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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.
| 51. |
An unknown frequency `x` produces 8 beats per seconds with a freuquency of `250` Hz and 12 beats with `270Hz`. Source then `x` isA. `258Hz`B. `242Hz`C. `262Hz`D. `282Hz` |
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Answer» Correct Answer - A Since source of frequency `x` gives 8 beats per second with frequency 250 , its possible frequency are 258 or 242. As source of frequency x gives 12 beats per second with a frequency 270 Hz. Its possible frequencies 282 or 258 Hz. The only possible frequency of x which gives 8 beats with frequency 250 Hz also 12 beats per second with 258 Hz. |
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| 52. |
A sound wave of frequency 440 Hz is passing through air. An `O_2` molecule (mass`=5.3xx10^(-26)Kg)` is set in oscillation with an amplitude of `10^-6m` speed at the centre of its oscillation is :A. `1.70xx10^-5m//s`B. `17.0xx10^-5m//s`C. `2.76xx10^-5m//s`D. `2.77xx10^-5m//s` |
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Answer» Correct Answer - C `v_(max)=omega_nA` `=(2pif)A=(2pi)(440)(10^-6)` `=2.76xx10^-5m//s` |
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| 53. |
The length of the wire shown in Fig. Between the pulley and fixed support is `1.5m` and mass is `12.0g` the frequency of vibration with which the wire vibrate two loops leaving the middle point of the wire between the pulleys at rest isA. `10Hz`B. `30Hz`C. `100Hz`D. `70Hz` |
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Answer» Correct Answer - C `f=(2)/(2l)sqrt((T)/(mu))` or `f=(1)/(l)sqrt((T)/(mu))` or `f=sqrt((Tl)/(Ml^(2))` or `f=sqrt((T)/(Ml))` or `v=sqrt((18xx10xx1000)/(12xx1.5))Hz=100Hz`. |
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| 54. |
The equation of a wave is represented by `y=10^-4sin[100t-(x)/(10)]`. The velocity of the wave will beA. `100m//s`B. `250m//s`C. `750m//s`D. `1000 m//s` |
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Answer» Correct Answer - D `v=(omega)/(k)=(100)/(1//10)=1000 m//s.` |
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| 55. |
A sine wave of wavelength `lamda` is travelling in a medium. The minimum distance between the two particles always having same speed is.A. `(lamda)/(4)`B. `(lamda)/(3)`C. `(lamda)/(2)`D. `lamda` |
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Answer» Correct Answer - C The minimum distance between the two particles having same speed is `(lamda)/(2)` |
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| 56. |
A wave travelling in positive X-direction with `A=0.2m` has a velocity of `360 m//sec` if `lamda=60m`, then correct exression for the wave isA. `y=0.2sin[2pi(6t+(x)/(60))]`B. `y=0.2sin[pi(6t+(x)/(60))]`C. `y=0.2sin[2pi(6t-(x)/(60))]`D. `y=0.2sin[pi(6t-(x)/(60))]` |
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Answer» Correct Answer - C A wave travelling in positive x-direction may be represnted as `y=Asin.(2pi)/(lamda)(vt-x)`. On putting values `y=0.2sin.(2pi)/(60)(360t-x)` `impliesy=0.2sin.2pi(6t-(x)/(60))` |
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| 57. |
A bir is singing on a tree and a man is hearing at a distance `r` from the bird. Calculate the displacement of the man towards the bird so that the loudness heard by man increases by `20dB`. [Assume that the motion of man is along the joining the bird and the man].A. `(3r)/(4)`B. `(3r)/(10)`C. `(2r)/(3)`D. `(9r)/(10)` |
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Answer» Correct Answer - D Loudness `beta=10log_(10).(I)/(I_0)` `:. beta_2-beta_1=10log_(10).(I_2)/(I_1)` and `I=(P)/(4pir^2)` `:. (I_2)/(I_1)=(r_1^2)/(r_2^2)` `:. (beta+20)-beta=10log_(10).(r^2)/(r_2^2)=20log_(10).(r )/(r_2)` `implies(r )/(r_2)=10impliesr_2=0.1r` `:.` Shift `=r-0.1r=0.9r=(9r)/(10)` |
