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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 car is moving along X-axis with a velocity `v=20m//s`. It sounds a whistle of frequency 660 Hz. If the speed of sound is `340m//s`. The apparent frequency heard by the observer `O` (shown in figure) is:A. `680Hz`B. `640 Hz`C. `700Hz`D. `720Hz` |
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Answer» Correct Answer - A `n=(v)/(v-v_Scos60^@)n` Here `v=340 m//s, v_S=20 m//s, n=660Hz` `n=(340)/(340-20(1)/(2))xx660=(340)/(330)xx660=680Hz` |
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| 2. |
A source 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 approaching him and after it crosses him will be source is approaching him and after it crosses him will beA. `9:8`B. `8:9`C. `1:1`D. `9:10` |
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Answer» Correct Answer - A When source is approaching the observer, the frequency heard `n_a=((v)/(v-v_S))xxn=((340)/(340-20))xx1000=1063Hz` When source is receding the frequency heard `n_r=((v)/(v+v_S))xxn=(340)/(340-20)xx1000=944` `impliesn_a:n_r=9:8` Short tricks : `(n_a)/(n_r)=(v+v_S)/(v-v_S)=(340+20)/(340-20)=(9)/(8)` (speed of sound `v=340m//s`). |
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| 3. |
A string of length 2 m is fixed at both ends. If this string vibrates in its fourth normal mode with a frequency of 500 Hz then the waves wouldf travel on it is with a velocity ofA. `125 m//s`B. `25m//s`C. `500 m//s`D. `1000 m//s` |
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Answer» Correct Answer - C For string `lamda=(2l)/(p)` where `p=` No. of loops`=`Order of vibration Hence for fourth mode `p=4implieslamda=(l)/(2)` hence `v=nlamda=500xx(2)/(2)=500Hz` |
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| 4. |
A fork A has frequency `2%` more than the standard fork and B has a frequency `3%` less than the frequency of same standard frok. The forks A and B when sounded together produced 6 beats/s. The frequency of fork A isA. `116.4 Hz`B. `120 Hz`C. `122.4 Hz`D. `238.8 Hz` |
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Answer» Correct Answer - C The frequency of `A,n_A=n+(2)/(100)n` And the frequency of `B,n_B=n-(3)/(100)n` According to question `n_A-n_B=6` `:. (n+(2)/(100)n)-(n-(3)/(100)n)=6` or `(5)/(100)n=6impliesn=(600)/(5)=120Hz` The frequency of A `n_A=(n+(2)/(100)n)=120+(2)/(100)xx120=122.4Hz` |
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| 5. |
The freuquency of tuning forks A and B are respectively `3%` more and `2%` less than the frequency of tuning fork `C`. When A and B are simultaneously excited, 5 beats per second are produced. Then the frequency of the tuning fork `A` (in Hz)` IsA. `98`B. `100`C. `103`D. `105` |
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Answer» Correct Answer - C Let `n` the frequency of fork `C` then `n_A=n+(3n)/(100)=(103n)/(100)` and `n_B=n-(2n)/(100)=(98)/(100)` but `n_A-n_B=5implies(5n)/(100)=5impliesn=100Hz` `:. n_A=((103)(100))/(100)=103Hz` |
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| 6. |
A wire of length l having tension T and radius `r` vibrates with fundamental frequency `f`. Another wire of the same metal with length `2l` having tension `2T` and radius `2r` will vibrate with fundamental frequency :A. `f`B. `2f`C. `(f)/(2sqrt2)`D. `(f)/(2)sqrt2` |
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Answer» Correct Answer - C `f=(1)/(2l)sqrt((T)/(mu))` If radius is doubled and length is doubled, mass per unit length will become four times. Hence `f=(1)/(2xx2l)sqrt((2T)/(4mu))=(f)/(2sqrt2)` |
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| 7. |
