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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. |
Assume that the sun rotates about an axis through its centre and perpendicular to the plane of rotatin of the earth about the sun. The appearange of the sun, from any one pont on the earth, is shown. Light belonging to a particular spectral line, as received from the points `A, B, C` and `D` on the edge of the sun, are analyzedA. Light from all four points have the same wavelength.B. Light from `C` has greater wavelength than the light from `D`.C. Light from `D` has greater wavelength than the light from `C`.D. Light from `A` has the same wavelength as the lighht from `B` |
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Answer» Correct Answer - C::D |
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| 2. |
A police car moving at `22 m//s`, chase a motoclist. The police man has horn at `176 H_(Z)`, While both of them move towards a stationary siran of frequency `165 H_(Z)`. Calulate the speed of the motorcyclist, if he does not observer any beats. (velocity of sound in air `= 330 m//s)` A. `33 m//s`B. `22 m//s`C. zeroD. `11 m//s` |
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Answer» Correct Answer - B :. `f_(2) = f_(2)` :. `176 ((330 - nu)/(330 - 22)) = 165 ((330 + nu)/(330))` Solving this equation we get, `nu = 22 m//s` |
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| 3. |
When a stretched string of length `L` vibrating in a particular mode, the distance between two nodes on the string is `l`. The sound produced in this mode of vibration constitutes the `nth` overtone of the fundamental frequency of the string.A. `L=(n+1)l`B. `L-(n-1)l`C. `L=nl`D. `L=(n+1//2)l` |
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Answer» Correct Answer - A |
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| 4. |
A cylindrical tube, open at both ends, has a fundamental frequency `f` in air . The tube is dipped vertically in water so that half of its length is in water. The fundamental frequency of the air column is now (a) `f // 2` (b) `3 f // 4` (C ) `f` (d) `2f` |
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Answer» Initially , the tube was open at both ends and then it is closed. `f_(0) = (nu)/2l_(0)` and `f _(C) = (nu)/(4l_(C))` Since, tube is half dipped in water , `l_(C) = (l_(0))/(2)` `f _(C) = (nu)/(4((l_(0))/(2))) = (nu)/(2l_(0)) = f_(0) =f` Hence, the correct option is (C ). |
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| 5. |
A cylinderical tube open at both ends, has a fundamental frequency f in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of air column is nowA. `4F`B. `2F`C. `F`D. `F//2` |
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Answer» Correct Answer - C |
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| 6. |
A source `S` and a detector `D` high frequency waves are a distance `d` apart on the ground. The direct wave from `S` is found to be in phase at `D` with the wave from `S` that is reflected from horizontal layer at an altitude `H`. The incident and reflected rayes make the same angle with the reflecting layer. When the layer rises a distance `h`, no signal is detected at `D` . Negle ct absorption in the atmosphere and find the relation between d,h, `H` and the wavelength `lambda` of the waves. |
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Answer» Correct Answer - A::B::D `2 sqrt(H^(2) + d^(2)//4) - d = n lambda` ..(i) Now, `2 sqrt((H + h)^(2) + d^(2)//4) - d = n lambda + (lambda)/(2)` ..(ii) Solving these two equations, we get `lambda = 2 sqrt(4(H + h)^(2) + d^(2)) - 2sqrt(4 H^(2) + d^(2))` |
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| 7. |
