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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.
| 201. |
When a tuning fork vibrates with `1.0 m` or `1.05 m` long wire of a , `5` beats per second are produced in each case. If the frequency of the tuning fork is `5f` (in `Hz`) find `f`. |
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Answer» Correct Answer - 41 Let frequency tuning fork `= f` frequency of wave in wire `= (v)/(2l)` when `l = 1 m` `f_(1) = (v)/(2)` when `l = 1.05m` `f_(2) = (v)/(2 xx (1.05))` `f_(1) - f = 5` , and `f - f_(2) = 5 rArr f_(1) - f_(2) = 10` `rArr (v)/(2)(1 - (1)/(1.05)) = 10` or `v = (20 xx 1.05)/(0.05) = 420 m//s`. `f_(1) = (v)/(2) = 210 Hz`. `f = f_(1) - 5 = 205 Hz`. |
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| 202. |
An electrically maintained tuning fork vibrates with constant frequency and constant amplitude. If the temperature of the surrounding air increases but pressure remains constant, the sound produced will have (i) larger wavelength (ii) larger frequency (iii) larger velocity (iv) larger time periodA. `(i), (iv)`B. `(i), (iii)`C. `(iii), (iv)`D. `(ii), (iv)` |
| Answer» Correct Answer - B | |
| 203. |
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. `6%`B. `12%`C. `18%`D. `24%` |
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Answer» Correct Answer - B `f_("before crossing") = f_(0)((c)/(c - v_(S))) = 1000((320)/(320 - 20))` `f_("after crossing") = f_(0)((c)/(c + v_(S))) = 1000((320)/(320 + 20))` `Deltaf = f_(0)"((2cv_(S))/(c^(2) - v_(S)^(2))) , (Deltaf)/(f) xx 100% = (2 xx 320 xx 20)/(300 xx 340) xx 100 = 12.54% approx 12%` |
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| 204. |
A listener is at rest with respect to the souce of sound. A wind starts blowing along the line joining the source and the observer. Which of the following quantities do not change?A. FrequencyB. Velocity of soundC. wavelengthD. time period |
| Answer» Correct Answer - A::D | |
| 205. |
A hospital uses an ultrasonic scanner to locate tumour in a tissue. What is the wavelength of sound in a tissue in which the speed of sound is `1.7 km//s` ? The operating frequency of the scanner is `4.2 MHz`. |
| Answer» Correct Answer - `4.1 xx 10^(-4)m` | |
| 206. |
A bat emitting an ultrasonic wave of frequecy `4.5 xx 10^4` Hz flies at a speed of `6ms^(-1)` between two parallel walls. Find the two frequecies heard by the bat and the beat frequecy between the two. The speed of sound is `330ms^(-1)` |
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Answer» Correct Answer - A::B::C::D Let the bat be flying between three walls `W_1W_2.` so it will listen two frequecny reflecting from walls `W_2 and W_1` So, appernt frequency, as received by well W, `W_2 = ((330 + 0+0)/(330 - 0.6))xx f = (330)/(324)` There, fore app. frequency received by the bat from wall `W_2` is given by `F_(B2) at wall w_1` `=((330 + 0- (-6))/(330 + 0 +0))f_1 W_2` `=((336)/(330))xx((330)/(324))f` Similarly, the apperent frequency recevied by the bat from wall `W_1` is, `f_(B1) = ((324)/(336))f` so the beat frequency head by the bat `= 4.67 xx 10^4 - 4.3430 xx 10^4` =3270 Hz. |
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| 207. |
A source on a swing which is covering an angle `theta` from the vertical is producing a frequency `v`. The source is distant `d` from the place of support of swing. If velocity of sound is `c`, acceleration due to gravity is `g`. then the maximum and minimum frequency hearfd by a listener in froun of swing is A. `(cv)/(sqrt(2gd-c)),(cv)/(sqrt(2gd+c))`B. `(cv)/(sqrt(2gd(1-costheta))-c), (cv)/(sqrt(2gd(1-costheta))+c)`C. `(cv)/(c-sqrt(2gd(1-costheta))), (cv)/(c+sqrt(2gd(1-costheta)))`D. `(cv)/(c-sqrt(2gd(1-sintheta))), (cv)/(c+sqrt(2gd(1-sintheta)))` |
| Answer» Correct Answer - C | |
| 208. |
The wavelength of light coming from a distant galaxy is found to be `0.5%` more than that coming from a source on earth. Calculate the velocity of galaxy. |