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| 58. |
A is singing a note and at the same time B is singing a note with exactly one eighth the frequency of the note of A. The energies of two sounds are equal, the amplitude of the note of B isA. same that of AB. twice as that of AC. four times as that of AD. eight times as that of A |
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Answer» Correct Answer - D It is given that energy remains the same Hence `E_A=E_B` energy `propa^2n^2implies(a_B)/(a_A)=(n_A)/(n_B)" "( ":.energy same ")` `:. ((a_A)/(a_B))^2=((n_B)/(n_A))^2` Given, `n_A=n,n_B=(n)/(8)` `:. (a_A)/(a_B)=(n//8)/(n)=(1) rArr a_(B)=8a_(A)=8a`. |
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| 59. |
The tension in a piano wire is `10 N`. The tension ina piano wire to produce a node of double frequency isA. `20 N`B. `120 N`C. `10 N`D. `40 N` |
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Answer» Correct Answer - D The frequency of note emitted by the wire, `n=(1)/(2l)sqrt((T)/(m))` `m=` mass `m` per unit length of wire and `T=` tension, and `l=` length of wire. `(n_1)/(n_2)=sqrt((T_1)/(T_2))` Given `T_1=10N,n_1=n` and `n_2=2n` `implies(n)/(2n)=sqrt((10)/(T_2))impliesT_2=10xx4=40N` |
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| 60. |
A transverse wave is represented by `y=Asin(omegat-kx)`. For what value of the wavelength is the wave velocity equal to the maximum particle velocity?A. `piA//2`B. `piA`C. `2piA`D. `A` |
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Answer» Correct Answer - C Wave velocity `v=(lamda)/(T)=(omegalamda)/(2pi)` Maximum particle velocity `(v_(max))_(rho)=Aomega` Given `v=(v_(max))_(rho)` `(omegalamda)/(2pi)=A omega rArr lambda=2piA` |
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| 61. |
A wave travelling in the `+ve` x-direction having displacement along y-direction as 1`m`, wavelength `2pi` m and frequency of `1//pi` Hz is represented byA. `y=sin(x-2t)`B. `y=sin(2pix-2pit)`C. `y=sin(10pix-20pit)`D. `y=sin(2pix+2pit)` |
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Answer» Correct Answer - A `y=asin(kx-omegat)` `=sin[(2pi)/(2pi)x-2pixx(1)/(pi)t]=sin(x-2t)` |
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| 62. |
A wave in a string has an amplitude of `2 cm`. The wave travels in the `+ve` direction of x axis with a speed of 1`28 ms^-1` and it is noted that `5` complete waves fit in `4 m` length of the string. The equation describing the wave isA. `y=(0.02)msin(7.85x+1005t)`B. `y=(0.02)msin(15.7x-2010t)`C. `y=(0.02)msin(15.7x+2010t)`D. `y=(0.02)msin(7.85x-1005t)` |
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Answer» Correct Answer - D Find the parameters and put in the general wave equation. Here, `A=2cm` direction `=+ve` x-direction `v=128ms^-1` and `5lamda=4` Now, `k=(2pi)/(lamda)=(2pixx5)/(4)=7.85` and `v=(omega)/(k)=128ms^-1` `impliesomega=vxxk=128xx7.28=1005` As `y=Asin(kx-omegat)` `:. y=2sin(7.85x-1005t)` `=(0.02)msin(7.85x-1005t)` |
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| 63. |
Three sound waves of equal amplitudes have frequencies `(v - 1), v, (v + 1)`. They superpose to give beats. The number of beats produced per second will be :A. `3`B. `2`C. `1`D. `4` |
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Answer» Correct Answer - B net beat frequency `=LCM` of individual beat frequencies `=LCM` of `"[(n,n-1),(n,n+1),(n-1,n+1)]"` `=LCM ` of `(1,1,2)` `=2Hz` So, no. of beats per second`=2` |
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| 64. |
Two sinusoidal waves with same wavelengths and amplitudes travel in opposite directions along a string with a speed `10ms^-1`. If the minimum time interval between two instant when the string is flat is `0.5s`, the wavelength of the waves isA. `25m`B. `20m`C. `15m`D. `10m` |