An open organ pipe of length L resonates at fundamental frequency with closed organ pipe. Find the length of closed organ pipe.A. `2L`B. `L//2`C. `Lsqrt2`D. `Lsqrt3//2` |
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Answer» Correct Answer - B `f_0=f_cimplies(v)/(2L)=(v)/(4L_c)impliesL_c=L//2` |
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| 8. |
A closed organ pipe and an open organ pipe are tuned to the same fundamental frequency. The ratio of their lengths isA. `1:2`B. `2:1`C. `2:3`D. `4:3` |
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Answer» Correct Answer - A Let `l_1` and `l_2` be the lengths of closed and open pipes respectively. (neglecting end correction) `l_1=(lamda_1)/(4)implieslamda_1=4l_1` and `l_2=(lamda_2)/(2)implieslamda_2=2l_2` Given `n_1=n_2` so `(v)/(lamda_1)=(v)/(lamda_2)implies(v)/(4l_1)=(v)/(2l_2)=(l_1)/(l_2)=(1)/(2)` |
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| 9. |
Thw two nearest harmonics of a tube closed at one end and open at other end are 220 Hz and 260 Hz. What is the fundamental frequency of the system?A. `20 Hz`B. `30 Hz`C. `40 Hz`D. `10 Hz` |
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Answer» Correct Answer - A Two successive frequencies of closed pipe `(nv)/(4l)=220` `((n+2)v)/(4l)=260` Dividing (ii) by (i) we get `((n+2))/(n)=(260)/(220)=(13)/(11)impliesn=11` so from (i), `(11xxv)/(4l)=220` `implies(v)/(4l)=20Hz` So fundamental frequency `=20Hz` |
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| 10. |
The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm, the length of the open organ pipe isA. `13.2 cm`B. `8 cm`C. `12.5 cm`D. `16 cm` |
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Answer» Correct Answer - B Fundamental frequency of open organ pipe `f_0=(v)/(2l_1)` third harmonic of a closed organ pipe `f_3=(3v)/(4l_2)` It is given `f_0=f_3` `(v)/(2l_1)=(3v)/(4l_2)` or `I_1=(2)/(3)l_2=(40)/(3)cm=13.3cm` |
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| 11. |
Tube `A` has both ends open while tube `B` has one closed, otherwise they are identical. The ratio of fundamental frequency of tube `A and B` isA. `1:2`B. `1:4`C. `2:1`D. `4:1` |
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Answer» Correct Answer - C `n_A=(v)/(2l),n_B=(v)/(4l)implies(n_A)/(n_B)=2:1` |
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| 12. |
The same progressive wave is represented by two graphs I and II. Graph I shows how the displacement `y` varies with the distance x along the wave at a given time. Graph II shows how y varies with time t at a given point on the wave. The ratio of measurements AB to CD, marked on the curves represents:A. wave number kB. wave speed vC. frequency vD. angular frequency `omega` |
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Answer» Correct Answer - B `AB=lamda` `CD=T` `(AB)/(CD)=(lamda)/(T)=flamda=v` i.e., wave speed |
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| 13. |
A tuning fork is used to produce resonance in glass tuve. The length of the air column in the tube can be adjusted by a variable piston. At room temperature of `27^@C` two succesive resonance are produced at 20 cm and 73 cm column length. If the frequency of the tuning fork is 320 Hz. the velocity of sound is air at `27^(@)C` isA. `300 m//s`B. `330m//s`C. `350m//s`D. `339 m//s` |
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Answer» Correct Answer - D Velocity of sound `v=2f_0(l_2-l_1)` `=2xx320xx(0.73-0.20)` `=640xx0.53=339.2 m//s`. |
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| 14. |
The displacement `y` of a partcle in a medium can be expressed as, `y = 10^(-6) sin ((100t + 20x + (pi)/(4))m` where `t` is in second and `x` in meter. The speed of the wave isA. `2000m//s`B. `5m//s`C. `20m//s`D. `5pim//s` |