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Answer» `(Delta lambda)/(lambda) = (v)/(c )` `(Delta lambda)/(lambda) - (0.5)/(100) = 5 xx 10^(-3)` `v = (Delta lambda)/(lambda) c = 5 xx 10^(-3) xx 3 xx 10^(8)` `= 15 xx 10^(5) m//s` |
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| 209. |
A person blows into open- end of a long pipe. As a result,a high pressure pulse of air travel down the pipe. When this pulse reaches the other end of the pipe,A. a high-pressure pulse starts traveling up the pipe, if the other end of the pipe is open.B. a low-pressure pulse starts traveling up the pipe, if the other end of the pipe is open.C. a low-pressure pulse starts traveling up the pipe, if the other end of the pipe is closed.D. a high-pressure pulse starts traveling up the pipe, if the other end of the pipe is closed. |
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Answer» Correct Answer - B::D At open end phase of pressure wave charge by `pi` so compression returns as rarefraction. While at closed end phase of pressure wave does not change so compression return as compression. |
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| 210. |
A source an an observer are at rest `w.r.t` ground. Which of the following quantities will remain same, if wind blows from source to observer ?A. FrequencyB. speed of soundC. wavelengthD. Time period |
| Answer» Correct Answer - A::C | |
| 211. |
A source and an observer are at rest `w.r.t` ground. Which of the following quantities will remain same, if wind blows from source to oberserver ?A. FrequencyB. Speed of soundC. wavelengthD. Time period |
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Answer» Correct Answer - A::D Dopplar effect in Frequency appears when there is relative motion between source and observer |
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| 212. |
Consider the following statements about and passing through a gas. A. the pressure of the gas at a point oscillates with time. B. The position of a small layer of the gas oscillates with time.A. Both `(i)` and `(ii)` are correct.B. `i` is correct but `(ii)` is wrongC. `(ii)` is correct but `(i)` is wrongD. Both `(i)` and `(ii)` are wrong |
| Answer» Correct Answer - A | |
| 213. |
Consider the situation shown in figure. The wire which has a mass of 4.00 g oscillates in its second harmonic and sets the air column in the tube into vibrations in its fundamental mode. Assuming that the speed of sound in air is `40 m s^-1`, find the tension in the wire. . |
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Answer» Frequency of wire `f_(2) = (2)/(2 l) sqrt((T)/(mu))` `= (l)/(l) sqrt((T)/(M//l)) = (1)/(0.4) sqrt((T)/(4 xx 10^(-3)// 0.4)) = 2.5 sqrt(100 T)` `= 25 sqrt(T)` Fundamental frequency of closed pipe `f_(1) = (v)/(4L) = (340)/(4 xx 1) = 85` `f_(2) = f_(1)` `25sqrt(T) = 85` `T = ((17)/(5))^(2) = (3.4)^(2) = 11.56 N` |
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| 214. |
A travelling wave in a stretched string is described by the equation `y = A sin (kx - omegat)` the maximum particle velocity isA. `Aomega`B. `omega//k`C. `domega//dk`D. `x//t` |
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Answer» Correct Answer - A |
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| 215. |
When two tuning forks (fork `1` and fork `2`) are sounded simultaneously, `4` beats per second are heard. Now some tape is attached on the prong of the fork `2`. When the tuning fork are sounded again, `6` beats per second are heard. If the frequency of fork `1` is `200 Hz`, then what was the original frequency of fork `2` ?A. `200 Hz`B. `202 Hz`C. `196 Hz`D. `204 Hz` |
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Answer» Correct Answer - C The frequency of fork `2` is `= 200 +- 4 = 196` or `204 Hz` Since, on attaching the tape the prong of fork `2`, its frequency decrease, but now the number of beats per second, is `6` i.e., the frequency difference now increases. It is possible only when before attaching the tape, the frequency of fork `2` is less thanthe frequency of tunin fork `1`. Hence, the frequency of fork `2 = 196 Hz`. |