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Answer» Correct Answer - D Minimum time interval between two instant when the string is flat`=(T)/(2)=0.5secimpliesT=1sec` Hence `lamda=vxxT=10xx1=10m`. |
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| 65. |
A sinusoidal wave moving along a string is shown twice in the figure. As crest A travels in the positive direction of an x axis by distance `d=6.0` cm in `4.0m`. The tick marks along the axis are separated by `10cm`, height `H=6.00mm`. The wave equation isA. `y=(3mm)sin[16x-2.4xx10^2t]`B. `y=(3mm)sin[16x+2.4xx10^2t]`C. `y=(3mm)sin[8x+2.4xx10^2t]`D. `y=(3mm)sin[8x-2.4xx10^2t]` |
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Answer» Correct Answer - A The amplitude A is half of the 6.00 mm vertical range shown in the figure, that is ,`A=3.0mm` The speed of the wave is `v=d//t=15m//s`, where `d=0.060m` and `t=0.0040s`. The angular wave number is `k=(2pi)/(lamda)` where `lamda=0.40m`. Thus, `k=(2pi)//(lamda)=16(rad)//(m)`. The angular frequency is found from `omega=kv=(16(rad)/(m))(15(m)/(s))=2.4xx10^2 rad//s`. We choose the minus sign (between kx and `omegat`) in the argument of the sine function because the wave is shown traveling to the right (in the `+x` direction). Therefore, with SI units understood, we obtain `y=Asin[kx-omegat]=(3mm)sin[16x-2.4xx10^2t]` |
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| 66. |
A `100 Hz` sinusoidal wave is travelling in the posotive x-direction along a string with a linear mass density of `3.5 xx10^(-3) kg//m` and a tension of `35 N`. At time `t= 0`, the point `x=0` , has maximum displacement in the positive y-direction. Next when this point has zero displacement, the slope of the string is `pi//20`. Which of the following expression represent (s) the displacement of string as a function of `x`(in metre) and `t` (in second)A. `y=0.025sin(200pit-2pix+(pi)/(4))`B. `y=0.025sin(2pix-200pit+(pi)/(2))`C. `y=0.025sin(2pix-200pit+(pi)/(4))`D. `y=0.025sin(200pit-2pix+(pi)/(2))` |
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Answer» Correct Answer - D The speed of wave in string is `V=sqrt((T)/(mu))=sqrt((35)/(3.5xx10^-3))=100 m//s` `lamda=(v)/(f)=(100)/(100)=1m` Let us assume that `t=0`,`x=0` and `y=+A` `because` `theta=(pi)/(2)` Hence `y=Asin(200pit-2pix+(pi)/(2))` It is given that `(dely)/(delx)=(pi)/(20)` at `t=(T)/(4)=(1)/(400)sec` `(dely)/(delx)]_(t=(1)/(400)S)=A2picos(200pit-2pix+(pi)/(2))` or `2Api=(pi)/(20)impliesA=(1)/(40)=0.025m` Wave equation is `y=0.025sin(200pit-2pix+(pi)/(2))` |
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| 67. |
When two tuning forks A and B are sounded together x `beats//s` are heard. Frequency A is `n`. Now when one prong of B is loaded with a little wax, the number of beats/s decreases. The frequency of fork B isA. `n+x`B. `n-x`C. `n-x^2`D. `n-2x` |
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Answer» Correct Answer - A There aare two possible frequencies `n+x` and `n-x` But only `n+x` satisfied the given conditions. |
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| 68. |
If two tuning fork A and B are sounded together they produce 4 beats per second. A is then slightly loaded with wax, they produce 2 beats when sounded again. The frequency of A is 256. The frequency of B will beA. `250`B. `252`C. `260`D. `262` |
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Answer» Correct Answer - B `n_a=`known frequency `=256Hz`,`n_B=?` `x=4bps`, which is decreasing after loading (i.e., `xdarr`) Also known tuning fork is loaded so `n_Adarr` Hence `n_Adarr-n_B=xdarr` .(i) correct `n_B-n_Adarr=xdarr` .(ii) Wrong `impliesn_B=n_A-x=256-4=252Hz` |
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| 69. |
An organ pipe is closed at one end has fundamental frequency of 1500 Hz. The maximum number of overtones generated by this pipe which a normal person can hear isA. `14`B. `13`C. `6`D. `9` |
