rad//sec` in the horizontal plane. Then the range of frequencies heard by an observer stationed at a large distance from the whistle will `(v=330 m//s)` |
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Answer» `v_S=romega=1.5xx20=30(m)/(s)` `n_(min)=(v)/(v+v_S)n=(330)/(330+30)xx440Hz=403Hz` `n_(max)=(330)/(330-30)xx440=484Hz` |
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| 45. |
Two tuning forks A and B having a frequency of 500 Hz each are placed with B to the right of A. An observer in between the forks in moving towards B with a speed of 25 m/s. The speed of sound is 345 m/s and the wind speed is 5 m/s from A to B. Calculate the difference in the two frequencies heard by observer. |
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Answer» Let the fork at A be source `S_1` and the fork at B source `S_2`. For the sound from `S_1`: Apparent frequency`=f_1=f((c-V_0)/(c-V_S))` Here `c=345+5=350(m)/(s)` as wind and sound are in same direction `V_0=+25(m)/(s)` ,`V_S0(m)/(s)` `impliesf_1=500((350-25)/(350-0))=464.3Hz` for sound from source `S_2`:`c=345-5=340(m)/(s)` as wind is in oppsite direction to that of sound. `V_S=0(m)/(s)`,`V_0=-25(m)/(s)` `impliesf_2=f((c-V_0)/(c-V_S))=500((c-(-25))/(c-0))=(500xx365)/(340)` `=536.80Hz` `impliesf_2-f_1=(536.80-464.30)=72.5Hz` |
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| 46. |
The intensity level of two sounds are 100 dB and 50 dB. What is the ratio of their intensities?A. `10^1`B. `10^3`C. `10^5`D. `10^10` |
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Answer» Correct Answer - C `100=10log_(10)((I_1)/(I_0))` `50=10log_(10)((I_2)/(I_0))` or `(I_1)/(I_0)=10^10` and `(I_2)/(I_0)=10^5` Dividing, `(I_1)/(I_2)=10^5` |
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| 47. |
(a) The power of sound from the speaker of a radio is 20 mW. By turning the knob of volume control the power of sound is increased to 400 mW, What is the power increase in dB as compared to original power? (b) How much more intense is an 80 dB sound than a 20 dB whisper? |
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Answer» a. As intensity is power per unit area, for a given source `PmuI`. If `L_1` and `L_2` are the initial and final loudness levels, Then we have `L_2-L_1=10log((I_2)/(I_1))` The increase in loudness level is `triangleL=10log((P_2)/(P_1))=10log((400)/(20))` `impliestriangleL=10[log20]=13dB` b. Increase in loudness level of sound is `L_2-L_2=10log((I_2)/(I_1))` `80-20=10log((I_2)/(I_1))` `implies6=log((I_2)/(I_1))` `implies((I_2)/(I_1))=10^6` |
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| 48. |
The faintest sound the human ear can detect at a frequency of 1000 Hz correspond to an intensity of about `1.00xx10^-12(W)/(m^2)`, which is called threshold of hearing. The loudest sounds the ear can tolerate at this frequency correspond to an intensity of about `1.00(W)/(m^2)`, the threshold of pain. Detemine the pressure amplitude and displacement amplitude associated with these two limits. Take speed of sound`=342(m)/(s)` and density of air`=1.20(kg)/(m^3)` |
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Answer» Think about the quietest environment you have ever experienced. It is likely that the intensity of sound in even this quietest environment is higher than the threshold of hearing. Because we are given intesities and asked to calculate pressure and displacement amplitudes, this problem requires the concepts discussed in this section. To find the pressure amplitude at the threshold of hearing, `trinangleP_(max)=sqrt(2rhovI)` `=sqrt(2(1.20(kg)/(m^3))(342(m)/(s))(1.00xx10^(-12)(W)/(m^2))` `=2.87xx10^-5(W)/(m^2)` Calculate the corresponding displacement amplitude using `S_(max)=(triangleP_(max))/(rhovomega)=(2.87xx10^-5(N)/(m^2))/((1.20(kg)/(m^3))(342(m)/(s))(2pixx1000Hz))` `=1.11xx10^(-11)m` In a similar manner, one finds that the loudest sounds the human ear can tolerate correspond to a pressure amplitude of `28.7(N)/(m^2)` and a displacement amplitude equal to `1.11xx10^-5m`. Because atmospheric pressure is about `10^5(N)/(m^2)`, the result for the pressure amplitude tells us that the ear is sensitive to pressure fluctuations as small as 3 parts in `10^(10)`. The displacement amplitude is also a remarkably small number. If we compare this result for `S_(max)` to the size of an atom (about `10^(-10)`m), we see that the ear is an extremely sensitive detector of sound waves. |
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| 49. |
An increase in intensity level of 1 dB implies an increase in intensity of (given anti `log_(10)0.1=1.2589`)A. `1%`B. `3.01%`C. `26%`D. `0.1%` |
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Answer» Correct Answer - C Intensity level is decibel is given by `L=10log_(10)((I)/(I_0))` `L+1=10log_(10)((I_1)/(I_0))` Subtracting, `1=10log_(10)((I_1)/(I_0))-10log_(10)((I)/(I_0))` or `(1)/(10)=log_(10)((I_1)/(I))` or `0.1=log_(10)((I_1)/(I))` or `(I_1)/(I)=1.26` |
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| 50. |
Calculate the velocity of sound in air at NTP. The density of air at NTP is `(1.29g)/(L)`. Assume air to be diatomic with `gamma=1.4`. Also calculate the velocity of sound in air at `27^@C`. |
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Answer» Velocity of sound in air given by `v=sqrt((gammaP)/(rho))=sqrt((1.4xx1.013xx10^5(N)/(m^2))/(1.29(kg)/(m^3)))=331.6(m)/(s)` We can see that the velocity of sound is proportional to the square root of absolute temperature. Hence, `(v_2)/(v_1)=sqrt((T_2)/(T_1))impliesv_2=v_1sqrt((T_2)/(T_1))=331.6sqrt((273+27)/(273))=347.6(m)/(s)` |
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