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| 216. |
Two sound waves move in the same direction in the same medium. The pressure amplitudes of the waves are equal but the wavelength of the first wave is double the second. Let the average power transmitted across a cross-section by the first wave be `P_(1)` and that by the second wave be `P_(2)`. ThenA. `P_(1)//P_(2) = 1`B. `P_(1)//P_(2) = 2`C. `s_(1)//s_(2) = 1//2`D. `s_(1)//s_(2) = 2//1` |
| Answer» Correct Answer - A::D | |
| 217. |
Two idential straight wires are stretched so as to produce `6` beats per second when vibrating simultaneously. On changing the tension slightly in one of them, the beat frequency remains unchanged. Denoting by `T_(1)`, `T_(2)` the higher and the lower initial tension in the strings, then it could be said that while making the above changes in tension,A. `T_(2)` was decreasedB. `T_(2)` was increasedC. `T_(1)` was increasedD. `T_(1)` was decreased |
| Answer» Correct Answer - B::D | |
| 218. |
Two wires are fixed in a sanometer. Their tension are in the ratio `8:1` The lengths are in the ratio `36:35` The diameter are in the ratio `4:1` Densities of the materials are in the ratio `1:2` if the lower frequency in the setting is `360Hz`. The beat frequency when the two wires are sounded together isA. `5`B. `8`C. `6`D. `10` |
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Answer» Correct Answer - D `(T_(1))/(T_(2)) =(8)/(1), (l_(1))/(l_(2)) = (36)/(35), (r_(1))/(r_(2)) = (4)/(1), (rho_(1))/(rho_(2)) = (1)/(2)` `(f_(1))/(f_(2)) = (l_(1))/(l_(2)) sqrt(T_(1)/(T_(2))(rho_(2))/(rho_(1)).(r_(2)^(2))/(r_(1)^(2)))` `= (35)/(36)sqrt((8)/(1).(2)/(1).((1)/(4))^(2)) = (35)/(36)` `f_(1) gt f_(2), f_(2) = 360` `f_(1) = (360 xx 36)/(35) = 370` Beat frequency `= f_(1)-f_(2) =10 Hz` |
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| 219. |
A tuning fork of `512 H_(Z)` is used to produce resonance in a resonance tube expertiment. The level of water at first resonance is `30.7 cm` and at second resonance is `63.2 cm` . The error in calculating velocity of sound is (a) `204.1 cm//s` (b) `110 cm//s` (c) `58 cm//s` (d) `280 cm//s` |
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Answer» Actual speed of sound in air is `330 m//s` `(lambda)/(2)=(l_(2)-l_(1))=(63.2-30.7)cm= 32.5 cm = 0.325m` or `lambda = 0.65 m` `:.` Speed of sound observed, `upsilon_(0) = f lambda= 512xx 0.65 = 332.8 m//s` `:.` Error in calculating velocity of sound = `2.8 m//s = 280 cm//s` `:.` The correct option is (d). |
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| 220. |
A circular plate of area `0.4 cm^(2)` is kept at distance of `2m` source of power `piW`. Find the amount of energy received by plate in `5` secs. |
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Answer» The energy emitted by the speaker in one second is `pi J`. Let us consider a sphere of radius `2.0 m` centered at the speaker. The energy `pi J` falls normally on the total surface of this sphere in one second. The energy fallinf on the area `0.4 cm^(2)` of the microphone of the second `= (0.4 xx 10^(-4))/(4 pi 2^(2)) xx pi = 2.5 xx 10^(-6)J` The energy falling on the microphone in one second `= 2.5 xx 10^(-6)J xx 5 = 12.5 muJ`. |
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| 221. |
Find the speed of sound in `H_(2)` at temperature `T`, if the speed of sound in `O_(2)` is `450 m//s` at this temperature, |
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Answer» Correct Answer - `1800 m//s` `v = sqrt((gammaRT)/(M))` since termperature, `T` is constant, `(v_(H_(2)))/(v_(H_(1))) = sqrt((M_(O_(2)))/(M_(O_(2)))) = sqrt((32)/(2)) = 4` `v(H_(2)) = 4 xx 450 = 1800 m//s` Aliter : The speed of sound in a gas is given by `u = sqrt((gammaP)/(rho))`. At `STP, 22.4` liters of oxygen has a mass of `32 g` whereas the same volume of hydrogen has a mass of `2 g`. Thus, the density of oxygen is `16` times the density of hydrogen at the same temperature and pressure. As `gamma` is same for both the gases, `v_(("hydrogen"))/(v_(("oxygen"))) = sqrt((rho_(("oxygen")))/(rho_(("hydrogen"))))` or, `v_(("hydrogen") = 4v_(("oxygen"))` `= 4 xx 450 m//s = 1800 m//s` |
