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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.
| 151. |
In a liquid with density `900 kg//m^(3)`, lonfitudinal waves with frequency `250 H_(Z)` are found to have wavelength `8.0 m`. Calculate the bulk modulus of the liquid. |
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Answer» Correct Answer - A::C `nu = f lambda = sqrt((B)/(rho))` `:. B = rho (f lambda)^(2)` `= (900)(250 xx 8)^(2)` `= 3.6 xx 10^(9) N//m^(2)` |
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| 152. |
Calculate the difference in the speeds of sound in air at `- 3^(@) c`,`60 cm` pressure of mercury and `30^(@) c`, `75 cm` pressure of mercury. The speed of sound in air at `0^(@)C` is `332 m//s`. |
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Answer» Correct Answer - A `nu prop sqrt(T)` `:. nu_(2) = (sqrt(T_(2)/(T_(1))))nu_(1)` `nu_(-3^(@)C) = (sqrt(270)/(273))(332)` `= 330.17 m//s` `nu_(3^(@)C) = (sqrt((270 + 30)/(273 )))(332)` `= 349.77 m//s` The difference in these two speed is approximately `19.6 m//s`. |
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| 153. |
Helium is a monatomic gas that has a density of `0.179 kg//m^(3)` at a pressure of `76 cm` of mercury and a temperature of `0^(@) C` . Find the speed of compressional waves (sound) in helium at this temperature and pressure. |
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Answer» Correct Answer - B `nu = sqrt((gamma P)/(rho)) = sqrt (((5//3) xx 0.76 xx 13.6 xx 10^(3) xx 9.8)/(0.179))` Where `P = h rho g` :. `nu = 972 m//s` |
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| 154. |
The same notes being played on sitar and veena differ inA. qualityB. pitchC. both quality and pitchD. neither quality nor pitch |
| Answer» Correct Answer - A | |
| 155. |
Choose the correct statement.A. Sound waves are transverse wavesB. Sound travels faster through vacuumC. Sound travels faster in solids than in gasesD. Sound travels faster in gases than in liquids |
| Answer» Correct Answer - C | |
| 156. |
Pitch of a sound depends on :A. FrequencyB. wavelengthC. amplitudeD. speed |
| Answer» Correct Answer - A | |
| 157. |
Velocity of sound in air is 332 `ms^-1`. Its velocity in vacuum will beA. `gt 332 m//s`B. `= 332 m//s`C. `lt332 m//s`D. meaningless |
| Answer» Correct Answer - D | |
| 158. |
Consider the following statements: Assertion (A) The velocity of sound in air increases due to the presence of moisture in it. Reason (R ): The presence of moisture in air lowers the density of air. Of these statements-A. both `A` and `R` are true and `R` is the correct explanation of `A`B. both `A` and `R` are true but `R` is not the correct explanation of `A`C. `A` is true but `R` is falseD. `A` is false but `R` is true |
| Answer» Correct Answer - A | |
| 159. |
The speed of sound in a medium depends onA. the elastic property but not on the inertia propertyB. the inertia property but not on the elastic propertyC. the elastic property as well as the inertia propertyD. neither the elastic property nor the inertia property |
| Answer» Correct Answer - C | |
| 160. |
A light pointer fixed to one prong of a tuning fork touches gnetly a smoked vertical plate. The fork is set vibrating and the plate is allowed to fall freely. 8 complete oscilllations are counted when the plate falls through 10cm.What is the frequency of the tuning fork?A. `65 Hz`B. `56 Hz`C. `46 Hz`D. `64 Hz` |
| Answer» Correct Answer - B | |
| 161. |
A light pointer fixed to one prong of a tuning fork touches a vertical plate. The fork is set vibrating at a frequency of 56 Hz and allowed to free fall. Calculate how many complete oscillation are couhnted when plate falls st 10 cmA. 10B. 9C. 8D. 7 |
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Answer» Correct Answer - C Let n = frequency and N= number of oscillations Time taken to full freely to a distance 10 cm `" s=1/2g.t^(2)rArrt=sqrt((2xx10)/(g))` The frequency of oscillation during this time `n=N/t rArrN=nt rArrN=56xxsqrt((2xx10)/(980))` `N=7.999=8` |