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Answer» Correct Answer - C Critical hearing frequency in Nth mode then frequency of vibration `n=((2N-1)v)/(4l)=(2N-1)n_1` (where `n_1=`fundamental frequency of vibration) Hence `20,00=(2N-1)xx1500impliesN=7.1=7` Also, in closed pipe Number of over tones`=`(No. of mode of vibration)`-1=7-1=6`. |
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| 70. |
A wave representing by the equation `y = a cos(kx - omegat)` is suerposed with another wave to form a stationary wave such that point `x = 0` is a node. The equation for the other wave isA. `y=asin(kx-omegat)`B. `y=-acos(kx+omegat)`C. `y=-acos(kx-omegat)`D. `y=-asin(kx-omegat)` |
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Answer» Correct Answer - B Since the point `x=0` is a node and reflection is taking place from point `x=0` This means that reflection must be taking place from the fixed end and hence the reflected ray must suffer an additional phase change of `pi` or a path change of `(lamda)/(2)` So, if `y_(Incident)=acos(kx-omegat)` `rArr y_("reflected")=acos(-kx-omegat+pi)` `=-acos(omegat+kx)`. |
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| 71. |
A wave representing by the equation `y = a cos(kx - omegat)` is suerposed with another wave to form a stationary wave such that point `x = 0` is a node. The equation for the other wave isA. `y=asin(kx-omegat)`B. `y=-acos(kx+omegat)`C. `y=-cos(kx-omegat)`D. `y=-asin(kx-omegat)` |
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Answer» Correct Answer - B Since the point `x=0` is a node and reflection is taking place from point `x=0` This means that reflection nust be taking place from the fixed end and hence the reflected ray must suffer an additional phase change of `pi` or a path change of `(lamda)/(2)`. So , if `y_(i ncident)=acos(kx-omegat)` `impliesy_(refl ected)=acos(-kx-omegat+pi)=-acos(omegat+kx)` |
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| 72. |
A wire is stretched between two rigid supports. It is observed that wire resonates at the frequencies `f_1,f_2,f_3` and `f_4(f_4gtf_3,f_2gtf_1)` forming 2,3,4 and 5 loops respectively. The ratio of any two resonance frequencies will be minimum for difference in the loops to be:A. `1`B. `2`C. `3`D. zero |
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Answer» Correct Answer - C Let `f_1=(mv)/(2L)=(2v)/(2L)`. Similarly `f_4=((n+3)v)/(2L)=(5v)/(2L)` `f=(f_1)/(f_4)` will be minimum `f=(f_1)/(f_4)=(n)/(n+3)` `f` will be minimum for difference in loops `=3`. |
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| 73. |
The paritcle displacement (in cm) in a stationary wave is given by `y(x,t)=2sin(0.1pix)cos(100pit)`. The distance between a node and the next antinode isA. `2.5 cm`B. `7.5 cm`C. `5 cm`D. `10 cm` |
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Answer» Correct Answer - C `y(x,t)=2sin(0.1pix)cos(100pit)` compare with `y=Asin(Kx)cosomegat` `K=0.1pi=(2pi)/(lamda)implieslamda=2cm` The distance between a node and the next antinode `=(pi)/(4)=(20)/(4)=5cm` |
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| 74. |
A wire having a linear mass density `5.0xx10^(3) kg//m` is stretched between two rigid supports with tension of 450 N. The wire resonates at a frequency of `420 Hz`. The next higher frequency at which the same wire resonates is` 480 N`. The length of the wire isA. `2.0 m`B. `2.1 m`C. `2.5 m`D. `3 m` |
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Answer» Correct Answer - B Two consecutive frequencies are 420 Hz and 480 Hz. So the fundamental frequency will be `60 Hz`. `60=(1)/(2xxl)sqrt((450)/(5xx10^-3))impliesl=2.1m` Hence (b). |
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| 75. |
For the stationary wave `y=4sin((pix)/(15))cos(96pit)`, the distance between a node and the next antinode isA. `7.5`B. `15`C. `22.5`D. `30` |
