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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.
| 101. |
It is desired to increse the fundamental resonance frequency in a tube which is closed at one end. This can be achieved byA. replacing the air in the tube by hydrogen gasB. increasing the length of the tubeC. decreasing the length of the tubeD. opening the closed end of the tube |
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Answer» Correct Answer - A::C::D `n prop v//l` To incrase fundamental resonance frequency of tube. Velocity of sound in hydrogen can be increased. Length of tube can be decreased. For tube closed from one end fundamental frequency is `v//4l` For tube open from both ends fundamental frequency is `v//2l` |
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| 102. |
At SPT, The speed of sound in hydrogen is 1324 m/s then the speed of sound in airA. 331 m/sB. 220 m/sC. 340 m/sD. 230 m/s |
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Answer» Correct Answer - A Molecular weight of hydrogen = 2 Molecular weigh of oxygen = 32 Both are diatomic `v_(air)=sqrt(((gamma_(air)P_(air))/(p_(air)))),v_(H)=sqrt((gamma_(H)P_(H))/(P_(H)))` `(P_(air))/(p_(H))=(M_(air))/(M_(H))=(32)/(2)=16/1` `(v_(air))/(v_(H))=sqrt(((gamma_(air)Pp_(H))/(gamma_(H)Pp_(air))))rArr(v_(air))/(1324)=sqrt(((P_(H))/(P_(air))))` `" "v_(air)=(1324)/4=331 m//s` |
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| 103. |
Show that if the room temperature changes by a small amount from `T to T +/_T`, the fundamental frequency of an organ pipe changes from `v to v + /_v, where `(/_v)/v=1/2(/_T)/T`. |
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Answer» Correct Answer - A We know that `fpropsqrtT` According to the questioin `f+/_fprop sqrt(/_T+T)` `rarr (f+/_f)/f=sqrt(/_T+T)/T)` `rarr 1+(/_f)/f=(1+(/_T)/T)^(1/2)` `=1+1/2 (/_T)/T` `……. (neglecting other terms) ` `rarr (/_f)/f=(1/2)(/_T)/T` |
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| 104. |
Two loudspeakers `L_1` and `L_2` driven by a common oscillator and amplifier, are arranged as shown. The frequency of the oscillator is gradually increased from zero and the detector at D records a series of maxima and minima. If the speed of sound is `330ms^-1` then the frequency at which the first maximum is observed isA. `165 Hz`B. `330 Hz`C. `496 Hz`D. `660 Hz` |
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Answer» Correct Answer - B `Delta x = L_(2)D - L_(1)D = sqrt((99)^(2) + (40)^(2)) - 40` `= 41 - 40 = 1 m` For maximum intensity `Delta x = n lambda = n(v)/(f)` `f = (n v)/(Delta x) = (n xx 330)/(1) = 330 n` `n = 1, f = 330 Hz` |
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| 105. |
Two point sound sources `A` and `B` each to power `25pi w` and frequency `850 Hz` are `1 m` apart. The sources are in phase (a) Determine the phase difference between the waves emitting from `A` and `B` received by detector `D` as shown in figure. (b) Also determine the intensity of the resultant sound wave as recorded by detector `D`. Velocity of sound `= 340 m//s`. |
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Answer» Correct Answer - (a) `pi` , (b) `l = (sqrt(l_(A)) - sqrt(l_(B)))^(2) = (25//312)^(2)` (a) `lambda = (340)/(850) Deltaphi = (2pi)/(lambda) DeltaS` `= (2pi xx 850)/(340) xx [sqrt(1^(2) + 2.4^(2)) - 2.4]` `= (2pi xx 850)/(340) xx [2.6 - 2.4] = (2pi xx 850)/(340) xx 0.2 = pi` (b) `l = l_(A) + l_(B) + 2 sqrt(l_(A) + l_(B)) cos pi` `= (sqrt(l_(A)) - sqrt(l_(B )))^(2) ` `= [sqrt((P)/(4pi(2.4)^(2)))=sqrt((P)/(4pi(2.6)^(2)))]^(2)=[sqrt((25pi)/(4xxpixx(2.4)^(2)))-sqrt((25pi)/(4pi(2.6)^(2)))]^(2)=((5)/(2 xx 2.4) - (5)/(2 xx 2.63))^(2)` `[(5)/(2)((2.6 - 2.4)/(2.4 xx 2.6))]^(2)=((5)/(2)xx(0.2)/(2.4xx2.6))^(2)=((25)/(12xx26))^(2)=((25)/(312))^(2)` |
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| 106. |
