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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.
| 251. |
When beats are produced by two progressive waves of same amplitude and of nearly same frequencies then the maximum loudness of the resulting sound is `n` times the loudness of each of the component wave trains. The value of `n` isA. `1`B. `2`C. `4`D. `8` |
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Answer» Correct Answer - C `I_(1) = I_(2) = I_(0)` `I_(max)/I_(1)=((sqrt(I_(0))+sqrt(I_(0)))^(2))/I_(0)=4` |
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| 252. |
Maximum number of beats frequency heard by a human being isA. `10`B. `4`C. `20`D. `6` |
| Answer» Correct Answer - A | |
| 253. |
A tuning fork of frequency 512 Hz is vibrated with sonometer wire and 6 beats per seconds are heard. The beat frequency reduces if the tension in the string is slightly increased. The original freqency of vibration of the string isA. 506 HzB. 512 HzC. 518 HzD. 524 Hz |
| Answer» Correct Answer - A | |
| 254. |
A man standing in front of a mountain beats a drum at regular intervals. The drumming rate is gradually increased and he finds that echo is not heard distinctly when the rate becomes `40` per minute. He then moves near to the mountain by `90` metres and finds that echo is again not heard distinctly when the drumming rate becomes `60` per minute. Calculate (a) the distance between the mountain and the initial position of the man and (b) the velocity of sound. |
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Answer» Correct Answer - `270 m` Time interval between beats (initially) `= (60)/(40) = 1.5` sec Since `t = (2S)/(v)` `1.5 = (2S)/(2v)` where `S` is the initial distance and `v` is the speed of sound . As the man approches `90m` towards the mountain `(2(S - 90))/(v) = 1` since drumming rate is `60//min` Solving for `S` from the two relations we get `(2S)/(2(S - 90)) = 1.5 , (S - 90) /(S) = (2)/(3)` `rArr 3S - 3 xx 90 = 2S , S = 270 m` |
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| 255. |
Two wires A and B are of same length, redius and same material are in unison. If tendion in A is increassed by 4% 4 beats are heard, then the frequency of the note produced when they ware in unison, will beA. 50 HzB. 100 HzC. 150 HzD. 200 Hz |
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Answer» Correct Answer - D `(104)/(100)=((n+4)/(n))^(2)or sqrt((104)/(100))= (n+4)/nrArrn=200 Hz` |
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| 256. |
Why no beats can be heard if the frequencies of the two interfering waves differ by more then ten ? |
| Answer» Correct Answer - This due to persistence of hearing. | |
| 257. |
The audible frequency for a normal human being is `20 Hz` to `20 kHz`. Fuind the corresponding wavelengths if the speed of sound in air `320 m//s`. (b) Find the phase difference between two position separated by `20 cm` at a particular instant. |
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Answer» Correct Answer - `16 mm, 16 m` `lambda = (V)/(eta)` `lambda_(max) = (320)/(20 Hz) = 16 m , lambda_(min) = (320)/(20 KHz) = 16 mm`. |
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| 258. |
Two monoatomic ideal gases `1` and `2` of molecular masses `m_(1)` and `m_(2)` respectively are enclosed in separate containers kept at the same temperature. The ratio of the of sound in gas `1` to that in gas `2` is given byA. `sqrt((m_(1))/(m_(2)))`B. `sqrt((m_(2))/(m_(1)))`C. `(m_(1))/(m_(2))`D. `(m_(2))/(m_(1))` |
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Answer» Correct Answer - B Speed of sound in a gas given by `v=sqrt((gammaRT)/M)` `v prop 1/(sqrt(M))` `:. (v_(1))/(v_(2))=sqrt((M_(2))/(M_(1)))=sqrt((m_(2))/(m_(1))` Here `gamma=(C_(p))/(C_(v))=5/3` for both the gases `(gamma_("monoatomic")=5/3)` |