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Answer» Correct Answer - A Comparing given equation with standard equation `y=2asin.(2pix)/(lamda)cos.(2pivt)/(lamda)` gives us `(2pi)/(lamda)=(pi)/(15)implieslamda=30` distance between nearest node and antinodes `=(lamda)/(4)=(30)/(4)=7.5`. |
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| 76. |
In a resonance pipe the first and second resonance are obtained at depths 22.7 cm and 70.2 respectively. What will be the end correction?A. `1.05 cm`B. `115.5 cm`C. `92.5 cm`D. `113.5 cm` |
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Answer» Correct Answer - A For end correction `x`, `(l_2+x)/(l_1+x)=(3lamda//4)/(lamda//4)=3` `x=(l_2-3l_1)/(2)=(70.2-3xx22.7)/(2)=1.05cm` |
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| 77. |
A transverse wave is propagating along `+x` direction. At `t=2` sec the particle at `x=4`m is at `y=2` mm. With the passage of time its `y` coordinate increases and reaches to a maximum of 4 mm. The wave equation is (using `omega` and `k` with their usual meanings)A. `y=4sin[omega(t+2)+k(x-2)+(pi)/(6)]`B. `y=4sin[omega(t+2)+k(x)+(pi)/(6)]`C. `y=4sin[omega(t-2)-k(x-4)+(5pi)/(6)]`D. `y=4sin[omega(t-2)-k(x-4)+(pi)/(6)]` |
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Answer» Correct Answer - D SHM equation of the particle at `x=4` is `y=4sin[omega(t+2)+(pi)/(6)]` Wave equation, replacing t by `[t-((x-4)/(v))]` is `y=4sin(omega(t-((x-4))/(v)-2)+(pi)/(6))` `=4sin[omega(t-2)-k(x-4)+(pi)/(6)]` |
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| 78. |
`S_(1)` and `S_(2)` are two coherent sources of sound of frequency `110Hz` each they have no initial phase difference. The intensity at a point P due to `S_(1)` is `I_0` and due to `S_2` is `4I_0`. If the velocity of sound is `330m//s` then the resultant intensity at P isA. `I_0`B. `9I_0`C. `3I_0`D. `8I_0` |
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Answer» Correct Answer - C The wavelength of sound source`=(330)/(110)=3` metre. The phase difference between interfering waves at P is `=Delta phi=(2pi)/(lamda)(S_2P-S_1P)=(2pi)/(3)(5-4)=(2pi)/(3)` `:.` Resultant intensity at `P=I_0+4I_0+2sqrt(I_0)sqrt(4I_0)cos(2pi)/(3)=3I_0` |
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| 79. |
The minimum intensity of sound is zero at a point due to two sources of nearly equal frequencie4s whenA. two sources are vibrating in opposite phaseB. The amplitude of two sources are equalC. at the point of observation, the amplitudes of two S.H.M. produced by two sources are equal and both the S.H.M. are slong the same straight lineD. Both the sources are in the same phase |
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Answer» Correct Answer - C If two waves of nearly equal frequency superpose, they give beats if they both travel in straight line and `I_(min)=0` if they have equal amplitudes. |
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| 80. |
There are `10` sound sources each producing intensity I at a point independently. The are incoherent. Average intensity of sound at that point will be:A. `I`B. `10I`C. `100I`D. `0` |
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Answer» Correct Answer - B since the sources are incoherent there will be no sustained interference. Hence at the given point, Average intensity `=I_0=I_1+I_2+I_3+`….`+I_(10)=10I` |
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| 81. |
A table is revolving on its axis at `5` revolutions per second. A sound source of frequency `1000 Hz `is fixed on the table at `70 cm` from the axis. The minimum frequency heard by a listener standing at a distance from the table will be (speed of sound `=352m//s`).A. `1000Hz`B. `1066Hz`C. `941Hz`D. `352Hz` |