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| 222. |
On a day when the speed is `345 m//s`, the fundamental frequency of a closed organ pipe is `220 H_(Z)` . (a) How long is this closed pipe? (b) The second overtone of this pipe has the same wavelength as the third harmonic of an open pipe . How long is the open pipe ? |
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Answer» Correct Answer - A::B::C::D (a) `f = (nu)/(4l)` `:. l = (nu) / (4 f) = (345)/(4 xx 220)` `= 0.392 m` (b) `5((nu)/(4l_(c))) = 3((nu)/(2l_(o)))` `:. l_(o) = (6)/(5) l_(c) = (6)/(5) (0.392)` `= 0.470 m` |
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| 223. |
In a wave motion `y = a sin (kx - omegat)`, `y` can representA. electric fieldB. magnatic fieldC. displacementD. pressure change |
| Answer» Correct Answer - A::B::C::D | |
| 224. |
An organ pipe has two successive harmonics with frequencies `400` and `560 H_(Z)`. The second of sound in air is `344 m//s`. (a) Is the an open or a closed pipe? (b) What two harmonics are three? (c ) What is the length of the pipe? |
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Answer» Correct Answer - A::B::C::D (a) `(400)/(560) = (5)/(7)` Since, these are odd harmonic (`5` and `7`). Hence, pipe is closed. (c ) Given frequencies are integer multiple of `80 H_(Z)`. Hence, fundamental is `80 H_(Z)`. `(nu)/(4l) = 80` `:. l = (nu)/(320) = (344)/(320)` `= 1.075 m` |
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| 225. |
An organ pipe of length `L` is open at one end and closed at other end. The wavelengths of the three lowest resonating frequencies that can be produced by this pipe areA. 4L,2L,LB. 2L,L,L/2C. 2L,L,2L/3D. 4L,4L/3,4L/5 |
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Answer» Correct Answer - D `L=(2n-1) (lambda)/4` `lambda=(4L)/((2n-1)), lambda_(1)=(4L)/(2(1)-1)=4L` `lambda_(2)=(4L)/(4-1) =(4L)/3` , `lambda_(3)=(4L)/(6-1) =(4L)/5` |
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| 226. |
A closed organ pipe and an open organ pipe of same length produce `4` beats when they are set into vibrations simultaneously. If the length of each of them were twice their initial lengths, the number of beats produced will beA. `2`B. `4`C. `1`D. `8` |
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Answer» Correct Answer - A `f prop (1)/ (l)` Both frequency will becomes half. Hence, `f_(b) = f_(1) - f_(2)` will also become half. |
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| 227. |
If `T` is the reverberation time of an auditorium of volume `V` thenA. `T prop (1)/(V)`B. `T prop (1)/(V^(2))`C. `T prop V^(2)`D. `T prop V` |
| Answer» Correct Answer - D | |
| 228. |
A railway engine whistling at a constant frequency moves with a constant speed. It goes past a stationary observer standing beside the railway track. The frequency `(n)` of the sound heard by the observer is plotted agains time `(t)`. Which of the following best represents the resulting curve?A. B. C. D. |
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Answer» Correct Answer - D |
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| 229. |
A resonance tube is resonated with tuning fork of frequency `256 Hz`. If the length of first and second resonating air columns are `32 cm` and `100 cm`, then end correction will beA. `1 cm`B. `2 cm`C. `4 cm`D. `6 cm` |
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Answer» Correct Answer - B `(lambda)/(4) = l_(1) + e` …….(1) `(3lambda)/(4) = l_(2) + e` ……(2) from `(1)` and `(2) , e = 2 cm` |
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| 230. |
A source of sound revloving in a circle of radius `5 m` is emitting a signal of frequency `320 Hz`. It completes one revolution in `(pi)/(2)` seconds. If the difference between maximum and minimum frequencies of the signal heard at a point `30 m` from the centre of the circle `= (25)/(7) P`. Find `P`. (Given speed of sound `= 330 ms^(-1)`)] |