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| 259. |
The sound intensity is `0.008W//m^( 2)` at a distance of 10 m from an isotropic point source of sound. The power of the source isA. `2.5` wattB. `0.8` wattC. `8` wattD. `10` watt |
| Answer» Correct Answer - D | |
| 260. |
The sound intensity is `0.008W//m^( 2)` at a distance of 10 m from an isotropic point source of sound. The power of the source isA. 2.5 wB. 0.8 wC. 8 wD. 10 w |
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Answer» Correct Answer - D `P/(4pir^(2))`=1 for an isotropic point sound source, `rArr F=L 4pir^(2)` `=(0.008w//m^(2))(4.pi.10^(2))` =10 watt |
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| 261. |
Loudness of sound from an isotropic point source at a distance of `10m` is `20dB`. If the distance at which it is not heard is `10^(K)` in meters find `K`. |
| Answer» Correct Answer - 2 | |
| 262. |
A sound of intensity `I` is greater by `3.0103 dB` from anoterh sound of intersity `10 nW cm^(-2)`. The absoulte value of intensity of sound level `I` is `Wm^(-2)` is :A. `2.5 xx 10^(-4)`B. `2 xx 10^(-4)`C. `2.0 xx 10^(-2)`D. `2.5 xx 10^(-2)` |
| Answer» Correct Answer - B | |
| 263. |
(a) What is the intensity of a `60 dB` sound ? (b) If the sound level is `60 dB` close to a speaker that has an area of `120 cm ^(2)` . What is the acoustic power output of the speaker? |
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Answer» Correct Answer - A::B (a) `L = 10 log_(10) ((I)/(I_(o)))` `60 = 10 log_(10) ((I)/(10^(-12)))` Solving we get, `I = 10^(-6) W//m^(2)` (b) `I = (P)/(S)` :. `P = (I) (S)` = `(10^(-6)) (120 xx10^(-4))` `= 1.2 xx 10^(-8) W` |
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| 264. |
Most people interpret a `9.0 dB` increase in sound intensity level as a doubling in loudness. By what factor must the sound intensity be increase to double the loudness?A. `1 m//s`B. `2 m//s`C. `3 m//s`D. `4 m//s` |
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Answer» Correct Answer - B `f_(b) = f_(1) - f_(2)` or `10 = 1700 ((340)/(340 + nu_(2))) - 1700 ((340)/(340 + nu_(1)))` `= 1700(1 + (nu_(2))/(340))^(-1) - 1700 (1 + (nu_(1))/(340))^(-1)` Using Binomial therefore, we get `10 = 1700(1 - (nu_(2))/(340)) - 1700 (1 - (nu_(1))/(340))` `= (nu_(1) - nu_(2)) ((1700)/(340))` :. `nu_(1) - nu_(2) = 2 m//s` |
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| 265. |
(a) Sound with intensity larger than `120 dB` appears painful to a person. A small speaker delivers `2.0 W` of audio output. How close can the person get to the speaker without hurting his ears? (b) If the sound level in a room is increased from `50 dB` to `60 dB`, by what factor level is the pressure amplitude increased? |
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Answer» (a) `L = 10 log_(10)((I)/(I_ (0)))` `120 = 10 log_(10)((I)/(10^(-12)))` `(I)/(10^(12)) = 10^(12) rArr I = 1 W//m^(2)` For a point source of power `P`, intensity at distance `r` `I = (P)/(4pi r^(2))` `r = sqrt ((P)/(4pi I)) = sqrt ((2)/(4pi xx 1)) = sqrt ((1)/(2pi)) = 0.4 m` (b) `L_(2) - L_(1) = 10 log_(10)(I_(2)/(I_(1)))` `60 - 50 = 10 log_(10)(I_(2)/(I_(1))) rArr log_(10)(I_(2)/(I_(1))) = 1` `I_(2)/(I_(1)) = 10` `I prop p_(0)^(2)` `I_(2)/(I_(1)) = [(p_(0))_(2)^(2)/(p_(0))_(1)^(2)] = 10` `(p_(0))_(2)/(p_(0))_(1) = sqrt (10)` |
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| 266. |
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. `60 dB`B. `68 dB`C. `74 dB`D. `81 dB` |
| Answer» Correct Answer - B | |
| 267. |
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. 60 dBB. 69 dBC. 74 dBD. 81 dB |