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Answer» Correct Answer - C For source `v_S=romega=0.70xx2pixx5=22m//sec` Minimum frequency is heard when the source is receding the man. It is given by `n_(min)=n(v)/(v+v_S)` `1000xx(352)/(352+22)=941Hz` |
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| 82. |
The equation of a wave is given by (all quantity expressed in S.I units) `y=5sin10pi(t-0.01x)` along the x-axis. The magnitude of phase difference between the points separated by a distance of 10 m along x-axis isA. `pi//2`B. `pi`C. `2pi`D. `pi//4` |
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Answer» Correct Answer - B The magnitude of phase difference between the points separated by distance 10 metres. `=kxx10=[10pixx0.01]xx10=pi` |
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| 83. |
In order to double the frequency of the fundamental note emitted by a stretched string, the length is reduced to `(3)/(4)`th of the original length and the tension is changed. The factor by which the tenstion is to be changed isA. `3//8`B. `2//3`C. `8//9`D. `9//4` |
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Answer» Correct Answer - D `n=(1)/(2l)sqrt((T)/(m))impliesnprop(sqrtT)/(l)` `implies(T_2)/(T_1)=((n_2)/(n_1))^2((l_2)/(l_1))^2=(2)^2((3)/(4))^2=(9)/(4)` |
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| 84. |
The equation of a stationary wave is given by `y=6sin(pi)//(x)cos40pit` Where `y` and `x` are given in cm and time `t` in second. Then the amplitude of progressive wave isA. `6cm`B. `3cm`C. `12cm`D. `2cm` |
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Answer» Correct Answer - B `2A=6` or `A=3cm` |
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| 85. |
If amplitude of a wave is represented by `A=(c )/(a+b-c)`. Then the resonance will occur when.A. `b=-c//2`B. `b=0` and `a=c`C. `b=(-a)/(2)`D. none |
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Answer» Correct Answer - B For resonance,t he amplitude should be infinitely large. The combination `(b=0` and `a=c`) best satisfies this condition. |
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| 86. |
Assertion : The basic of Laplace correction was that, exchange of heat between the region of compression and rarefaction in air is not possible. Reason : Air is a bad conductor of heat and velocity of soundin air is large.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 - C Accordin to Laplace, the changes in pressure and volume of a gas, when sound waves propagated through it, are not isothermal, but adiabatic. A gas is a bad conductor of heat. It does not allow the free exchange of heat between compressed layer, rarefied layer and surrounding. |
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| 87. |
Assertion : When we start filling an empty bucket with water, the pitch of sound produced goes on decreasing. Reason : The frequency of man voice is usually higher than of woman.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 - D A bucket can be treated as a pipe closed at one end. The frequency of the note produced `=(v)/(4L)`, here L the bucked is filled with water L decreases, hence frequency increases. Therefore, frequency of pitch of sound produced goes on increasing. Also, the frequency of woman voice is usually higher than of man. |
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| 88. |
At a point, beats frequenc of `n` Hz is observed it means:A. medium particles. At the point, are vibrating with frequency `n` Hz.B. amplitude of vibrations changes simple Harmonically with frequency `n` Hz at that point onlyC. at that zero intensity is observed `2n` times per second.D. none of these |