| Answer» Correct Answer - 12 | |
| 231. |
A fixed source of sound emitting a certain frequency appears as `f_(a)` when the observer is apporoaching the source with `upsilon_(0)` and `f_(r)` when the observer recedes from the source with the same speed. The frequency of source isA. `(f_(r) + f_(a))/(2)`B. `(f_(a) - f_(r))/(2)`C. `sqrt(f_(a)f_(r))`D. `(2f_(r)f_(a))/(f_(r) + f_(a))` |
| Answer» Correct Answer - A | |
| 232. |
A fixed source of sound emitting a certain frequency appears as `f_(a)` when the observer is apporoaching the source with `upsilon_(0)` and `f_(r)` when the observer recedes from the source with the same speed. The frequency of source isA. (a) `(f_(r) + f_(a))/(2)`B. (b) `(f_(r) - f_(a))/(2)`C. (c ) `sqrt(f_(a) f_(r)`D. (d) `(2f_(r)f_(a))/(f_(a) + f_(a))` |
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Answer» Correct Answer - A `f_(a) = f(nu + nu_(o)/(nu))` `:. nu_(o)/(nu )= f_(a)/(f) - 1` …(i) `f_(r) = f ((nu - nu_(o))/(nu))` `:. nu_(o)/(nu) = 1 - (f_(r))/(f)` …(ii) From Eqs. (i) and (ii), we get `f = (f_(a) + f_(r))/(2)` |
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| 233. |
A detector is released from rest over a source of sound of frequency `f_(o) = 10^(3) H_(Z)`. The frequency observer by the decector at time `t` is plotted in the graph. The speed of sound in air `(g = 10 m//s^(2))` A. `330 m//s`B. `350 m//s`C. `300 m//s`D. `310 m//s` |
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Answer» Correct Answer - C `f = f_(o) ((nu + nu_(o))/(nu))` `= 10^(3) (1 + (10 t)/(nu))` (as `nu_(o) = g t`) Hence, `f` versus `t` graph is straight line of slope `(10^(4))/(nu)`. `:. (10^(4))/(nu) =` slope `= (100)/(3)` `:. nu = 300 m//s` |
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| 234. |
A glass tube of `1.0 m` length is filled with water. The water can be drained out slowly at the bottom of the tube. If a vibrating tuning fork of frequency `500 c//s` is brought at the upper end of the tube and the velocity of sound is `300 m//s`, then the total number of resonances obtained will beA. `4`B. `3`C. `2`D. `1` |
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Answer» Correct Answer - B `lambda = (v)/(f) = (300)/(500) = 0.6 m` Resonance will occur at lengths `(lambda)/(4) , (3 lambda)/(4) , (5 lambda)/(4) , (7 lambda)/(4)` ,… `0.15 m, 0.45 m, 0.75 m, 1.05 m`, … `1.05 m` is not possible Total number of resonances `= 3` |
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| 235. |
How do we identify our friend from his voice while sitting in a dark room ? |
| Answer» Correct Answer - The quality of sound helps us to identify the sound. | |
| 236. |
When you speak of 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 | |
| 237. |
A small source of sound moves on a circle as shown in figure and an observer is sitting at O. Let `v_1,v_2,v_3` be the frequencies heard when the source is at A,B and C respectively. .A. `v_1gtv_2ltv_3`B. `v_1=v_2gtv_3`C. `v_2gtv_3gtv_1`D. `v_1gtv_3gtv_2` |
| Answer» Correct Answer - C | |
| 238. |
The intensity of a plane progressive wave of frequency `1000 Hz` is `10^(-10) Wm^(-2)`. Given that the speed of sound is `330 m//s` and density of air is `1.293 kg//m^(3)`. Then the maximum change in pressure in `N//m^(2)` isA. `3 xx 10^(-4)`B. `3 xx 10^(-5)`C. `3 xx 10^(-3)`D. `3 xx 10^(-2)` |
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Answer» Correct Answer - A `I = (p_(0)^(2))/(2 rho v)` `p_(0) = sqrt(2 pv I) = sqrt(2 xx 1.293 xx 330 xx 10^(-10))` `= 30 xx 10^(-3) = 3 xx 10^(-4) N//m^(2)` |
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| 239. |
If the fundamental frequency of a pipe closed at one is `512 H_(Z)` . The frequency of a pipe of the same dimension but open at both ends will beA. `1024 h_(Z)`B. `512 H_(Z)`C. `256 H_(Z)`D. `128 H_(Z)` |
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Answer» Correct Answer - A `f_(0) = (nu)/ (2l)` and `f_(c) = (nu) / (4l)` `:. f_(o) = 2 f_(c)` |