Two identical loudspeakers are located at point `A` & `B, 2 m` apart. The loudspeakers are driven by the same amplifier (coherent and are in phase). A small detector is moved out from point `B` along a line frequency below which there will be no position |
| Answer» Correct Answer - `83 Hz` | |
| 107. |
A tunnig fork of frequency 256 Hz produces 4 beats per second with a wire of length 25 cm vibrating in its fundamental mode. The beat frequency decreases when the length is slightly shortened. What could be the minimum length by which the wire be shortened so that it produces no beats with the tuning fork ? |
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Answer» Correct Answer - C Given that `I = 25 cm = 25 xx 10^(-2)m` By shortening the wire the frequency increses `[f = ((1)/(2I)) sqrt((T)/(m))]` As the vibrating wire produces 4 beats with 256 Hz, its frequency must be 252 Hz or 260 Hz. Its frequecy must be 252 Hz, beacuse beat frequecy decreases by shortening the wire. So, `252 = (1)/(2 xx 25 xx 10^(-2)) sqrt((T)/(M)) ... (1)` Let length of the wire be I, after is slightly shortened, `rArr 252 = (1)/(2 xx I_1) sqrt((T)/(M))......(2)` Dividing (1) by (2), we get, `(252)/(256) = (I_1)/(2xx 25 xx 10^(-2))` `rArr I_1 = (252 xx 2 xx 25 xx 10^(-2))/(256)` = 0.24609 m So it should be shorten by (25 - 24.61) = 0.39 cm. |
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| 108. |
Two identical straight wires are stretched so as to products `6 beats// s` when vibrating simultaneously. On changing the tension slightly in one of them, the beat frequency remains unchanged. Denoting by `T_(2)` , `T_(2)` the higher and the lower initial tension in the strings, then it could be said that that while making the above changes in tension (a) `T_(2)` was decreased (b) `T_(2)` was decreased (c) `T_(1)` was decreased (d) `T_(1)`was decreased |
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Answer» `T_(1)gt T_(2)` `:. upsilon_(1) gt upsilon_(2)` or `f_(1) gt f_(2)` and `f_(1) - f_(2) = 6 H_(z)` Now, if `T_(1)` is increased, `f_(1)` will increase or `f_(1) - f_(2)` will increase . Therefore, (d) option is wrong. If `T_(1)` is decreased, `f_(1)` will decrease and it may be possible that now `f_(2) - f_(1)` become `6 H_(Z)` . Therefore, (c ) option is correct. Similarly , when `T_(2)` will increase and again `f_(2) - f_(1)` may because aqual to `6H_(Z)` . So , (b) is also correct . But (a) is wrong. |
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| 109. |
Two tuning forks with natural frequencies `340 H_(Z)` each move relative to a stationary observer . One forks moves away from the oberver while the other moves towards him at the same speed . The observer hearts beats of frequency `3 H_(Z)` . Find the speed the of the tuning fork (velocity of sound in air is `340 m//s)` . |
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Answer» Correct Answer - A::B::C::D Given, `f_(1) - f_(2) = 3` or `((nu)/(nu - nu_(s))) f - ((nu) / (nu +nu_(s))) f = 3` or `[(1)/((nu - nu_(s) /(nu)))- (1) / ((1 +nu_(s) / (nu)))] f= 3` or `[(1-nu_(s)/(nu))^(-1) -(1+nu_(s)/(nu))^(-1)] f= 3` or `[(1+nu_(s)/(nu)) -(1-nu_(s)/(nu))] f= 3` or `(2nu_(s)f)/ (nu) = 3` or Speed of tuning fork, `nu_(s) = (3nu)/(2f)` subsituting the values, we get `nu_(S)= ((3)(340))/((2)(340))= 1.5 m//S` |
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| 110. |
When an engine passes near to a stationary observer then its apparent frequencies occurs in the ratio 5/3. if the velocity of sound is 340 m/s then speed of engine isA. 540 m/sB. 270 m/sC. 85 m/sD. 52.5 m/s |
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Answer» Correct Answer - C `(f_(1))/(f_(2))=(V+V_(s))/(V-V_(s))` `V_(s)=85 m//s` |
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| 111. |