Two speakers `A` and `B`, placed `1m` apart, each produce sound waves of frequency `1800 Hz` in phase. A detector moving parallel to line of speakers distant `2.4 m` away detects a maximum intensity at `ɵ` and then at `P`. Speed of sound wave is A. `330 ms^(-1)`B. `360 ms^(-1)`C. `350 ms^(-1)`D. `340 ms^(-1)` |
| Answer» Correct Answer - B | |
| 8. |
Two audio speakers are kept some distance apart and are driven by the same amplifier system. A person is sitting at a place 6.0 m from one of the speakers and 6.4 m from the other. If the sound signal is continuously varied from 500 Hz to 5000 Hz, what are the frequencies for which there is a destructive interference at the place of the listener? Speed of sound in air` = 320 ms^-1` |
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Answer» Correct Answer - A::B::C::D The path difference of the two sound waves is given by `/_L=6.4-6.0=0.4m` The wavelength of either wave `lamda=V/p=320/p (m/s)` For destructive inteference `/_L=(2n+1)lamda/2` where n is an integers `or 0.4m =2+1/2xx320/p` `rarr p=n=320/0.4` `=800 (2n+1)/2 Hz` `=(2n+1)400Hz` Thus the frequency within the specified range which cause detructive inteference are 1200 Hz, 2000 Hz, 2800Hz, 3600Hz and 4400 Hz.` |
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| 9. |
The first overtone of an open orgen pipe beats with the first ouertone of a closed orgen pipe with a beat frequency of `2.2 H_(Z)` . The fundamental frequency of the closed organ pipe is `110 H_(Z)` . Find the lengths of the pipes . Speed of sound in air `u = 330 m//s` . |
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Answer» Let `l_(1)` and `l_(2)` are lengths of closed and open pipe -respectively. Fundamental frequency of closed pipe `= 110 Hz` `(v)/(4 l_(1)) = 110` `l_(1) = (v)/(4 xx 110) = (330)/(4 xx 110) = (3)/(4) m = 0.75 m` `2 . (v)/(2 l_(2)) - 3 . (v)/(4 l_(2)) = +-5` `(v)/(l_(2)) - 3 xx 110 = +-5` `(v)/(l_(2)) = 330 +- 5 = 335 or 325` `(v)/(l_(2)) = 335 rArr l_(2) = (v)/(335) = (330)/(335) = 0.98 m` `(v)/(l_(2)) = 325 rArr l_(2) = (v)/(325) = (330)/(325) = 1.02 m` Length of closed pipe `= 0.75 m` Length of open pipe `= 0.98 m or 1.02 m` |
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| 10. |
A source of sound operates at 2.0 kHz, 20 W emitting sound uniformly in all direction. The speed of sound in air is `340 m s^(-1)` at a distance of air is `1.2kgm ^(-3)` (a) What is the intensity at a distance of 6.0 m from the source ? (b) What will be the pressure amplitude at this point ? (c ) What will be the displacement amplitude at this point ? |
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Answer» Correct Answer - A::B::C::D (a) Here given, `V_(air) = 340 m//s` power ` =(e )/(t) = 20 W` `f =2,000 Hz, p = 1.2 kg//m^3` So, intensity, `I = (e )/(t.A) = (20)/(4pir^2) = (20)/(4 xx pi xx 6^2)` ` = 44 mw//m^2 [because r = 8m]` (b) We know that, `I = (p_0^2)/(2p V_(air))` `rArr P_0 = sqrt(1 xx 2p V_(air))` `= sqrt(2xx 1.2 xx 340 xx 44 xx 10^(-3))` `=6.0 N//m^2` (c ) We know that, `I = 2pi^2 S_0^2 upsilon^2 P V` where `S_0` = displacement amplitude `rArr S_0 = sqrt((I)/(2pi^2 P^2 p V_(air))` Putting the value we get, `S_0 = 1.2 xx 10^(-6) m` |
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| 11. |
A sound wave of frequency 10 kHz is travellilng in air with a speed of `340 ms^-1`. Find the minimum separation between two points where the phase difference is `60^@`. |
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Answer» The wavelength of the wave is `lamda=v/v=(340ms^-1)/(10xx10^3s^-1)=3.4cm` The wave number is `k=(2pi)/lamda=(2pi)/3.4cm^-1` The phase of the wave is `(kx-omegat)`. At any given instant the phase difference between two points at separatin d is kd. If this phase difference is `60^@` i.e. pi/3` radian, `pi/3=((2pi)/3.4cm^-1)d or d=3.4/6cm=0.57cm. |
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| 12. |