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| 162. |
Dentermine the speed of sound waves in water , and find the wavelength of a wave having a frequency of `242 H_(Z)` . Take `B_(water) = 2xx10^(9) Pa`. |
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Answer» Correct Answer - A::D Speed of sound wave, `nu= sqrt((B)/(rho))= sqrt((2 xx 10^(9))/(10^(3))) = 1414 m//s` Wavelength, `lambda = (nu)/(f) = 5.84m` |
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| 163. |
Using the fact that hydrogen gas consister of diatomic molecules with `M = 2 kg//K - mol`. Find the speed of sound in hydrogen at `27^(@) C` . |
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Answer» Correct Answer - A::B::C `nu = sqrt((gammaRt) /(M))` = `sqrt((1.40 xx8.31 xx300)/(2 xx10^(-3))) = 1321 m//s` |
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| 164. |
Define Mach numberA. It is the ratio of the stress to the strainB. It is the ratio of the strain to stressC. It is the ratio of the velocity of an object to the velocity of soundD. It is the ratio of the velocity of sound to the velocity of an object |
| Answer» Correct Answer - C | |
| 165. |
The time of reverberation of a room A is one second. What will be the time (in seconds) of reverberation of room, having all the dimensions double of those of room A?A. `1//2`B. `1`C. `2`D. `4` |
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Answer» Correct Answer - C Reverberation time `T = (0.61 V)/(a S)` `V`: volume, `S`: surface area `a`: coefficient of absorption `T_(1)/T_(2)=(V_(1)/V_(2))(S_(2)/S_(1))=(V/(8V))((4S)/S)=1/2` `T_(2)=2T_(1)=2xx1=2s` |
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| 166. |
For the wave described in exercise `27` plot the displacement `(y)` versus `(t)` graphs for `x = 0, 2` and `4 m`. What are the shaps of these graphs ? In which aspects does the oscillatory motion in travelling wave differ from one point to another amplitude, frequency or phase? |
| Answer» Correct Answer - All the graphs are sinusoidal. They have same amplitude and frequency, but different initial phases. | |
| 167. |
The change in frequency due to Doppler effect does not depend onA. the speed of the sourceB. the speed of the observorC. the frequency of the sourceD. separation between the source and the observer |
| Answer» Correct Answer - D | |
| 168. |
When you speak to your friend, which of the following parameters have a unique value in the sound produced?A. FrequencyB. WavelengthC. AmplitudeD. Wave velocity |
| Answer» Correct Answer - D | |
| 169. |
Why the church bell has large surface area ? |
| Answer» Correct Answer - Larger the surface area of the vibrating body, more is the amplitude and hence intensity of sound produced. | |
| 170. |
A train is moving on a straight track with speed `20ms^(-1)`. It is blowing its whistle at the frequency of `1000 Hz`. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound `= 320 ms^(-1)`) close to :A. `12%`B. `6%`C. `18%`D. `24%` |
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Answer» Correct Answer - A Observer is stationary and source is moving. During approach, `f_(1) = f((nu)/(nu - nu_(s)))` `= 1000((320)/(320 - 20))` `= 1066.67 H_(Z)` During recede, `f_(2) = f((nu)/(nu + nu_(s)))` `= 1000((320)/(320 + 20)) = 941.18 H_(Z)` `|%` change in frequency| `= ((f_(1) - f_(2))/(f_(1))) xx 100` `~~ 12%` |
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| 171. |
A sound wave of frequency 500 Hz covers a distance of 1000 m is 5 sec between the points X and Y. Then the number of waves between X and Y isA. 500B. 1000C. 2500D. 5000 |
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Answer» Correct Answer - C `V=1000/5=200 m//s` `V=n lambdarArr lambda=V/n=200/500` No. of wave `=L/(lambda)=(1000xx500)/200=2500` |
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| 172. |