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| 240. |
Three sound sources `A`, `B` and `C` have frequencies `400`, `401` and `402 H_(Z)`, respectively. Cacluated the number of beats noted per second. |
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Answer» Correct Answer - A::B::C Let as make the following table. Beat time period for `A` and `B` is `1` s. It implies that `A` and `B` are in phase at time `t = 0`, they are again in phase after `1` s. Same is the case with `B` and `C` . But beat time period for `A` and `C` is `0.5`s. Therefore , beat time period for all together `A, B` and `C` will be `1` s. Because if , at `t = 0, A, B` and `C` all are in phase then after `1` s. (`A` and `B`) and (`B` and `C`) will again be in phase for the first time while (`A` and `C`) will be in phase for the second time. Or we can that all `A, B` and `C` are again in phase after `1` s. `:.` Beat time period , `T_(b) = 1 s` or Beat frequency , `f_(b) = (1)/(T_(b)) = H_(Z)` Suppose at time `t`, the equations of waves are `y_(1) = A_(1) sin 2 pif_(A)t` `(omega = 2pi f)` `y_(2) = A_(2) sin 2pi f_(B) t` and `y_(3) = A_(3) sin 2pi f_(C) t` If they are beat time at some given instant of time `t`, then `2 pif_(A)t = 2 pi f_(B)t = 2pi f_(C)t` ...(i) Let `T_(b)` be the beat time period, i.e. after time `T_(b)` they all are again in phase. As `f_(C) gt f_(B) gt f_(A)` , so `2pi f_(C) (t + T_(b)) = 2 pi f_(A) (t + T_(b)) + 2 mpi` ...(ii) and `2pi f_(B) (t + T_(b)) = 2 pi f_(A) (t + T_(b)) + 2 npi` ....(iii) Here, `m` and `n (lt m)` are positive integers. From Eqs. (i) and (ii) `(f_(C) - f_(A)) T_(b) = m` ...(iv) Similary, From Eqs. (i) and (ii) `(f_(B) - f_(A)) T_(b) = n` ...(v) Dividing Eq. (iv)by Eq. (v), `(m)/(n) = (f_(C) - f_(A))/(f_(B) -f_(A))= (402 - 400)/(401 - 400)= (2)/(1)` Thus, letting `m = 2` and `n = 1` `T_(b)= (m)/(f_(C) - f_(A))` [from Eq. (iv)] = `(2)/(2) = 1 H_(Z)` `:.` Beat frequency , `f_(b) = (1)/(T_(b)) = 1 H_(Z)` |
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| 241. |
A source `S` emitting sound of `300 Hz` is fixed on block `A` which is attache to free end of a spring `S_(A)` as shown in figure. The detector `D` fixed on block `B` attached to free end of spring `S_(B)` detects this sound. The blocks `A` and `B` simultaneously displaced towards each other thrugh a distance of `2.0m` and then left to vibrate. If the product of maximum and minimum freuquencies of sound detected by `D` is `K + 10^(4) sec^(-2)`. Find `K`. Given the vibrational frequencies of each block is `5//pi Hz`. speed of sound in air `= 300 m//s` |
| Answer» Correct Answer - 9 | |
| 242. |
A stationary observer receives sonic oscillations from two tuning forks one of which approaches and the other recedes with the same velocity. As this takes place, the observer hears the beats of frequency `f = 2.0 H_(Z)`. Find the velocity of each tuning fork if their oscillation frequency is `f_(o) = 680 H_(Z)` and the velocity of sound in air is `upsilon = 340 m//s`. |
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Answer» `f_(b) = f_(1) - f_(2) ` `2 = 680 ((340)/(340 - nu_(s))) - 680 ((340)/(340 +nu_(s)))` `= 680 (1- (nu_(s))/(340))^(-1) - 680 (1+ (nu_(s))/(340))^(-1)` Using Binomial expansion, we have `2 = 680 [(2 nu_(s))/(340)] rArr nu_(s) = 0.5 m//s` |
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| 243. |
One train is approaching an observer at rest and another train is receding from him with the same velocity `4 m//s` . Both trains blow whistles of same frequency of `243 H_(Z)` . The beat frequency in `H_(Z)` as heard by observer is (speed of sound in air = `320 m//s`)A. `10`B. `6`C. `4`D. `1` |
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Answer» Correct Answer - B `f_(b) = 243 ((320) / (320 - 4 ))- 243 ((320)/ (320 + 4)) = 6 H_(Z)` |
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| 244. |