The fundamental frequency of a closed pipe is `220 H_(Z)`. (a) Find the length of this pipe. (b) The second overtone of this pipe has the same frequency as the third harmonic of an open pipe. Find the length of this open pipe. Take speed of sound in air `345 m//s`. |
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Answer» Correct Answer - A::B::C::D (a) `f_(1) = (nu)/(4l)` `:. l = (nu)/(4 f_(1)) = (345)/(4 xx 220)` `= 0.392 m` (b) `5((nu)/(4l_(c))) = 3 ((nu)/(2l_(0)))` `:. l_(0) = (6)/(5) l_(c) = ((6)/(5)) (0.392)` `= 0.47 m` |
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| 13. |
The separation between a node and the next antinode in a vibrating air column is 25 cm. If the speed of sound in air is `340 m s^-1`, find the frequency of vibration of the air column. |
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Answer» Correct Answer - C::D Distance betweenn two nodes or antinodes `rarr lamda/4=25cm` `rarr lamda=100cm=1m` `rarr n=v/lamda=340Hz` |
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| 14. |
`v_(rms), v_(av)` and `v_(mp)` are root mean square average and most probable speeds of molecules of a gas obeying Maxwellian velocity distribution. Which of the following statements is correct ?A. `v_(rms) gt v_(av) gt v_(mp) gt v_(s)`B. `v_(av) gt v_(mp) gt v_(rms) gt v_(s)`C. `v_(mp) gt v_(av) gt v_(rms) gt v_(s)`D. `v_(rms) gt v_(av) gt v_(s) gt v_(mp)` |
| Answer» Correct Answer - A | |
| 15. |
Under simuliar conditions of temperature and pressure, in which of the following gases the velocity of sound will be largest?A. `H_(2)`B. `N_(2)`C. `He`D. `CO_(2)` |
| Answer» Correct Answer - A | |
| 16. |
If `v_(m)` is the velocity of sound in moist air, `v_(d)` is the velocity of sound in dry air, under identical conditions of pressure and temperatureA. `v_(m) gt v_(d)`B. `v_(m) lt v_(d)`C. `v_(m) = v_(d)`D. `v_(m) . v_(d)=1` |
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Answer» Correct Answer - A Under indentical pressure and temperture condition, speed of sound in moist air is more than that in dry air i.e. `" "v_(m) gt v_(d)` |
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| 17. |
The velocity of sound in dry air is `V_(d)`, and in moist air it is `V_(m)`. The velocities are measured under the same conditions of temperature and pressure. Which of the following statements is fully correct?A. `V_(d)gtV_(m)` because dry air has lower density than moist air.B. `V_(d)ltV_(m)` because moist air has lower density than dry air.C. `V_(d) gt V_(m)` because the bulk modulus of dry air is greater than that of moist air.D. `V_(d) lt V_(m)` because the bulk modulus of moist air is greater than that of dry air. |
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Answer» Correct Answer - B |
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| 18. |
A man stands before a large wall at a distance of 50.0 m and claps his hands at regular intervals. Initially, the interval is large. He gradually reduces the interval and fixes it at a value when the echo of a clap merges with the next clap. If he has to clap 10 times during every 3 seconds, find the velocity of sound in air. |
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Answer» Correct Answer - A::C He has to clap 10 times in 3 seconds. so, time interval between two clap `=(3/(10 second))` so, the time taken to go the wall `=(v=(3/(20second))` We know `s=vt` `rarrV=s/t=50/(3/20)=333m/s` |
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| 19. |
A closed organ pipe resonates in its fundamental mode at a frequency of `200 H_(Z)` with `O_(2)` in the pipe at a certain temperature. If the pipe now contains `2` moles of `O_(2)` and `3` moles of ozone, then what will be fundamental frequency of same pipe at same temperature?A. `268.23 H_(Z)`B. `175.4 H_(Z)`C. `149.45 H_(Z)`D. none of these |