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^(-6) m//s`B. `17.0xx10^(-6) m//s`C. `2.76 xx10^(-3) m//s`D. `2.77xx10^(-6) m//s` |
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Answer» Correct Answer - C `v_(max)=omega_(0)A` `=(2pif)A=(2pi)(440)(10^(-6))` `=2.76xx10^(-3) m//sec` |
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| 173. |
Calculate the frequency of beats produced in air when two sources of sound are activated, one emitting a wavelength of 32 cm and the other of 32.2 cm. The speed of sound in air is `350 ms^-1`. |
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Answer» Correct Answer - C::D Group I, Given `V=350, I_1=32cm =32x10^-2m` So `eta_1 =frequency` 350/(32xx10^-2)=1093Hz` Group II `V=350, lamda_2=32.2cm` `=32.2xx10^-2m` `eta_2=350/(32.2xx10^-2)` `=1086Hz` So, beat frequency =1093=7Hz |
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| 174. |
The wavelength of two sound waves are `49cm` and `50cm` respectively. If the room temperature is `30^(@)C` then the number of beats produced by them is approximentely (velocity of sound in air at `0^(0)C = 332 m//s`).A. `6`B. `10`C. `14`D. `18` |
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Answer» Correct Answer - A::D `v = 332sqrt((303)/(273)) rArr` Beat frequency `= f_(1) - f_(2) = v((1)/(lambda_(1)) - (1)/(lambda_(2)))` `= 332 sqrt((303)/(273))((1)/(49) - (1)/(50)) xx 100 = 14` |
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| 175. |
A machine gun is mounted on an armored car moving with a speed of `20 ms^(-1)`. The gun point against the direction of motion of car. The muzzle speed of bullet is equal to speed of sound in air i.e., `340 ms^(-1)`. The time difference between bullet actually reaching and sound of firing reaching at a target `544 m` away from car at the instant of firing is reaching at a target `544 m` away from car at the instant of firing is A. `1.2 s`B. `0.1 s`C. `1 s`D. `10 s` |
| Answer» Correct Answer - B | |
| 176. |
The ratio of speed of sound in monomatomic gas to that in water vapours at any temperature is. (when molecular weight of gas is `40 gm//mol` and for water vapours is `18 gm//mol`)A. `0.75`B. `0.73`C. `0.68`D. None of these |
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Answer» Correct Answer - A `(V_(1))/(V_(2)) = sqrt((gamma_(1))/(m_(1))xx(M_(2))/(gamma_(2))) = sqrt((5)/(3 xx 40) xx (16 xx 3)/(4)) = 0.75` |
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| 177. |
A gas mixture has `24%` of Argon, `32%` of oxygen, and `44%` of `CO_(2)` by mass. Find the velocity of sound in the gas mixture at `27^(@)C`. Given `R = 8.4 S.I.` units. Molecular weight of `Ar = 40, O_(2) = 32, CO_(2) = 44, gamma_(Ar) = 5//3 gamma_(O2) = 7//5, gamma_(CO_(2)) = 4//3`. |
| Answer» Correct Answer - `V cong 303.5 m//s` | |
| 178. |
Find the speed of sound in a mixture of 1 mole of helium and 2 moles of oxygen at `27^(@)C`. If the temperature is raised by 1K from 300K, find the percentage change in the speed of sound in the gaseous mixture. Take `R=8.31Jmo l e^(-1)K^(-1)`. |
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Answer» Correct Answer - (a) `approx 409.9 m//s` , (b) `(1)/(6) %` (a) `gamma_(mix) = (n_(1)C_(P_(1)) + n_(2)C_(P_(2)))/ (n_(1)C_(V_(1)) + n_(2)C_(V_(2))) = (1xx(5)/(2)R+2xx(7)/(2)R)/(1xx(3)/(2)R+2xx(5)/(2)R) = (19)/(13)` `m_(mix) = (n_(1)m_(1) + n_(2)m_(2))/(n_(1) + n_(2)) = (1 xx 4 + 2 xx 32)/(1 + 2) = (68)/(3)` `V = sqrt((gamma_(mix)RT)/(m_(mix))) = sqrt((19 xx 25 xx 300 xx 3)/(13 xx 3 xx 68 xx 10^(-1))) = 400.9 m//sec` (b) `V = sqrt((gammaRT)/(M)) rArr ln V = (1)/(2)ln"(gammaR)/(M) + (1)/(2) ln T` `(1)/(V) (dV)/(dT) = 0 + (1)/(2T)` `(dV)/(V) xx 100 = (1)/(2) (dT)/(T) xx 100 = (1)/(2 xx 300) xx 100 = (1)/(6)%` |
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| 179. |
An open organ pipe has a length of 5 cm. (a) Find the fundamental frequency of vibration of this pipe. (b) What is the highest harmonic of such a tube that is in the audible range ? Speed of sound in air is `340 ms^-1` and the audible range is 20-20,000 Hz. |