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Answer» Correct Answer - B `v=(co-efficent of t)/(co-efficent of x)=(omega)/(k)=(100)/(20)=5 m//s` |
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| 15. |
Assertion: Transverse waves are not produced in liquids and gases. Reason: Light waves are transverse waves.A. if both the assertion and reason are true and reson is a true explanation of the assertion.B. if both the assertion and reason are true but the reason is not the correct explanation of assertion.C. If the assertion is true but reason is false.D. If both the assertion and reason are false. |
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Answer» Correct Answer - A Transverse waves travel in the form of crest and troughs involving change in shape of the medium. As liquids and gases do not possess the elasticity of shape, therefore transverse waves cannot be produced in liquid and gases. Also light wave is one example of transverse wave. |
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| 16. |
Two closed organ pipes of length 100 cm and 101 cm 16 beats is 20 sec. When each pipe is sounded in its fundamental mode calculate the velocity of sound `A. `303ms^(-1)`B. `332ms^(-1)`C. `323.2ms^(-1)`D. `300ms^(-1)` |
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Answer» Correct Answer - C Number of beats per second `n=(16)/(20)=(4)/(5)impliesn=n_1-n_2=(v)/(4)((1)/(l_1)-(1)/(l_2))` `implies(4)/(5)=(v)/(4)((1)/(1)-(1)/(1.01))=(0.01v)/(4xx1.01)` `v=(16xx101)/(5)=323.2ms^-1` |
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| 17. |
The second overtone of an open pipe A and closed pipe B have the same frequencies at a given temperature. Both pipes contain air. The ratio of fundamental frequency of a to the fundamental frequency of B isA. `3:5`B. `5:3`C. `5:6`D. `6:5` |
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Answer» Correct Answer - B Second overtone of open pipe `=(3V)/(2l_1)` Second overtone of closed pipe `=(5V)/(4l_2)` Since, ratio of frequency are same `(3V)/(2l_1)=(5V)/(4l_2)implies(l_1)/(l_2)=(4xx3)/(2xx5)=(6)/(5)` Now, the ratio of fundamental frequencies: `(f_1)/(f_2)=((V)/(2l_1))/((V)/(4l_2))implies(2l_2)/(l_1)=10:6=5:3` |
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| 18. |
A source of sound gives five beats per second when sounded with another source of frequency `100s^-1`. The second harmonic of the source together with a source of frequency `205s^-1` gives five beats per second. What is the ferquency of the source?A. `105s^-1`B. `205s^-1`C. `95s^-1`D. `100s^-1` |
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Answer» Correct Answer - A Frequency of the source `=100+-5=105Hz` or `95Hz`. Second harmonic of the source `=210Hz` or `190Hz`. As the second harmonic gives 5 beats/sec with sound of frequency 205 Hz, the sound harmonic should be 210 Hz. `implies`frequency of the source `=105Hz` |
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| 19. |
The two pipes are submerged in sea water, arranged as shown in figure. Pipe A with length `L_A=1.5m` and one open end, contains a small sound source that sets up the standing wave with the second lowest resonant frequency of that pipe. Sound from pipe A sets up resonance in pipe B, which has both ends open. The resonance is at the second lowest resonant frequency of pipe B. The length of the pipe B isA. 1mB. 1.5 mC. 2 mD. 3 m |
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Answer» Correct Answer - C For pipe A, second resonant frequency is third harmonic thus `f=(3V)/(4L_A)` For pipe B, second resonant frequency is second harmonic thus `f=(2V)/(2L_B)` Equating `(3V)/(4L_A)=(2V)/(2L_B)` `impliesL_B=(4)/(3)L_A=(4)/(3)xx(1.5)=2m` |
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| 20. |