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Answer» Correct Answer - D Beats are observed when intensity at a point varies with time and beat frequency is equal to the frequency of oscillations of intensity at the point. Intensity at a point not only depends upon the frequency of medium particle also. Therefore, beat are observed when amolitude of oscillation of medium particles varies with time. If the beat frequency at a point is equal to `n`, it means, at that point amplitude of oscillation of medium particles varies with frequency `n`. Amplitude of vibrations changes not only at the point of observation, but at all the point. Therefore, (a) and (b) are wrong. Since the frequency of variation of intensity is observed `n` times per second therefore, the maximum of intensity is observed `n` times per second and the intensity become zero `n` times per second. Hence (c ) is also wrong. Obviously, only option (d) is correct. |
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| 89. |
Two tunig forks of frequency `250Hz` and `256Hz` produce beats. If a maximum is observed just now, after how much time the next maximum is observed at the same place?A. `1//18sec`B. `1//24s`C. `1//6sec`D. `1//12sec` |
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Answer» Correct Answer - C Beat frequency `=256-250=6s^-1` Thus maximum is heard after every `(1)/(6)` second. |
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| 90. |
If we study the vibration of a pipe open at both ends, then the following statements is not trueA. Open end will be antinodeB. Odd harmonics of the fundamental frequency will be generatedC. All harmonics of the fundamental frequency will be generatedD. Pressure change will be maximum at both ends |
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Answer» Correct Answer - D Statement (d) is not true, because the open ends of the pipe are open to atmosphere hence there will be no pressure change. |
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| 91. |
In a stationary wave all the particlesA. on either side of a node vibrate in same phaseB. in the region between two nodes vibrate in same phaseC. in the region between two antinodes vibrate in same phase.D. Of the medium vibrate in same phase. |
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Answer» Correct Answer - B In statinary wave all the particle in one particular segment (i.e., between two nodes) vibrates in the same phase. |
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| 92. |
When a stationary wave is formed then its frequency isA. Same as that of the individual wavesB. Twice that of the individual wavesC. Half that of the individual wavesD. none of the above |
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Answer» Correct Answer - A If `y_("Incident")=asin(omega-kx)` and `y_("stationary")=asin(omegat)coskx` then it is crease that frequency of both is same `(omega)`. |
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| 93. |
The string fixed at both ends has standing wave nodes for which distance between adjacent nodes is `x_1`. The same string has another standing wave nodes for which distance between adjacent nodes is `x_2`. If `l` is the length of the string then `x_2//x_1=l(l+2x_1)`. What is the difference in numbers of the loops in the two cases?A. `1`B. `2`C. `3`D. `4` |
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Answer» Correct Answer - B Let no. of loops formed in first case `=n` `x_1n=l` .(i) Let no. of loops formed in second case `=(n+k)` `x_2(n+k)l` .(ii) From (i) and (ii) `x_2[(l)/(x_1)+k]=l,(x_2)/(x_1)=(l)/(l+kx_1)` Comparing with `(x_2)/(x_1)=(l)/(l+2x_1),k=2` |
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| 94. |
A string is fixed at both ends. The tension in the string and density of the string are accurately known but the length and the radius of cross section of the string are known with some errorl If maximum errors made in the measurements of length and radius are `1%` and `0.5%` respectively them what is the maximum possible percentage error in the calculation of fundamental frequencyof the that string ?A. `1.5%`B. `2%`C. `2.5%`D. `1%` |