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Answer» Correct Answer - A::B::C::D a. Here given I=5xm=5xx10^=2m` `v=340m/s` `rarr n=V/(2I)=340/(2xx5xx10^-2)` `=3.4kHz` b. If the fundamental frequency =3.1kHz then the highest Harmonic in the audible range `(20Hz-20kHz)` `=20000/3400=5.8=5` (Integral multiple of 3.4 kHz)` |
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| 180. |
An electronically driven loudspeaker is placed near the open end of a resonance column apparatus. The length of air column in the tube is 80 cm. The frequency of the loudspeaker can be varied between 20 Hz and 2 kHz. Find the frequencies at which the column will resonate. Speed of sound in air `= 320 m s^-1` |
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Answer» Correct Answer - A::B::C the resoN/Ance colun apparatus is equivalent to a closed organ pipe. Here `I=80cm=80xx10^-2` mv=320m/s` `rarr n_0=v/(4I)=320/(4xx50x10^-2)=100Hz` So the frequency of the other harmonies are odd multiple of `n_0` `=(2n+1)100Hz` According to the question the harmonic should be betwen 20 Hz nd2kHz` So, n=(0,1,2,3,4,5,.....9)` |
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| 181. |
The fimdamental frequency of a closed organ pipe is equal to second of an open organ pipe is the length of closed organ pipe is 15 cm. The length of open organ pipe isA. 90 cmB. 30 cmC. 15 cmD. 20 cm |
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Answer» Correct Answer - A For closed organ pipe `f_(1)=v/(4l_(1))" "["for fundamental"]` For open organ pipe, `f_(2)=(3v)/(2l_(2))" "["second overlone"]` `f_(1)=f_(2)thereforev/(4l_(1))=(3v)/(2l_(2))rArr(l_(2))/(l_(1))=6` `therefore" "l_(2)=6l_(1)=6xx15=90cm` |
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| 182. |
The velocity of sound in air at NTP is 331 m/s. Find its velocity when temperature eises to `91^(@)C` and its pressure is doubled.A. 372 m/sB. 382.1 m/sC. 423 m/sD. 392.5 m/s |
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Answer» Correct Answer - B Given velocity of sound in air `v_(o)` = 331 m/s `" "t=91^(@)C` `v_(t)=v_(o)sqrt((273+t)/(273))=331 sqrt((273+91)/(273))=331sqrt(1+91/273)` `v_(t)=331sqrt(1+1/3)=331xx2/sqrt3` `" "=382.1m//s` |
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| 183. |
The explosion of a fire cracker in the air at the a heigth of `40 m` produced a `100 dB` sound level at ground below. What is the instantaneous total radiated power? Assuming that it radiates as a point source. |
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Answer» Correct Answer - A::B `l_(1) = 10 log_(10) ((I)/(I_(o)))` `100 = 10 log_(10) ((I)/(10^(-12)))` Solving we get, `I = 10^(-2) W//m^(2)` Now, `I = (P)/(4pi r ^(2))` :. `P = I (4 pi r^(2))` = `(10^(-2)) (4 pi) (40)^(2)` = `201 W` |
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| 184. |
How can we tell in a completely dark room that a particular soun is produced by a violin or a flute ? |
| Answer» Correct Answer - The distinction is possible on account on account of the qually of sound. | |
| 185. |
At the closed end of an organ pipe :A. the displacement is zeroB. the displacement amplitude is maximumC. the pressure amplitude is zeroD. the pressure amplitude is maximum |
| Answer» Correct Answer - A::D | |
| 186. |
The energy per unit area associated with a progressive sound wave be doubled if :A. the amplitude of the wave is doubledB. the amplitude of the wave is increased by `50%`C. the amplitude of the wave is increased by `41%`D. the frequency of the wave is increased by `41%` |
| Answer» Correct Answer - C::D | |
| 187. |
Two plane harmonic sound waves are expressed by the equations. `y_(1)(x,t) = A cos(0.5 pix - 100 pit)` `y_(2)(x,t) = A cos (0.46 pix - 92 pit)` (All parameter are in `MKS`) : What is the speed of the sound ?A. `200 m//s`B. `180 m//s`C. `192 m//s`D. `96 m//s` |