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Answer» Correct Answer - B Let intensity due to a single person =I then `10 log I//I_(0)=60` Also, intensity due to 8 person =8L `:.` final decibal level `=10 log((8I)/(I_(0)))=10(log.(I)/(I_(0))+log8)` `=60+10 log 8=60 +30 log2=60+30(.3010)` `=69` |
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| 268. |
Two tuning forks `A` and `B` vibrating simultaneously produce `5 beats//s`. Frequency of `B` is `512 Hz`. If one arm of `A` is filed, the number of beats per second increases. Frequency of `A` isA. `502 Hz`B. `507 Hz`C. `517 Hz`D. `522 Hz` |
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Answer» Correct Answer - C `f_(B) = 512 Hz, f_(A) = 512+-5=517` or `507 Hz` If arms of `A` is filed, `f_(A): uparrow` Beat frequency: `uparrow`. `f_(A) = 517 Hz` |
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| 269. |
There is a set of four tuning forks , one with the lowest frequency vibrating at `550 Hz`. By using any two tuning forks at a time , the following beat frequencies are heard : `1 , 2, 3, 5 , 7 , 8`. The possible frequencies of the other three forks areA. `552, 553, 560`B. `557, 556, 560`C. `552, 553, 558`D. `551, 553, 558` |
| Answer» Correct Answer - D | |
| 270. |
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 is |
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Answer» Correct Answer - A Length of the organ pipe is same in both the cases. Fundamental grequency of open pipe is `f_(1) = (nu)/(2l)` and frequency of third harmonic of closed pipe will be `f_(2) = 3((nu)/(4l))` Given that, `f_(2) = f_(1) + 100` or `f_(2) - f_(1) = 100` or `(3)/(4) ((nu)/(l)) - ((1)/(2)) ((nu)/(l)) = 100` rArr `(nu)/(4l) = 100 H_(Z)` :. `(nu)/(2l)` or `f_(1) = 200 H_(Z)` Therfore, fundamental frequency of the open pipe is `200 H_(Z)`. |
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| 271. |
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. `200Hz`B. `300Hz`C. `240Hz`D. `480Hz` |
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Answer» Correct Answer - A |
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| 272. |
About how many times more intense will the normal ear perceiver a sound of `10 ^(6) W //m^(2)` than one of `10^(9) W//m^(2)` ? |
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Answer» `L_(1) = 10 log_(10) ((I_(1))/(I_(o)))` = `10 log_(10) ((10^(-6))/(10^(-12))) = 60 dB` `L_(2) = 10 log_(10) ((I_(2))/(I_(o)))` = `10 log_(10) ((10^(-9))/(10^(-12))) = 30 dB` `L_(2) = 2l_(2)` |
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| 273. |
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. `4 m`B. `5 m`C. `10 m`D. `20 m` |
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Answer» Correct Answer - C `L_(1) - L_(2) = 10 log_(10) ((I_(1))/(I_(2))) =10 log_(10) ((r_(2)^(2))/(r_(1)^(2)))` `40 - 20 = 10 log_(10) ((r_(2))/(r_(1)^(2)))` `20 = 20 log_(10) ((r_(2))/(r_(1)))` `log_(10)((r_(2))/(r_(1))) = 1 rArr (r_(2))/(r_(1)) = 10` ` r_(2) = 10 r_(1) = 10 m` |
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| 274. |
How many times more intense is a `90 dB` sound than a `40 dB` sound?A. `2.5`B. `5`C. `50`D. `10^(5)` |
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Answer» Correct Answer - D `L_(2) - L_(1) = 10log_(10) ((I_(2))/(I_(1)))` `(90 - 40) = 10 log_(10) ((I_(2))/(I_(1)))` `(I_(2))/(I_(1)) = 10^(5)` |
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| 275. |
A sample of oxygen at `NTP` has volume `V` and a sample of hydrogen at `NTP` has volume `4 V`. Both the gases are mixed and the mixture is maintained at `NTP` if the speed of sound in hydrogen at `NTP` is `1270 m//s`, that in the mixture will beA. `317 m//s`B. `635 m//s`C. `830 m//s`D. `950 m//s` |