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Answer» Correct Answer - B `M = (n_(1) M_(1) + n_(2) M_(2))/( n_(1) + n_(2))` `= ((2) (32) + (3) (48))/(5) = 41.6` `f = (nu)/(4l)` or `f prop nu` But, `nu = sqrt((gammaRT)/(M))` or `nu prop (1)/sqrt(M)` :. `f prop (1)/sqrt(M)` :. `(f_(2))/(f_(1)) = sqrt((M_(1))/(M_(2)))` or `f_(2) = (sqrt((M_(1))/(M_(2)))) f_(1)` `= (sqrt((32)/(48))) (200) = 175.4 H_(Z)` |
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| 20. |
When the open organ pipe resonates in its fundamental mode then at the centre of the pipeA. the gas molecule undergo vitrations of maximum amplitude.B. the gas molecule are at restsC. the pessure of the gas is constantD. the pressure of the gas undergoes maximum variation |
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Answer» Correct Answer - B::D |
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| 21. |
A person riding a car moving at `72 km h^-1` sounds a whistle emitting a wave of frequency 1250 Hz. What frequency will be heard by another person standing on the road (a) in front of the car (b) behind the car ? Speed of sound in air `= 340 m s^-1` . |
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Answer» Correct Answer - A::B::C a. Given `v_s=72km/hour` `=20 m/s, F=1250Hz` Apparent frequency `=(340+0+0)/(340+0-20)xx1250` `=1328Hz` For second case Apparent frequency `=((340+0+0)/(340+0-(-20))xx1250` `=340/360xx1250=1181Hz |
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| 22. |
A train approaching a platform at a speed of `54 km h^-1` sounds a whistle. An observer on the platform finds its frequency to be 1620 Hz. The train passes the platform keeping the whistle on and without slowing down. What frequency will the observer hear after the train has crossed the platfrom ? The speed of sound in air `= 332 ms^-1` |
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Answer» Correct Answer - A::D Here given Apparent frequency =1620Hz so origiN/Al frequency of the train is given by `120=((332+0+0()/(332-16))f` `rarr =((1620xx317)/332))Hz` `(where p=origiN/Al frequency) So, apparent frequency of the tran observed by the observer in f `=((332+0+0)/332+15)xx(1620x317)/332)` `=317/347xx1620=1480Hz. |
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| 23. |
(i) A person hums in a wall and finds strong resonance at frequencies `60 Hz, 100Hz` and `140 Hz`. What is the fundamental frequency of the well? Explain? How deep is the wall ? (velocity of sound `= 344 m//`). (ii) A violin string `33cm` long vibrates at a fundamental frequency of `440 Hz`. Where shold a finger be pressed on the string board so that its decreased length causes it to vibrate in its fundamental mode at five fourth of its original frequnency. |
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Answer» Correct Answer - (i) fundamental frequercy `n = 20 Hz`, depth of the weil `= 4.3`, (ii) at `26.4 cm` from the end. (i) The well can be considered as an organ pipe closed at one end and open at the other. At the fundamental mode, `lambda//4 = l` (depth of the well in this case). The given frequencies happen to be `3rd, 5th` & `7th` harmonics of the fundamental frequency `n = 20 Hz`. Now use `v = nlambda` so as to obtain depth of the well `= 4.3 m`. (ii) Use the law of vibrating string ni `=` constant. To have the frequency `5//4` th of original, the length requireed `4//5` of `33cm`. Thus, the finger must be pressed at a point will divid the length of the string in a ratio `1:4` is at `26.4 cm` from the end. |
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| 24. |
A sound wave of frequency 100 Hz is travelling in air. The speed of sound in air is `350 ms^-1` (a) By how much is the phase changed at a given point in 2.5 ms ? (b) What is the phase difference at a given instant between two points separated by a distance of 10.0 cm along the direction of propagation ? |