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Answer» Correct Answer - A `f=(1)/(2l)sqrt((T)/(mu))=(1)/(2l)sqrt((T)/(rhos))=(1)/(2l)sqrt((T)/(rhopir^2))=(1)/(2//r ) sqrt((T)/(rhopi))` `(Deltaf)/(f)=-(Delta)/(l)-(Deltaf)/(r )` `((Deltaf)/(f))_(max)=1_0.5=1.5%` |
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| 95. |
A tunig fork whose frequency as given by mufacturer is `512 Hz` is being tested with an accurate oscillator it is found that the fork produces a beat of 2 Hz when oscillator reads 514 Hz but produces a beat of 6 Hz when oscillator reads 510 Hz. The actual frequency of fork isA. `508Hz`B. `512Hz`C. `516Hz`D. `518Hz` |
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Answer» Correct Answer - C The tuning fork whose frequency is being tested produces 2 beats with oscillator at `514 Hz`, therefore, frequency of tuning fork may either be `512` or `516`. with oscillator frequency of tuning fork may be either `516` or `504`. Therefore the actual frequency is 516 Hz which gives `2beats//sec` with 514 and `6 beats//sec` with `510 Hz`. |
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| 96. |
A person speaking normally produces a sound intensity of `40 dB` at a distance of `1 m`. If the threshold intensity for reasonable audibility is `20 dB`, the maximum distance at which he can be heard cleary is.A. `10 m`B. `5 m`C. `4 m`D. `20 m` |
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Answer» Correct Answer - A `dB=10log_(10)[(I)/(I_0)]` where `I_0=10^(-12)wm^-2` Since `40=10log_(10)[(I_1)/(I_0)]implies(I_1)/(I_0)=10^4` Also, `20log_(10)[(I_2)/(I_0)]implies(I_2)/(I_1)=10^2` `implies(I_2)/(I_1)=10^-2=(r_1^2)/(r_2^2)` `impliesr_2^2=100r_1^2impliesr_2=10m` |
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| 97. |
For a sound source of intensity `IW//m^(2)`, corresponding sound level is `B_0` decibel If the intensity is increased to `4I`, new sound level becomes.A. `2B_(0)dB`B. `(B_(0)+3)dB`C. `(B_(0)+6)dB`D. `4B_(0)dB` |
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Answer» Correct Answer - C If intensity is doubled the sound level is increased by `3.0dB`. i.e., it becomes `(B_0+3)` decibels. Hence when intensity is increased four times, level becomes `(B_0+6)` |
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| 98. |
In expressing sound intensity we take `10^(-12)(W)/(m^2)` as the reference level. For ordinary conversation the intensity level is about `10^-6(W)/(m^2)`. Expressed in decibel, this isA. `10^(6)`B. `6`C. `60`D. `log_(e)(10^6)` |
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Answer» Correct Answer - C Bel is `log_(10)((I)/(I_0))`. Here it is 6. In decibel It is 60. |
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| 99. |
A person is talking in a small room and the sound intensity level is `60dB` everywhere within the room. If there are eight people talking simultaneously in the room, what is the sound intensity level?A. `60dB`B. `69dB`C. `74dB`D. `81dB` |
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Answer» Correct Answer - B Let intensity due to a single person `=I` then `10log I//I_0=60` Also, intensity due to 8 person `=8I` final decibel level `=10log.((8I)/(I_0))=10(log.(I)/(I_0)+log8)` `=60+10log8=60+30log2=60+30(.3010)` `=69dB` |
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| 100. |
An open pipe is suddenly closed at one end with the result that the frequency of third harmonic of the closed pipe is found to be higher by `100 Hz` then the fundamental frequency of the open pipe. The fundamental frequency of the open pipe isA. `100 Hz`B. `300 Hz`C. `150 hz`D. `200 Hz` |
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Answer» Correct Answer - D `5xx(v)/(4l)-(v)/(l)=100` or `(v)/(l)((5)/(4)-1)=100` or `(v)/(4l)=100` or `(v)/(2l)=200` or `v=200Hz` |
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