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Answer» Correct Answer - B `rho_(max) = (rho_(O_(2)) V_(O_(2)) + rho_(H_(2)) V_(N_(2)))/(V_(O_(2)) + V_(H_(2)))` `= (16 xx V + 1 xx 4V)/(V + 4V)` `(v_("mix"))/(v_(H_(2))) = sqrt((rho_(H_(2)))/(rho_("mix"))) = sqrt((1)/(4)) = (1)/(2)` `v_("mix") = (v_(H_(2)))/(2) = (1270)/(2) = 635 m//s` |
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| 276. |
The ratio of the velocity of sound in hydrogen gas to that in helium gas at the same temperature isA. `sqrt(21)//5`B. `sqrt(42)//5`C. `5//42`D. `5//sqrt(21)` |
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Answer» Correct Answer - B `v = sqrt((gamma RT)/M_(0))` `v_(H_(2))/v_(He)=sqrt(gamma_(H_(2))/gamma_(He) ((M_(0))_(He))/((M_(0))_(H_(2)))=sqrt(((7//5))/((5//3)).4/2)` `=sqrt(42/5)` |
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| 277. |
If `c_(0)` and `c` denote the sound velocity and the rms velocity of the molecules in a gas, thenA. `c_(0) gt c`B. `c_(0) = c`C. `c_(0) = c(gamma//3)^(1//2)`D. no relation |
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Answer» Correct Answer - C `c_(0) = sqrt((gamma R T)/(M_(0))), c = sqrt((3 R T)/(M_(0)))` `(c_(0))/(c ) = sqrt((gamma)/(3)) rArr c_(0) = c ((gamma)/(3))^(1//2)` |
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| 278. |
A source emits sound of frequency 600Hz inside water. The frequency heard in air will be equal to (velocity of sound in water`=1500(m)/(s)`, velocity of sound in air=300(m)/(s))A. `200 Hz`B. `3000 Hz`C. `120 Hz`D. `600 Hz` |
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Answer» Correct Answer - D When a wave goes from one medium to another, its frequency remains same. |
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| 279. |
A source of sound S and a detector D are placed at some distance from one another. A big cardboard is placed near the detector and perpendicular to the line SD as shown in figure. It is gradually moved away and it is found that the intensity changes from a maximum to a minimum as the board is moved through a distance of 20 cm. Find the frequency of the sound emitted. Velocity of sound in air is `336 m s^-1`. |
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Answer» Correct Answer - B::D According to the given data, `V=336 m/s` `lamda/4=` distance between maximum and minimum intensilty `=20cm` `rarr lamda=80 cm` `rarr n=frequency =V/lamda` `=336/)80x10^-2)=420Hz` |
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| 280. |
The sound produced in a big hall repeats itself. Why ? |
| Answer» Correct Answer - The sound repeats itself because of successive reflections. | |
| 281. |
A star is moving away from the earth with a velocity of `100 km//s`. If the velocity of light is `3 xx 10^(8) m//s` then the shift of its spectral line of wavelength `5700 A` due to Doppler effect isA. `0.63Å`B. `1.90 Å`C. `3.80 Å`D. `5.70 Å` |
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Answer» Correct Answer - B `(Delta lambda)/(lambda) = (v)/(c )` `Delta lambda = (v)/( c) lambda = (100 xx 10^(3))/(3 xx 10^(8)) xx 5700` `= 1.9 Å` |
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| 282. |
The apparent wavelength of the light from a star, moving away from the earth is `0.01%` more than its real wavelength. The speed of the star with respect to earth isA. `10 km//s`B. `15 km//s`C. `30 km//s`D. `60 km//s` |
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Answer» Correct Answer - C `(Delta lambda)/(lambda) = (v)/(c ) rArr (0.01)/(100) = (v)/(3 xx 10^(8))` `v = 3 xx 10^(4) m//s = 30 km//s` |
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| 283. |