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Answer» Correct Answer - A::B::C a. Here given `n=100, v=350 m/s` `rarr lamda(v/n)=350/100=3.5m` In 2.5 ms, the distance travelled by the particle is given by `Dx=(350xx2.5xx10^-3)` So, phase difference `phi=(2pi)/lamdaxxDx` `=((2pixx350xx2.5xx10^-3)/3.5)` `=(pi/2)` b. In the second case Given `/_eta=10 cm=10^-1m` `So, phi=(2pi)/x/_x` `=(2pixx10^-1)/(350/100)=(2pi)/35` |
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| 25. |
The equation of a travelling sound wave is `y = 6.0 sin (600 t - 1.8 x) where y is measured in `10^-5m`, t in second and x in metre. (a) Find the ratio of the displacement amplitude of the particles to the wavelength of the wave. (b) Find the ratio of the velocity amplitude of the particles to the wave speed. |
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Answer» Correct Answer - A::B::D Here given `r_y=6.0xx10^-5m` a. Given `2 pi/lamda=1.8K` lamda=((2pi)/1.8)` so, `r_y/lamda=(6.0x(1.8)xx106-5m/s)/((2pi))` b. Let velocity Amplitude `=V_y` `V=(dy)/(dt)` `=3600cos(600t-1.8x)` `xx10^-5m/s` Here `V_y=3600xx10^-5m/s` Again` lamda=(2pi)/1.8` `and T=(2pi)/600` `rarr wave speed =v` `=lamda/T=600/1.8=100/3 m/s` So, the ratio of `(V_y/v)` ltbr. `=(3600xx3xx10^-5)/1000` |
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| 26. |
The transvers displacement of a string (clamped at its both ends) is given by `y(x,t) = 0.06 sin ((2pi)/(3)s) cos (120 pit)` Where `x` and `y` are in `m` and `t` in `s`. The length of the string `1.5 m` and its mass is `3.0 xx 10^(-2) kg`. Answer the following : (a) Does the funcation represent a travelling wave or a stational wave ? (b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength. Frequency and speed of each wave ? Datermine the tension in the string. |
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Answer» Correct Answer - (a) Stationary wave (b) `I = 3 m, n = 60 Hz`, and `v = 180 ms^(-1)` for each wave (c) `648 N` |
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| 27. |
A wave is represented by the equation `y=A sin 314[t/(0.5s)-x/(100m)]` The frequency is `n` and the wavelength is `lamda`A. `n=2Hz`B. `n=100Hz`C. `lamda=2m`D. `lamda=100m` |
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Answer» Correct Answer - B::C::D |
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| 28. |
(i) The transverse displacement of a string (clamped at its two ends ) is given by `y(x,t)=0.06 sin[ (2pi)/(3)x] cos 120pit,` where x, y are in m and t is in s. Do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude ? Explain your answers. (ii) What is the amplitude of a point `0.375m` away from one end? |
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Answer» Correct Answer - All the points except the nodes on the string have the same frequency and phase, but not the same amplitude. (b) `0.042 m` |
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| 29. |
When a sound wave of wavelength `lamda` is propagating in a medium, the maximum velocity of the particle is equal to the velocity. The ampilitude of wave isA. `lamda`B. `lamda/2`C. `lamda/(2lamda)`D. `lamda/ (4lamda)` |
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Answer» Correct Answer - C `v_(max)= v rArrAomega=v rArr Axx2piv=vlamdaor A=lamda/(2pi)` |
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| 30. |
The equation `y=A cos^(2) (2pi nt -2 pi (x)/(lambda))` represents a wave withA. amplitude `a`, frequency `n` and wavelenth `lamda`B. amplitude `a` frequency `2n` and wavelenth `2lamda`C. amplitude `a//2`, frequency `2n` and wavelength `lamda`D. amplitude `a//2`, frequency `2n` and wavelenth `lamda//2` |
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Answer» Correct Answer - D |
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| 31. |