A string is stretched between fixed points separated by `75.0cm`. It is observed to have resonant frequencies of `420 Hz` and `315 Hz`. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string isA. `105 Hz`B. `155 Hz`C. `205 hz`D. `10.5 hz` |
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Answer» Correct Answer - A Two consecutive resonant frequencies for a string fixed at both ends will be `(nv)/(2l)` and `((n+1)v)/(2l)` `implies((n+1)v)/(2l)-(nv)/(2l)=420-315` `implies(v)/(2l)=105Hz`. Which is the minimum resonant frequency. |
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| 21. |
Two sources of sound `A` and `B` produces the wave of `350 Hz`. They vibrate in the same phase. The particel `P` is vibrating under the influence of these two waves, if the amplitudes at the point P produced by the two waves is `0.3mm` and `0.4mm` then the resultant amplitude of the point `P` will be when `AP-BP=25cm` and the velocity of sound is `350m//sec`.A. `0.7mm`B. `0.1mm`C. `0.2mm`D. `0.5mm` |
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Answer» Correct Answer - D `lamda=(v)/(n)=(350)/(350)=1m=100cm` Also path difference `(Delta x)` between the waves at the point of observation is `AP-BP=25cm`. Hence `Delta phi=(2pi)/(lamda)(Delta x)=(2pi)/(1)xx((25)/(100))=(pi)/(2)` `impliesA=sqrt((a_1)^2+(a_2_^2))=sqrt((0.3)^2+(0.4)^2)=0.5mm` |
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| 22. |
A point source emits sound equally in all directions in a non-absorbing medium. Two point `P` and `Q` are at distance of `2 m` and `3 m` respectively from the source. The ratio of the intensities of the wave at `P` and `Q` is :A. `9:04:00 AM`B. `2:03:00 AM`C. `3:02:00 AM`D. `4:09:00 AM` |
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Answer» Correct Answer - A Intensity of sound `I=(P)/(4pir^2)` or `1prop(1)/(r^2)` or `(I_1)/(I_2)=((r_2)/(r_1))^2` Here `r_1=2m,r_2=3m` Substituting the values, we have `(I_1)/(I_2)=((3)/(2))^2=(9)/(4)` |
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| 23. |
Two small boats are `10m` apart on a lake. Each pops up and down with a period of 4.0 seconds due to wave motion on the surface of water. When one boat is at its highest point, the other boat is its lowest point. Both boats are always within a single cycle of the waves. The speed of the waves is:A. `2.5m//s`B. `5.0m//s`C. `14m//s`D. `40m//s` |
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Answer» Correct Answer - B Distance between boats`=10m` `implieslamda=20m` time period `T=4sec` `:. V=(lamda)/(T)=(20m)/(4sec)=5m//s` |
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| 24. |
Two identical sound `S_1` and `S_2` reach at a point P is phase. The resultant loudness at point P is `n` dB higher than the loudness of `S_1` the value of n is :A. `2`B. `4`C. `5`D. `6` |
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Answer» Correct Answer - D Let `a` be the amplitude due to `S_1` and `S_2` individually. Loudness due to `S_1=I_1=Ka^2` Loudness due to `S_(1)+S_(2)=I=K(2a)^(2)=4I_(I)` `:. n=10log_(10)((4l_1)/(l_1))` `=10log_(10)(4)=6` |
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| 25. |
`S_1` and `S_2` are two coherent sources of sound having no initial phase difference. The velocity of sound is `330m//s`. No minimal will be formed on the line passing through `S_2` and perpendicular to the line jouning `S_1` and `S_2`. If the frequency of both the sources is:A. `50Hz`B. `60Hz`C. `70Hz`D. `80Hz` |
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Answer» Correct Answer - A For minimum `Delta x=(2n-1)(lamda)/(2)` between maximum possible path difference `=`distance beween the sources `=3m` For no minimum `(lamda)/(2)gt3implieslamdagt6` `:. f=(V)/(lamda)lt(330)/(6)=55`. if `:. flt55Hz` no minimum will occur. |