A girl stops singing a pure note. She is surprised to hear an echo of higher frequency, i.e., a higher musical pitch. Then :A. there could be some warm air between the girl and the reflecting surfaceB. there could be two identical fixed reflecting surfaces, one half a wavelengths of the sound wave away from the otherC. the girl could be moving towards a fixed reflectorD. the reflector could be moving towards the girl |
| Answer» Correct Answer - C::D | |
| 112. |
The minimum distance to hear echo (speed of sound in air is `340 m//s`)A. `15 m`B. `16 m`C. `17 m`D. `18 m` |
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Answer» Correct Answer - C `(2d)/(v) ge (1)/(10) rArr d_(min) = (v)/(20) = (340)/(20) = 17 m` |
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| 113. |
The musical interval between two tones of frequencies `320 Hz` and `240 Hz` isA. `80`B. `(4//3)`C. `560`D. `320 xx 240` |
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Answer» Correct Answer - B Musical scale is ratio of frequencies `= (320)/(240) = (4)/(3)` |
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| 114. |
The speed of sound in an air column of `80 cm` closed at one end is `320 m//s`. Find the natural freqencies of air column between `20 Hz` and `2000 Hz`. |
| Answer» Correct Answer - `100(2n + 1) Hz` where `n = 0, 1, 2, 3, ……, 9` | |
| 115. |
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» Correct Answer - A Let `n_0=`frequency of the tuning fork, `T=`Tension of the string `L=40cm=0.4m, m=4g` `=4xx10^-3kg` `So, m=Mass/(Unit Length)=10^-2kg/m` `n_0=1/(2L) sqrt(T/m)` So, `2nd Harmonic 2n_0=(2/(2L))sqrt(T/m)` As it is tension with fundamental frequency of vibration in the air column. `rarr 2n_0=340/4xx1=85Hz` `rarr 85=2/(2xx0.4)sqrt(T/14)` `=T=(85)^2xx(0.4)^2` `=11.6Newton` |
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| 116. |
A wire of length `40 cm` which has a mass of `4` g oscillates in its second harmonic and sets the air column in the tube to vibrations in its funrations in its fundamental mode as shows in figure. Assuming the speed of sound in air as `340 m//s` . Find the tension in the wire. |
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Answer» Correct Answer - A `2((nu)/(2l_(1)))= ((nu_(2))/(4l_(2)))` or `sqrt(T//mu)/(l_(1) = ((nu_(2))/(4l_(2))))` `:. T = mu ((nu_(2) l_(1))/(4l_(2)))^(2)` `= ((4 xx 10^(-3))/(0.4)) ((340 xx40)/(4 xx100))^(2)` `= 11.56 N` |
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| 117. |
Consider the following statements about and pasing through a gas. A. the pressure of the gas ast a point oscillages in time. B. The position of a small layer of the gas oscillates in time.A. both A and B are correctB. A is correct but B is wrongC. B is correct but A is wrongD. Both A and B are wrong |
| Answer» Correct Answer - A | |
| 118. |
When we clap our hands, the sound produced is best described by Here p denotes the change in pressure from the equilibrium valueA. `p = p_(0) sin (k x - omega t)`B. `p = p_(0) sin k x cos omega t`C. `p = p_(0) cos k x sin omega t`D. `p = Sigma p_(0) sin (k_(n) x - omega_(n) t)` |
| Answer» Correct Answer - D | |
| 119. |
When we clap our hands, the sound produced is best described by Here p denotes the change in pressure from the equilibrium valueA. `p = p_(0) sin (kx - omegat)`B. `p = p_(0) sin kx cos omegat`C. `p = p_(0) cos kx sin omegat`D. `p = sump_(on) sin (k_(n)x - omega_(n)t)` |
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Answer» Correct Answer - D During clappin muliple wave are produced with difference wave paremeters so wave is resultant of all these wave |
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| 120. |
When we clap our hands, the sound produced is best described by Here p denotes the change in pressure from the equilibrium valueA. `p=p_0sin(kx-omegat)`B. `p=p_0sinkxcosomegat`C. `p=p_0coskxsinomegat`D. `p=sump_(0n)sin(k_nx-omega_nt)` |
| Answer» Correct Answer - D | |
| 121. |
(iv) A closed organ pipe of length 83.2 cm and 6 cm diameter is vibrated. The velocity of sound is 340 m/s. find the number of overtones in this tube having frequency below 1000 Hz ?A. 2B. 4C. 1D. 3 |