(a) A longitudinal wave propagating in a water-filled pipe has intensity `3.00xx10^(-6) W//m ^(2)` and frequency `3400 H_(Z)` . Find the amplitude `A` and wavelength `lambda` of the wave . Water has density `1000 kg//m ^(3)` and bulk modulus `2.18xx10^(9) Pa`. (b) If the pipe is filled with air at pressure `1.00 xx10^(5)` Pa and density `1.20 kg//m^(3)`, What will be the amplitude `A` and wavelength `lambda` of a longitudinal wave the same intensity and frequency as in part (a) ? (c ) In which fluid is the amplitude larger, water or air? What is the ratio of the two amplitude ? Why is this ratio so different from/ Conider air as diatomic. |
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Answer» Correct Answer - A::B::C::D (a) `nu = sqrt((B)/(rho)) = sqrt((2.18 xx 10^(9))/(1000))` `= 1476 m//s` `lambda = (nu)/(f) = (1476)/(3400) = 0.43 m` `I = (1)/(2) rho omega ^(2) A^(2) nu` `:. A = sqrt((2l)/(rho (2pi f)^(2) nu)` `= sqrt((2 xx3 xx10^(-6))/((1000) (2pixx3400)^(2)(1476)))` `= 9.44 xx10^(-11) m` (b) `nu = sqrt((gamma P)/(rho)) = sqrt(((1.4) (10^(5)))/((1.2)))` `= 341.56 m//s` `lambda = (nu)/(f) = (341.56)/(3400) = 0.1 m` `A = sqrt ((2l)/(rho (2 pi f)^(2) nu))` `= sqrt((2 xx3.0 xx10^(-6))/(1.2xx (2pixx3400)^(2)(341.56)))` (c) `(A_(air))/(A_(water)) = (5.66 xx 10^(-9))/(9.44 x10^(-11)) = 60` `I = (1)/(2) rho omega^(2) A^(2) nu` `rho` and `nu` are less in air . So , for same intensity `A` should be large. |
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| 32. |
The amplitude of a wave disturbance propagating in the positive x-direction is given by `y = (1)/((1 + x))^(2)` at time `t = 0` and by `y = (1)/([1+(x - 1)^(2)])` at `t = 2 seconds`, `x and y` are in meters. The shape of the wave disturbance does not change during the propagation. The velocity of the wave is ............... m//s`.A. `0.5`B. 1C. 2D. 4 |
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Answer» Correct Answer - A |
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| 33. |
A heavy unifrm rope hangs vertically from the ceiling, with its lower end free. A disturbance on the rope travelling upward from the lower end has a velocyt `v` at a distance `x` from the lower end.A. `v prop 1//x`B. `v prop x`C. `v prop sqrt(x)`D. `v prop 1//sqrt(x)` |
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Answer» Correct Answer - C |
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| 34. |
A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340Hz ) when the tube length is 25.5cm or 79.3cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects amy be neglected. |
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Answer» Correct Answer - `347 m s^(-1)` Hint : `v_(n) = ((2n - 1)v)/(4l) ; n = 1,2,3`…… for a pipe with one end closed |
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| 35. |
A long tube of length /= 25 cm and diameter equal to 2 cm is taken and at its mouth air is blown as shown in figure. The sound emitted by tube will have all the frequencies of the group (velocity of sound = 330 m/s ) A. 660, 1320, 2640 HzB. 660, 1000, 3300 HzC. 302, 684, 1320 HzD. 330, 990, 1690 Hz |
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Answer» Correct Answer - A `n=v/(2l)=(330)/(2xx25)xx100=660Hz` `" "n_(1)=2xx660=1320Hz` `" "n_(2)=3xx660=2460Hz` |
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| 36. |
An air column, closed at one end and open at the other end, resonates with a tuning fork of frequency `f` when its length is `45 cm, 99 cm` and at two other lengths in between these values. The wavelength of sound in the air column isA. `180 cm`B. `108 cm`C. `54 cm`D. `36 cm` |
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Answer» Correct Answer - D Resonance will occur at lengths `(lambda)/(4) , (3lambda)/(4) , (5lambda)/(4) , (7lambda)/(4)` ,… `(7lambda)/(4) - (lambda)/(4) = 99 - 45` `(3lambda)/(2) = 54 rArr lambda = 36 cm` |
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| 37. |