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| 26. |
The velocity of sound waves in air is `330m//s`. For a particluar sound in air, a path difference of `40 cm` is equivalent to a phase difference of `1.6pi`. The frequency of this wave isA. `165 hz`B. `150 Hz`C. `660 Hz`D. `330 Hz` |
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Answer» Correct Answer - C Phase difference `=(2pi)/(lamda)xx` Path difference `implies=1.6pi=(2pi)/(lamda)xx40implieslamda=50cm=0.5m` `impliesv=nlamdaimplies330=0.5xxnimpliesn=660Hz` |
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| 27. |
A point source of power `50pi` watts is producing sound waves of frequency `1875Hz`. The velocity of sound is `330m//s`, atmospheric pressure is `1.0xx10^(5)Nm^(-2)`, density of air is `(400)/(99pi)kgm^-3`. Then the displacement amplitude at `r=sqrt(330)`m from the point source isA. `0.5mum`B. `0.2mum`C. `1mum`D. `2mum` |
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Answer» Correct Answer - C `P_0=BKS`, `k=(2pi)/(lamda)`, `lamda=(v)/(f)`, `v=sqrt((B)/(rho))` Using above, we get `S_0=(P_0)/(2rhovpif)=(5)/(2xx1xx330xx3.14xx1875)` `=1mu` meter |
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| 28. |
A point source of power `50pi` watts is producint sound waves of frequency `1875Hz`. The velocity of sound is `330m//s`, atmospheric pressure is `1.0xx10^(5)Nm^(-2)` density of air is `1.0kgm^(-3)`. Then pressure amplitude at `r=sqrt(330)m` from the point source is (using `pi=22//7`:A. `5Nm^(-2)`B. `10Nm^(-2)`C. `15Nm^(-2)`D. `20Nm^(-2)` |
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Answer» Correct Answer - A `I=(P_0^2_)/(2rhoV)implies(P)/(4pir^2)=(P_0^2)/(2rhoV)`, where P,`P_0`,`V` are power pressure amplitude and velocity respectively. `impliesP_0=sqrt((PrhoV)/(2pir^2))=sqrt((50pixx1xx330)/(2pixx330))=5` |
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| 29. |
The diagram below shows the propagation of a wave. Which points are in same phase? .A. `F and G`B. C and FC. B and GD. B and F |
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Answer» Correct Answer - D Points B and F are in same phase as they are `lamda` distance apart. |
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| 30. |
If the speed of the wave shown in the figure is `330m//s` in the given medium then the equation of the wave propagating in the positive x-direction will be (all quantities are in M.K.S units)A. `y=0.05sin2pi(4000t-12.5x)`B. `y=0.05sin2pi(4000t-122.5x)`C. `y=0.05sin2pi(3300t-10x)`D. `y=0.05sin2pi(3300x-10t)` |
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Answer» Correct Answer - C Here `A=0.05m`,`(5lamda)/(2)=0.025implieslamda=0.1m` Now standard equation of wave `y=Asin(2pi)/(lamda)(vt-x)impliesy=0.5sin2pi(3300t-10x)` |
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| 31. |
A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode. Amplitude at the centre of the string is 4 mm. Minimum distance between the two points having amplitude 2 mm is:A. `1m`B. `75cm`C. `60cm`D. `50cm` |
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Answer» Correct Answer - B Equation of standing wave `y=Asinkxcosomegat` `y=A` as amplitude is 2A `A=2Asinkx` `(2pi)/(lamda)x=(pi)/(6)impliesx_1=(1)/(4)m` and `(2pi)/(lamda).x=(pi)/(2)+(pi)/(3)` `impliesx_1=1.25mimpliesx_2-x_1=1m` |
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| 32. |
In the experiment for the determination of the speed of sound in air using the resonance column method, the length of the air column that resonates in the fundamental mode, with a tuning fork is `0.1m`. When this length is changed to `0.35m`, the same tuning fork resonates with the first overtone. Calculate the end correction.A. `0.012 m`B. `0.025 m`C. `0.05 m`D. `0.024 m` |