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Answer» Correct Answer - A `v = (omega)/(k) = (100pi)/(0.5pi)` Speed `v = 200` |
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| 188. |
Two waves travelling in a medium in the x-direction are represented by `y_(1) = A sin (alpha t - beta x)` and `y_(2) = A cos (beta x + alpha t - (pi)/(4))`, where `y_(1)` and `y_(2)` are the displacements of the particles of the medium `t` is time and `alpha` and `beta` constants. The two have different :-A. speedsB. directions of propagationC. wavelengthsD. frequencies |
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Answer» Correct Answer - B |
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| 189. |
Which of the following statements is wrong?A. In an open pipe the fundamental frequency is `v//2 l`B. In a closed pipe, the closed end is a displacement nodeC. In an open pipe, only the odd harmonics of fundamental frequency are presentD. In a closed pipe, the fundamental frequency is `v//4 l` |
| Answer» Correct Answer - C | |
| 190. |
A source of sound moves towards an observe.A. The frequency of the sources increasedB. The velocity of sound in the medium is increasedC. The wavelength of sound in the medium towards the observer is decreasedD. The amplitude of vibration of the particles is increased |
| Answer» Correct Answer - C | |
| 191. |
How does the speed `v` of sound in air depend on the atmospheric pressure P?A. `v prop P^(-1)`B. `v prop P^(-2)`C. `v prop P^(1//2)`D. `v prop P^(0)` |
| Answer» Correct Answer - D | |
| 192. |
The change in frequency due to Doppler effect does not depend onA. the actual frequency of the waveB. the distance of the source from the listenerC. the velocity of the sourceD. the velocity of the observer |
| Answer» Correct Answer - B | |
| 193. |
A bullet passes past a person at a speed of `220 m s^-1`. Find the fractional change in the frequency of the whistling sound heard by the person as the bullet crosses the person. Speed of sound in air `= 330 m s^-1`. |
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Answer» Let the frequency of the bullet wil be f given `u=330 m/s, v_s=220m/s` a. Apparent frequency before crossing `rarr f^1=(330/(330-220))f=3f` b. Apparent frequency after crossing `rarr f^11=((330-(330+220))f=0.6f` So, `(f^11/f^1)=(0.6f)/(3f)=0.2` therefore FractioN/Al change `=1-0.2=0.8` |
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| 194. |
In previous question, what would be the apparent wavelengths in front of and behind the locomotive if a wind of speed `10 m//s` were blowing in the same direction as that in which the locomotive is travelling?A. `0.65 m, 0.73 m`B. `0.60 m, 0.73 m`C. `0.65 m, 0.78 m`D. `0.60 m, 0.71 m` |
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Answer» Correct Answer - C `f_(1) = f ((v+w)/(v+w-v_(s)))` `= 546.15 Hz` `lambda_(1) = (v+w)/(f_(1)) = (355)/(5+6.15) = 0.65 m` `f_(2) = f ((v+w)/(v+w+v_(s)))=500 ((345+10)/(345+10+30))=461 Hz` `lambda_(2)=(v+w)/f_(2)=(345+10)/461=0.77 m` |
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| 195. |
A point `A` is located at a distance `r = 1.5 m` from a point source of sound of frequency `600 H_(Z)`. The power of the source is `0.8 W`. Speed of sound in air is `340 m//s` and density of air is `1.29 kg//m^(3)`. Find at the point `A`, (a) the pressure oscillation amplitude`(Deltap)_(m)` (b) the displacement oscillation amplitude `A`. |
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Answer» Correct Answer - A::B::C::D (a) At a distance `r` from a point source of power `P`, the intensity of the sound is `I = (P)/(4 pi r^(2)) = (0.8)/((4 pi) (1.5)^(2))` or `I = 2.83 xx 10^(-2) W//m^(2)` ..(i) Further, the intensity of sound in terms of `(Delta p)_(m)`, `rho` and `nu` is given by `I = ((Delta p)^(2)m)/ (2 rho nu)` ...(ii) From Eqs. (i) and (ii), `(Deltap)_(m) = sqrt(2 xx 2.83 xx 10^(-2) xx 1.29 xx340)` `= 4.98 N//m^(2)` (b) Pressure oscillations amplitude `(Deltap)_(m)` and displacement oscillation amplitude `A` are related by the equation `(Delta p)_(m) = BAK` subsituting `B = pnu^(2)`, `k = (omega)/(nu)` and `omega = 2 pi f` We, ger, `(Delta p)_(m) = 2 pi Arho nu f` `:. A = ((Deltap)_(m))/(2 pi rho nu f) = (4.98)/(2 pi)(1.29)(340)(600)` `= 3.0 xx 10^(-6) m` |