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Answer» Correct Answer - B Effective length 83.2+0.3 d=85 cm `n=V/(4l) =340/(4xx0.85)=100` (fundamental frequency) other resonating frequency below 1000 Hz 100,300,500,700,900 so four overtone below 1000 Hz. |
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| 122. |
At what temperature will the speed of sound be double of its value at `[email protected] ` ? |
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Answer» Correct Answer - A::C `T_1=273, v_2=2V_1` `V_1=v, T_2=?` we know that `V a/T` `T_2/T_1=V_2^2/V_1^2` `rarr T_2=273xx(2)^2` `=4xx273K` So, temperature will be `4xx273=819^@C` So temperature will be `4xx273=819^@C` |
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| 123. |
At what temperature will the speed of sound be double of its value at `[email protected] ` ? |
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Answer» Correct Answer - A::C `T_1=273, v_2=2V_1` `V_1=v, T_2=?` we know that `V a/T` `T_2/T_1=V_2^2/V_1^2` `rarr T_2=273xx(2)^2` `=4xx273K` So, temperature will be `4xx273=819^@C` So temperature will be `4xx273=819^@C` |
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| 124. |
The speed of sound as measured by a student in the laboratory on a winter day is 340 m s when the rdoobmy temperature is `17^@C`. What speed will be measure another student repeating the experiment on a day when the room temperature is `[email protected]` ? |
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Answer» Correct Answer - A::C::D `V_1=340m/s, V_2=?` `T_1=273+17` `=290K` `T_2=273+32=305K` We know `V=a/sqrtT` `V_1/V_2=sqrtT_1/sqrtT_2` `rarr V_2=(sqrtV_1xxsqrtT_2)/sqrtT_1` `=340xxsqrt(350/290)` `=349m/s |
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| 125. |
The speed of sound as measured by a student in the laboratory on a winter day is 340 m s when the rdoobmy temperature is `17^@C`. What speed will be measure another student repeating the experiment on a day when the room temperature is `[email protected]` ? |
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Answer» Correct Answer - A::C::D `V_1=340m/s, V_2=?` `T_1=273+17` `=290K` `T_2=273+32=305K` We know `V=a/sqrtT` `V_1/V_2=sqrtT_1/sqrtT_2` `rarr V_2=(sqrtV_1xxsqrtT_2)/sqrtT_1` `=340xxsqrt(350/290)` `=349m/s |
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| 126. |
At certain instant the shape of a simple train of ple wave is `y = 12sin "(pix)/(50)` (`x` and `y` are in `cm`.). The velocity of the wave propagation is `100 cm//s` in a positive direction away from the origin. Find the equation giving the shape of the wave `0.25s` later. |
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Answer» Correct Answer - `y = 12 sin ((pix)/(50) - (pi)/(2))` `y = 12 sin "(pix)/(50)`, at `t = 0` `y = 12 sin "(pi)/(50)(x - vt)` `y = 12 sin "(pi)/(50)(x - 100 xx 0.25)` So equation of shape of wave `y = 12sin ((pix)/(50) - (pi)/(2))` |
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| 127. |
A train blowing its whistle moves with a constant velocity `v` away from an observer on the ground. The ratio of the natural frequency of the whistle to that measured by the observer is found to be `1.2`. If the train is not rest and the observer moves away from it at the same velocity, this ratio would be given byA. `0.51`B. `1.25`C. `1.52`D. `2.05` |
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Answer» Correct Answer - B `f_(1) = f_(0) (v_("sound"))/(v_("sound") + v_("train")) = (1)/(1.2)f_(0) rArr 1.2 v_("sound") = v_("sound") + v rArr v = (v_("sound"))/(5)` `f_(2) = f_(0) (v_("sound") - v_("man"))/(v_("sound")) = 0.8 rArr (f_(0))/(f_(2)) = 1.25` |
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| 128. |