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Answer» Correct Answer - B Let x the end correction then accoeding to question. `(v)/(4(l_1+x))=(3v)/(4(l_2+x))impliesx=2.5cm=0.025m`. |
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| 33. |
Assertion: Water waves in a river are not polarized. Reason: Water waves are longitudinal in nature.A. if both the assertion and reason are true and reson is a true explanation of the assertion.B. if both the assertion and reason are true but the reason is not the correct explanation of assertion.C. If the assertion is true but reason is false.D. If both the assertion and reason are false. |
| Answer» Correct Answer - A | |
| 34. |
The equation for the vibration of a string fixed at both ends vibrating in its third harmonic is given by `y=2cm` `sin[(0.6cm^-1)x]cos[(500pis^-1)t]`. The length of the string isA. `24.6 cm`B. `12.5 cm`C. `20.6 cm`D. `15.7 cm` |
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Answer» Correct Answer - D Wave number `K=(2pi)/(lamda)=0.6cm^-1` `:. (lamda)/(2)=(pi)/(0.6)cm` `:. l=(3lamda)/(2)3((pi)/(0.6))cm=15.7cm` |
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| 35. |
If the frequency of a wave is increased by `25%`, then the change in its wavelength will be: (medium not changed)A. `20%` increaseB. `20%` decreaseC. `25%` increaseD. `25%` decrease |
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Answer» Correct Answer - B Since, the medium has not changed, speed of wave remains same. `impliesv=flamda=`constant `f_1lamda_1=f_2lamda_2` `impliesf_1lamda_1=(1.25f_1)lamda_2` (`:.` frequency increased by `25%`) `lamda_2=(lamda_1)/(1.25)implies_2` decreases. `%` change in wavelength `=(lamda_1-lamda_2)/(lamda_1)xx100` `(lamda_1-(lamda_1)/(1.25))/(lamda_1)xx100=(0.25)/(1.25)xx100=(100)/(5)=20%` |
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| 36. |
A string of 7 m lengt has a mass of 0.035 kg. If tension in the string is 60.5 N, then speed of a wave on the string isA. `77m//s`B. `102 m//s`C. `110m//s`D. `165 m//s` |
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Answer» Correct Answer - C `v=sqrt((T)/(m))impliesv=sqrt((60.5)/((0.035//7))=110 m//s.` |
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| 37. |
A string of 7 m length has a mass of 0.035 kg. If tension in the string is 60.5 N. Then speed of a wave on the string isA. `77m//s`B. `102 m//s`C. `110m//s`D. `165 m//s` |
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Answer» Correct Answer - C `v=sqrt((T)/(m))impliesv=sqrt((60.5)/(((0.035)/(7))))=110m//s` |
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| 38. |
Beats are produced by two waves given by `y_1 = a sin2000 pit` and `y_2 =asin2008 pit`. The number of beats heard per second isA. zeroB. oneC. fourD. eight |
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Answer» Correct Answer - C Number of beats per second `=n_1-n_2` `omega_1=2000pi=2pin_1impliesn_1=1000` and `omega_2=2008pi=2pin_2impliesn_2=1004` Number of heats heard per sec`=1004-1000=4` |
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| 39. |
The frequencies of two sound sources are 256 Hz and 260 Hz, At `t=0` the intesinty of sound is maximum. Then the phase difference at the time `t=1//16` sec will beA. zeroB. `pi`C. `pi//2`D. `(pi)/(4)` |
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Answer» Correct Answer - C Time arrival between two consecutive beats `T=(1)/(n_1-n_2)=(1)/(260-250)=(1)/(4)`sec so, `(1)/(16)=(T)/(4)`sec By using time difference `=(T)/(2pi)xx`phase difference `implies(T)/(4)=(T)/(2pi)xxphiimpliesphi=(pi)/(2)` |
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| 40. |