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| 196. |
Speed of sound in air is `320 m//s` . A pipe closed at one end has a length of `1 m` and there is another pipe open at both ends having a length of `1.6 m` . Neglecting end corrections, both the air columns in the pipes can resonate for sound of frequencyA. `80 H_(Z)`B. `240 h_(Z)`C. `320 h_(Z)`D. `400 H_(Z)` |
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Answer» Correct Answer - D Closed pipe fundamental frequency is `f_(1) = (nu)/ (4l)= (320) / (4xx1) = 80 H_(Z)` other frequencies are `3 f_(1) : 5 f_(1)` etc. or `240 H_(Z)`, `400 H_(Z)` etc. Open pipe fundamental frequency is `f_(1) = (nu)/ (2l)= (320)/(2 xx 1.6) = 100 H_(Z)` Other frequencies are`f_(1) : 3 f_(1) : 4 f_(1)` etc . or `200 H_(Z)` , `300 H_(Z)` and `400 H_(Z)` etc. So, then resonate at `400 H_(Z)`. |
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| 197. |
Assertion : Sound level increases linearly with intensity of sound. Reason : If intensity of sound is doubled, sound level increases approximatel `3 dB` .A. If both Asseration and Reason are true an dthe Reason is correct explanation of the Asseration.B. If both Asseration and Reason are true but Reason is not the correct explanation of Asseration.C. If Asseration is true , but the Reason is false.D. If Asseration is false but the Reason is true. |
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Answer» Correct Answer - D `L = 10 log_(10)(I)/(I_(o))` Increase in the value of `L` is not linear with`I` . `Delta L = L_(2) - L_(1) = log .(I_(2)) / (I_(1))` `Delta L = 3 dB` if `I_(2) = 2I_(1)` |
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| 198. |
Assertion : A train is approachinng towards a hill . The driver of the train will hear beats. Reason : Apparent frequency of reflected sound observerd by driver will be more than the frequency of direct sound observered by him.A. If both Asseration and Reason are true an dthe Reason is correct explanation of the Asseration.B. If both Asseration and Reason are true but Reason is not the correct explanation of Asseration.C. If Asseration is true , but the Reason is false.D. If Asseration is false but the Reason is true. |
| Answer» Correct Answer - A | |
| 199. |
A 30.0-cm-long wire having a mass of 10.0 g is fixed at the two ends and is vibrated in its fundamental mode.A 50.0-cm-long closed organ pipe, placed with its open end near the wire, is set up into resonance in its fundamental mode by the vibrating wire. Find the tension in the wire. Speed of sound in air `= 340 m s^-1`. |
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Answer» Correct Answer - C::D Given `m=10gm` `=10xx10^-3kg` `I=30cm=0.3m` Let the tension in the string will be =T `m=Mass/(Unit length)` `=33x10^-3 kg/m` The fuN/Admental frequency `rarr n_0=1/(2L) sqrt(T/m)`………..1 `rarr n_0=(v/(4I))` `=340/(2xx30xx10^-2)` `=170Hz`.........2 `According to equation 1 and 2 we get `170=1/(2xx30xx10^-2)xxT/(33x10^-3)` `rarr T=347Newtons` |
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| 200. |
In an experiment, it was found that a tuning fork and sonometer wire gave 5 beats per second, both when the length of wire was 1m and 1.05m. Calculate the frequency of the fork.A. `420 Hz`B. `410 Hz`C. `210 Hz`D. `205 Hz` |
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Answer» Correct Answer - D `f prop (l)/(l) rArr f l =` constant `f_(1) l_(1) = f_(2) l_(2)` `(n + 5) (1) = (n - 5) (1.05)` `n + 5 = 1.05 n - 5.25` `0.05 n = 10.25` `n = 205 Hz` |
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