A string of length L is stretched along the x-axis and is rigidly clamped at its two ends. It undergoes transverse vibration. If n an integer, which of the following relations may represent the shape of the string at any time :-A. `y=Asin((npix)/L)cosomegat`B. `y=Asin((npix)/L)sin omegat`C. `y=Acos((npix)/L)cos omegat`D. `y=Acos((npix)/L) sin omegat` |
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Answer» Correct Answer - A::B |
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| 284. |
In an orchestra, the musical sounds of different instruments are distinguished from one another by which of the following characteristics.A. PitchB. LoudnessC. QualityD. Overtones |
| Answer» Correct Answer - C | |
| 285. |
Beats occur because ofA. interferenceB. reflectionC. refractionD. Doppler effect |
| Answer» Correct Answer - A | |
| 286. |
The tension of a string is inceased by `44%`. If its frequency of vibration is to remain unchanged its length must be increased byA. `44%`B. `sqrt(44)%`C. `22%`D. `20%` |
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Answer» Correct Answer - D |
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| 287. |
When sound wave is refracted from air to water, which of the following will remain unchanged?A. wave numberB. wavelengthC. wave velocityD. frequency |
| Answer» Correct Answer - D | |
| 288. |
when a wave is propagated from rarer medium ot denser medium. Which of the following will remain unchanged?A. Wave speedB. Propagation constantC. FrequencyD. None of these |
| Answer» Correct Answer - C | |
| 289. |
A physiscist points out that glass is rarer than water.A. This statement is correct in the case of soundB. This statement is always wrongC. This statement is correct in the case of lightD. This statement is always correct |
| Answer» Correct Answer - A | |
| 290. |
A tuning fork of unknows frequency makes three beats per second with a standard fork of frequency `384 H_(Z)`. The beat frequency decreases when a small piece of wax is put on a prong of the first fork. What is the frequency of this fork? |
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Answer» Correct Answer - C By putting wax on first tuning fork, its frequency will decrease. Beat frequency is also decreasing. Hence, `f_(1) gt f_(2)` `:. f_(1) - f_(2) = 3` or `f_(1) = 3 + f_(2) = 3 + 384` `= 387 H_(Z)` |
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| 291. |
Two strings X and Y of a sitar produce a beat frequency 4 Hz.When the tension of the string Y is slightly increased the beat frequency is found to be 2 Hz. If the frequency of X is 300 Hz,then the original frequency of Y wasA. `296 Hz`B. `298 Hz`C. `302 Hz`D. `304 Hz` |
| Answer» Correct Answer - A | |
| 292. |
When height increases, velocity of sound decreasesA. this is due to decrease of pressureB. this is due to decrease in temperatureC. this is due to both decrease in temperature and pressureD. statement is wrong |
| Answer» Correct Answer - B | |
| 293. |
A sound source of frequency `512 Hz` is producing `6` beats with a guitar. If the string of gultar is stretched slightly then beat frequency decreases. The original frequency of guitar isA. `506 Hz`B. `512 Hz`C. `518 Hz`D. `524 Hz` |
| Answer» Correct Answer - A | |
| 294. |
Velocity of sound in air (i) increases with temperature (ii) decreases with temperature (iii) increases with pressure (iv) is independent of pressureA. (i) and (ii)B. (i) and (iii)C. (ii) and (iii)D. (i) and (iv) |
| Answer» Correct Answer - D | |
| 295. |
Use the formula `v=sqrt((gamma P)/(rho))` to explain why the speed of sound in air (a) is independent of pressure, (b) increases with temperature, (c) increases with humidity. |