Two loudspeakers `L_1` and `L_2` driven by a common oscillator and amplifier, are arranged as shown. The frequency of the oscillator is gradually increased from zero and the detector at D records a series of maxima and minima. If the speed of sound is `330ms^-1` then the frequency at which the first maximum is observed isA. `165Hz`B. `330Hz`C. `496Hz`D. `660Hz` |
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Answer» Correct Answer - B Path difference between the wave reaching at D `Delta x=L_2P-L_1P=sqrt(40^2+9^(2))-40=41-40=1m` For maximum `Delta x=(2n)(lamda)/(2)` For first maximum `(n=1)implies1=2(1)(lamda)/(2)implieslamda=1m` `impliesn=(v)/(lamda)=330Hz` |
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| 41. |
The faintest sound the human ear can detect at a frequency of ` kHz` (for which ear is most sensitive) corresponds to an intensity of about `10^(-12)w//m^(2)`. Assuming the density of air `cong1.5kg//m^(3)` and velocity of sound in air `cong300m//s`, the pressure amplitude and displacement amplitude of the sound will be rspectively ____`N//m^(2)` and ____`m`.A. `3xx10^(-5)Pa, (1)/(3pi)xx10^(-10)m`B. `2xx10^(-5)Pa, (2)/(3pi)xx10^(-10)m`C. `5xx10^(-5)Pa, (1)/(pi)xx10^(-10)m`D. `5xx10^(-5)Pa, (4)/(3pi)xx10^(-10)m` |
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Answer» Correct Answer - A `I=(P_m^2)/(2rhoV)` `10^(-12)=(P_m^2)/(2(1.5)(300))` `P_m=3xx10^(-5)Pa` `P_m=BAK=(rhoV^2) A (2pi)/(v//f)` `P_m=(2pirhoVf)A` `3xx10^-5=(2pi)(1.5)(300)(10^3)A` `A=1/3pixx10^(-10)m` |
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| 42. |
A wave moving with constant speed on a uniform string passes the point `x=0` with amplitude `A_0`, angular frequency `omega_0` and average rate of energy transfer `P_0`. As the wave travels down the string it gradually loses energy and at the point `x=l`, the average rate of energy transfer becomes. At the point `x=l`. Angular frequency and amplitude are respectively:A. `omega_(0)` and` (A_(0))/(sqrt2)`B. `(omega_0)/(sqrt2)` and `A_0`C. less that `omega_(0)` and `A_(0)`D. `(omega_0)/(sqrt2)` and `(A_0)/(sqrt2)` |
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Answer» Correct Answer - A `P=(1)/(2)muomega^2A^2V` and `v=sqrt((T)/(mu))` will not change as both T and `mu` are constant. `omega` will also not change as it is property of the source only that is causing the wave motion. Hence to make power half the amplitude Becomes `(A_0)/(sqrt2)` |
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| 43. |
The ratio of intensities between two coherent sound sources is `4:1` the difference of loudness in decibels between maximum and minimum intensities, when they interfere in space, isA. `10 log 2`B. `20 log 3`C. `10 log 3`D. `20 log 2` |
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Answer» Correct Answer - B Given `(I_1)/(I_2)=(4)/(1)` We know `Ipropa^2` `:. (a_1^2)/(a_2^2)=(I_1)/(I_2)=(4)/(1)` or `:. (a_1)/(a_2)=(2)/(1)` `(I_(max))/(I_(min))=((a_1+a_2)^2)/((a_1-a_2)^2)=((2+1)/(2-1))^2` `=((3)/(1))^2=(9)/(1)` Therefore, difference of loudness is given by `L_1-L_2=10log_(10).(I_(max))/(I_(min))=10log(9)` `=10log3^2=20log3` |
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| 44. |
Assertion : It is not possible to have interference between the waves produced byu two violins. Reason : For interferene of two waves the phase differnce between the wave must remain constant.A. If both assertion and reason are true and reason is the correct expleanation of assertion.B. If both assertion and reason are true and reason is not the correct expleanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - A Since the initial phase difference between the two waves coming from different violins changes, therefore, the waves produced by two different violins does not interfere because two waves interfere only when the phase difference between them remain constant throughout. |