Two identical tubes `A` and `B` are kept in air and water respetively as shown. If the fundamental frequency of `A` is `f_(0)`, then thr fundamental frequency of `B` is A. `(f_(0))/(4)`B. `(f_(0))/(2)`C. `f_(0)`D. `2f_(0)` |
| Answer» Correct Answer - C | |
| 38. |
A tuning fork whose natural frequency is `440 H_(Z)` is placed just above the open end of a tube that contains water. The water is slowly drained from the tube while the tuning fork remains in place and is kapt vibrating. The sound is found to be echanced when the air column is `60 cm` long and when it is `100 cm` long . Find the speed of sound in air. |
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Answer» `(lambda)/(2) = (100 - 60) xx 10^(-2) m` `:. lambda = 0.8 m` Now, `nu = f lambda = 440 xx 0.8` `= 352 m//s` |
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| 39. |
A closed organ pipe of length `l = 100 cm` is cut into two unequal pieces. The fundamental frequency of the new closed organ pipe place is found to be same as the frequency of first overtone of the open organ pipe place. Datermine the length of the two piece and the fundamental tone of the open pipe piece. Take velocity of sound `= 320 m//s`. |
| Answer» Correct Answer - `20, 80 cm, 200 Hz` | |
| 40. |
`S_(1)`: Sound travel faster in water than in air. `S_(2)`: The fundamental frequency of closed organ pipe increases if its diameter is increased. `S_(3)`: Beats are created by superposition of two sound waves with the same amplitude moving in same directions with slightly different frequncies.A. TFTB. FTTC. FTFD. TFF |
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Answer» Correct Answer - A Beats are produced when wave move in the same direction . `upsilon_(f)=v/(4(l+epsilon))=v/(4(l+0.3d))` where d is diameter of pipe so, `upsilon` decreases. velocity `=sqrt(B/(rho))` where `rho` is a density of fluid |
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| 41. |
A tuning fork sends sound waves in air. If the temperature of the air increases, which of the following parameters will changes?A. Displacement amplitudeB. FrequencyC. WavelengthD. Time period |
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Answer» Correct Answer - C `v = f lambda," "v prop sqrt(T)` |
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| 42. |
Statement 1: In sound waves variation of pressure and density of a gas above and below average have maximum value at displacement node. Statement 2: When particles on opposite sides of displacement node approach each other, gas between them is compressed and pressure rises so that at displacement node gas undergoes maximum amount of compression.A. Statement-1 is true, Statement-2: is true, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is true, Statement-2: is true, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is true but statement-2 is falseD. Statement-1 is false, Statement-2 is true |
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Answer» Correct Answer - A Statament-1 is true, statement-2 is true, statement -2 is a correct explanation for statement-1 |
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| 43. |
Sound waves from a tuning fork `F` reach a point `P` by two separate routes `FAP` and `FBP` (when `FBP` is greater than `FAP` by `12 cm` there is silence at `P`). If the difference is `24 cm` the the sound becomes maximum at `P` but at `36 cm` there is silence again and so on. of velocity of sound in sir is `330 ms^(-1)`, then least frequency of tuning fork is :A. `1537 Hz`B. `1735 Hz`C. `1400 Hz`D. `1375 Hz` |
| Answer» Correct Answer - D | |
| 44. |
The pressure variation in a sound wave in air is given by `Deltap = 12 sin (8.18X - 2700 t + pi//4) N//m^(2)` find the displacement amplitude. Density of air `= 1.29 kg//m^(3)`. |