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Answer» Correct Answer - Assume ideal gas law : `P = (rhoRT)/(M)` where `rho` is the density, `M` is the molecular mass, and `T` is the temperature of the gas. This gives `v = sqrt((gammaRT)/(M))`. This shows that `v` is (a) Independent of pressure. (b) Increase as `sqrt(T)`. (c) The molecular mass of water `(18)` is less then that of `N_(2) (28)` and `O_(2)(32)`. Therefore as humidity increases, the effective molecular mass of air decrease and hence `v` increases. |
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| 296. |
A closed organ pipe of length `L` and an open organ pipe contain gass of densities `rho_(1)` and `rho_(2)`, respectively. The compressibility of gass are equal in both the pipes. Both the pipes are vibrating in their first overtone with same frequency . The length of the open orange pipe is (a) `(L)/(3)` `(4l) / (3)` (c ) `(4l)/(3)sqrt((rho_(1))/(rho_(2)))` (d) `(4l)/(3)sqrt((rho_(2))/(rho_(1)))` |
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Answer» `f _(C) = (nu)/(4((l_(0))/(2))) = (nu)/(2l_(0)) = f_(0) =f` (both first overtone) or `3((nu_(C))/(4L)) = 2((nu_(0))/(2l_(0)))` `:. l_(0) = (4)/(3) (nu_(0)/(nu_(C)))L = (4)/(3) (sqrtB//rho_(2))/(sqrtB//rho_(1)) L` = `(4)/(3) sqrt((rho_(1))/(rho_(2))) L` Hence, the correct option is (C). |
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| 297. |
The speed of sound waves having a frequency of `256 Hz` compared with the speed of sound waves having a frequency of `512 Hz` is :A. halfB. twiceC. four timesD. same |
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Answer» Correct Answer - D Speed depend on medium |
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| 298. |
Velocity of sound in air is `320m//s`. A pipe closed at one end has of `1m`. Neglecting end corrections, the air column in air pipe can resonate for sound of frequency :A. `80 Hz`B. `240 Hz`C. `320 Hz`D. `400 Hz` |
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Answer» Correct Answer - C Fundamental frequency `= (v)/(4l) = (320)/(4 xx 1) = 80 Hz` Frequency of closed pipe are `= (v)/(4l), (3v)/(4l), (5v)/(4l), …` `= 80, 240, 400,…` |
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| 299. |
A `3 m` long organ pipe open at both ends is driven to third harmonic standing wave. If the amplitude of pressure oscillations is `1` per cent of mean atmospheric pressure `(p_(o) = 10^(5) Nm^(2))`. Find the amplitude of particle displacement and density oscillations. Speed of sound `upsilon = 332 m//s` and density of air `rho = 1.03 kg//m^(3)`. |
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Answer» Correct Answer - (i) `(1)/(1089pi)m` , (ii) `(1)/(1089)kg//m^(3)` Given length of pipe `= 3m` Third harmonic Implies that `(3lambda)/(2) , lambda = (2l)/(3), = (2 xx 3)/(3) = 2 m` `k = (2pi)/(lambda) = pi` `BkS = P` `BpiS = 100` `B = rhoV^(2) = 330^(2) xx 1` `S = (100)/(piB)` `S = (100)/(pi330^(2))` `S = (1)/(1089pi)` Bulk moduls of elasticity `B = (-dp)/((dv//v))`, Volume `= (mass)/(density) = (m)/(rho)` `dv = (-m)/(rho^(2)) drho = (-Vdrho)/(rho) rArr (dv)/(V) = - (drho)/(rho) , dP = B(drho)/(rho)` `Deltarho_(max) = (rho)/(B)DeltaP_(max) = (DeltaP_(max))/(v^(2)) = (10^(2))/((330)^(2)) , (1)/(1089) kg//m^(3)` |
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| 300. |
An organ pipe `P_(1)` open at one end vibrating in its first harmonic and another pipe `P_(2)` open at ends vibrating in its third harmonic are in resonance with a given tuning fork. The ratio of the length of `P_(1)` to that `P_(2)` isA. `8 : 3`B. `3 : 8`C. `1 : 2`D. `1 : 3` |
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Answer» Correct Answer - B `(2 xx 1 + 1)(v)/(4l_(1)) = (3 + 1)(v)/(2l_(2)) rArr (3)/(4l_(1)) + (2)/(l_(2))` `(l_(1))/(l_(2)) =(3)/(8)` |
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