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| 45. |
Two sound waves have phase difference of `60^(@)`, then they will have the path difference of:A. `2lamda`B. `(lamda)/(2)`C. `(lamda)/(6)`D. `(lamda)/(3)` |
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Answer» Correct Answer - C Phase difference `=(2pi)/(lamda)xx` path difference path difference `Delta =(lamda)/(2pi)xxphi=(lamda)/(2pi)xx(pi)/(3)=(lamda)/(6)` |
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| 46. |
In the above question the ratio of the first overtone of A to first overtone of B is :A. `5:6`B. `6:5`C. `10:9`D. `9:10` |
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Answer» Correct Answer - C First overtone of open pipe `=(2v)/(2l_1)` , First overtone of closed pipe `=(3v)/(4l_2)` Required ratio `implies(v)/(l_1)xx(4l_2)/(3v)=(4)/(3)xx(l_2)/(l_1)` From above question `(l_2)/(l_1)=(5)/(6)` So ratio will be `(4)/(3)xx(5)/(6)=(10)/(9)=10:9` Hence (c ). |
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| 47. |
Two vibrating tuning forks produce progressive waves given by `y_(1)=sin500pit` and `y_(2)=2sin506pit`. Number of beats produced per minute is:A. `360`B. `180`C. `3`D. `60` |
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Answer» Correct Answer - B The reach the solution the given wave equation ust be compared with standard equation of progressive wave, so `y_1=4sin500pit` .(i) `y_2=2sin506pit` .(ii) Comparing Eqs. (i) and (ii) with `y=asinomegat` ..(iii) we have `omega_1=500pi` `impliesf_1=(500pi)/(2pi)=250beats//s` and `omega_2=506pi` `impliesf_2=(506pi)/(2pi)=253beats//s` Thus, number of beats produced `=f_2-f_1-253-250-3 beats//s` `=3xx60(beats)/(min)=180beats//min` |
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| 48. |
When beats are produced by two progressive waves of nearly the same frequency, which one of the following if correct?A. the particle vibrate simple harmonically with the frequency equal to the difference in the component frequencies.B. The amplitude of vibration at any point changes simple harmonically with a frequency equal to the difference in the frequencies of the two waves.C. The frequency of beats depends upon the position, where the observer is.D. The frequency of beats changes as the time progresses. |
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Answer» Correct Answer - B As `y=A_bsin(2pin_(av)t)` where `A_b=2Acos(2pin_At)` where `n_A=(n_1-n_2)/(2)` Hence, the amplitude f vibration at any point changes simple harmonically with a frequency equal to the difference in the frequencies of the two waves. |
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| 49. |
Wavelength of two notes in air are 1 m and `1(1)/(164)`m. Each note produces 1 beats/s with a third note of a fixed frequency. The speed of sound in air isA. `330 m//s`B. `340m//s`C. `350 m//s`D. `328 m//s` |
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Answer» Correct Answer - A `lamda_1=1m, lamda_2=(165)/(164)m` `f_1=(C )/(1)`,`f_2=(164C)/(165)` `f-f_2=2` `(C )/(165)=2` , `C=330 m//s` |
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| 50. |
The wavelength of two sound waves are 49 cm and 50 cm respectively. If the room temperature is `30^@C` then the number of beats produced by them is approximately (velocity of sound in air `0^@C=332m//s`A. 6B. 10C. 14D. 18 |
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Answer» Correct Answer - C `y=332sqrt((303)/(273))` Beat frequency `=f_1-f_2=v((1)/(lamda_1)-(1)/(lamda_2))` `=332sqrt((303)/(273))((1)/(49)-(1)/(50))xx100cong14` |
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