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Answer» Correct Answer - A::D In the above problem, we have found that `A = (Delta p_(max))/(2 pi f rho nu)` `= ((Delta p_(max)k))/(rho omega^(2))` (as `nu = (omega)/(k)` and `2 pi f = omega`) Now subsituting the value, we have `A= ((12)(8.18))/((1.29)(2700)^(2)) = 1.04 xx 10^(-5)m` |
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| 45. |
What happens when a sound wave interfers with another wave of same frequency and constant phase difference?A. Energy is gainedB. Energy is lostC. Redistribution of energy occurs changing with timeD. Redistribution of energy occurs not changing with time |
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Answer» Correct Answer - D In the interference the energy is redistributed and the distribution remains constant in time |
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| 46. |
If it wave possible to generate a sinusoidal `300 H_(Z)` sound wave in air that has a displacement amplitude of `0.200 mm` . What would be the sound level ? (Assume `upsilon = 330 m//s` and `rho_(air) = 1.29 kg//m^(3))` |
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Answer» Correct Answer - A::B::C::D `I = (1)/(2) rho omega^(2) A^(2)nu` `= 2rho pi^(2) f^(2) A^(2) nu` (as `omega = 2 pi f`) =`(2) (1.29) (pi)^(2) (300)^(2) (0.2 xx10^(-3))^(2) (330)` `= 30.27 W//m^(2)` `L = 10 log_(10) (I//I_(o))` where, `I_(o) = 10^(-2) W//m^(2)` Subsituting the value we get, `L = 134.4 dB` |
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| 47. |
`S_(1)` and `S_(2)` are two sources of sound emitting sine waves. The two sources are in phase. The sound emitted by the two sources interfers at point `F`. The waves of wavelength : A. `1 m` will result in constructive interferenceB. `(2)/(3) m` will result in constructive interferenceC. `2 m` will result in destructive interferenceD. `4 m` will result in destructive interference |
| Answer» Correct Answer - A::B::D | |
| 48. |
A sound wave in air has a frequency of `300 H_(Z)` and a displacement ampulitude of `6.0 xx10^(-3) mm`. For this sound waves calculate the (a) Pressure ampulitude (b) intensity (c ) Sound intensity level (in dB) Speed of sound `= 344 m//s` and density of air `= 1.2 kg//m^(3)`. |
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Answer» Correct Answer - A::B::C::D `Delta p_(max) = Bak` `= (rho nu)^(2) (A)((omega)/(nu))` `= (2 pi f A rho nu)` `= (2 pi) (300)(6.0 xx 10^(-3)(1.2)(344)` `= 4.67 Pa` (b) `I = (nu(Deltap)^(2)max)/(2B)` `= ((Deltap)^(2)max)/(2 rho nu)` (as `B = rho nu^(2))` `= (4.67)^(2)/((2)(1.2)(344)` `= 0.0264 w//m^(2)` `= 2.64 xx 10^(-2) W//m^(2)` (c ) `L = log_(10) ((I)/(I_(0)))` `= 10 log_(10) ((2.64 xx 10^(-2))/(10^(-12)))` `= 104 dB` |
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| 49. |
P, Q and R are three particles of a medium which lie on the x-axis. A sine wave of wavelength `lambda` is travelling through the medium in the x-direction. P and Q always have the same speed, while P and R always have the same velocity. The minimum distance between – (1) P and Q is `lambda` (2) P and Q is `lambda//2` (3) P and R is `lambda//2` (4) P and R is `lambda`A. `P` and `Q` is `lamda//2`B. `P` and `Q` is `lamda`C. `P` and `R` is `lamda//2`D. `P` and `R` is `lamda` |
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Answer» Correct Answer - A::D |
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| 50. |
The stationary waves set up on a string have the equation : `y = ( 2 mm) sin [ (6.28 m^(-1)) x] cos omega t` The stationary wave is created by two identical waves , of amplitude `A` each , moving in opposite directions along the string . Then :A. `A=2mm`B. `A=1mm`C. The smallest length of the string is `50cm`D. The smallest length of the string is `2m` |
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Answer» Correct Answer - B::C::D |
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