A stationary observer receives sonic oscillations from two tuning forks one of which approaches and the other recedes with the same velocity. As this takes place, the observer hears the beats of frequency `f = 2.0 H_(Z)`. Find the velocity of each tuning fork if their oscillation frequency is `f_(o) = 680 H_(Z)` and the velocity of sound in air is `upsilon = 340 m//s`.A. 1m/sB. 2m/sC. 0.5 m/sD. 1.5 m/s |
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Answer» Correct Answer - C `[(v/(v-v_(s)))-(v/(v+v_(s)))]f_(0)=2Hz` `v_(s)=0.5 m//s` |
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| 245. |
An observer standing on a railway crossing receives frequencies 2.2 kHz and 1.8 kHz when the tran approaches and recedes from the observer. Find the velocity of the train (speed of sound in air is 300 m/s). |
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Answer» `{:(rarr v_(s),,rarr v_(o) = 0),(overset(*)(S),,overset(*)(O)),(,rarrv,):}` When source is moving towards observer, frequency received by stationary observer `f_(1) = f((v)/(v - v_(s)))` (i) `{:(rarr v_(s)=0,,rarr v_(o)),(overset(*)(S),,overset(*)(O)),(,rarrv,):}` When source is moving away from observer, frequency received by stationary observer `f_(2) = f((v)/(v + v_(s)))` (ii) `(f_(1))/(f_(2)) = (v + v_(s))/(v - v_(s))` `(2.2)/(1.8) = (v + v_(s))/(v - v_(s))` `2.2 v - 2.2 v_(s) = 1.8 v + 1.8 v_(s)` `0.4 v = 4 v_(s)` `v_(s) = 0.1 v = 0.1 xx 300 = 30 m//s` |
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| 246. |
A source is approaching towards an observer with constant speed along the line joining them. After crossing the observer, source recedes from observer with same speed. Let `f` is apparent frequency heard by observer. Then,A. `f` will keep on increasing during approachingB. `f` will keep on decreasing during recedingC. `f` will remain constant during apporachingD. `f` will remain constant during receding |
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Answer» Correct Answer - C::D During approach, `f = f_(o) ((nu)/(nu - nu_(s)))` `gt f_(o)` but `f` is constant. During receding. `f = f_(o) ((nu)/(nu - nu_(s)))` `lt f_(o)` but `f` is again constant. |
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| 247. |
In an organ pipe (may be closed or open ) of `99 cm` length standing wave is set up , whose equation is given by longitudinal displacement. `xi = (0.1 mm) cos ( 2pi)/(0.8) (y + 1 cm) cos (400) t`where `y` is measured from the top of the tube in `metres` and `t "in" seconds` . Here `1 cm` is the end correction. Equation of the standing wave in terms of excess pressure is ( take bulk modulus ` = 5 xx 10^(5) N//m^(2)`))A. `P_(ex) = (125 piN//m^(2)) sin "(2pi)/(80)(y + 1 cm) cos 2pi(400t)`B. `P_(ex) = (125 piN//m^(2)) cos "(2pi)/(80)(y + 1 cm) sin 2pi(400t)`C. `P_(ex) = (225 piN//m^(2)) sin "(2pi)/(80)(y + 1 cm) cos 2pi(200t)`D. `P_(ex) = (225 piN//m^(2)) cos "(2pi)/(80)(y + 1 cm) sin 2pi(200t)` |
| Answer» Correct Answer - A | |
| 248. |
When beats are produced by two progressive waves of nearly the same frequency, which one of the following if correct?A. the beat frequency depends on the position where the beats are heardB. the beat frequency decreases as time passesC. the particles vibrate simple harmonically with a frequency equal to the difference of the two frequenciesD. the amplitude of vibration at any point changes simple harmonically with a frequency equal to the difference of the two frequencies |
| Answer» Correct Answer - D | |
| 249. |
In a stationary longitudinal wave, nodes are points ofA. maximum pressureB. minimum pressureC. minimum pressure variationD. maximum pressure variation |
| Answer» Correct Answer - A::D | |
| 250. |
A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium. (a) Does the pulse have a definite (i) wavelength, (ii) frequency, (iii) speed of propagation ? (b) If the pulse rate is `1` after every `20 s`, (i.e. the whistle is blown for a split second after every `20 s`) is the frequency of the note produced by the whistle equal to `(1)/(20) = 0.05 Hz` ? |
| Answer» Correct Answer - The pulse does not have a definite wavelength or frequency but has a definite speed of propagation (in a non-dispersive medium) (b) No. | |