The faintest sound the human ear can detect at a frequency of ` kHz` (for which ear is most sensitive) corresponds to an intensity of about `10^(-12)w//m^(2)`. Assuming the density of air `cong1.5kg//m^(3)` and velocity of sound in air `cong300m//s`, the pressure amplitude and displacement amplitude of the sound will be rspectively ____`N//m^(2)` and ____`m`.A. `3xx10^(-5) Pa`B. `2xx10^(-5) Pa`C. `5xx10^(-5) Pa`D. `4xx10^(-5) Pa` |
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Answer» Correct Answer - A `I=(P_(m)^(2))/(2rhoV)` `10^(-12)=(P_(m)^(2))/(2(1.5)(300))` `P_(m)=3xx10^(-4) P_(a)` |
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| 129. |
Figure shows a tube having sound source at one end and observe at other end. Source produces frequencies upto `10000 Hz`. Speed of sound is `400 m//s`. Find the frequencies at which person hears maximum intensity. |
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Answer» The sound wave bifurcates at the juction of the straight and the rectangular parts. The wave through the straight part travels a distance `p_(1) = 10 cm ` and the wave through the rectangular part travels a distance `p_(2) = 3 xx 10 xm = 30` before they mwwt agian and travel to the receiver. The path difference between the two wave received, therefore. `Deltap = p_(2) - p_(1) = 30 cm - 10 cm = 20 cm` The wavelength of either wave is `(v)/(upsilon) rArr (400 m//s)/(upsilon)`. For constructive interference, `Deltap = nlambda`, where `n` is an integer. or, `Deltap = n.(v)/(v) rArr v = (n.v)/(Deltap)` `rArr v = (400)/(0.1) = 4000 n` Thus, the frequencies within the specified range which cause maximum of intensity are `4000 xx 1 Hz, 4000 xx 2 Hz` |
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| 130. |
The speed of sound in oxygen `(O_2)` at a certain temperature is `460ms^-1`. The speed of sound in helium (He) at the same temperature will be (assume both gases to be ideal)A. `500 ms^(-1)`B. `650 ms^(-1)`C. `330 ms^(-1)`D. `460 ms^(-1)` |
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Answer» Correct Answer - B `V_(0_(2)) = sqrt((gammaRT)/(M)) = sqrt((7)/(5)(RT)/(32)) = 460` `V_(H_(e)) = sqrt((gammaRT)/(M)) = sqrt(((5)/(3)RT)/(4)) = sqrt((5)/(12)460xx460xx32xx(5)/(7)) = 1419 m//s` |
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| 131. |
A cylindrical metal tube has a length of 50 cm and is open at both ends. Find the frequencies between 1000 Hz and 2000 Hz at which the air column in the tube can resonate. Speed of sound in air is `340 m s^-1`. |
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Answer» Correct Answer - A::B::C::D Here given that `I=50cm, v=340m/s` `As it is an open organ pipe the fundamental freqency `f_1=(v/(2I)` `=340/(2xx50x10^-2)` So the harmonics are `f_3=3xx340=1020Hz` `f_5=5xx340=1700Hz` `f_6=6xx340=240Hz` so the possible frequencies are between 1000 Hz and 2000 Hz are 1020,1360,1700`. |
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| 132. |
In a resonance column experiment, a tuning fork of frequency 400 Hz is used. The first resonance is observed when the air column has a length of 20.0 cm and the second resonance is observed when the air column has a length of 62.0 cm. (a) Find the speed of sound in air.(b) How much distance above the open end does the pressure node form ? |
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Answer» Correct Answer - A::B::C Here given `I_2=0.67m, ` `I_1=0.2m, f=400Hz` we know that `lamda=2(I_2-I_1)` `rarr lamda=2(62-60)=84cm=0.84m` `rarrv=nlamda=0.84xx400=336m/s` we know from above that `I_1=d=lamda/4` `rarr d=lamda/4` `I_1=21-20=1cm` |
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| 133. |
An ultrasound signal of frequency 50 kHz is sent vertically into sea water. The signal gets reflected from the ocean bed and returns to the surface 0.80 s after it was emitted. The speed of sound in sea water is `1500ms^-1`. a. Find the depth of the sea. b. What is the wavelength of this signal in water. |
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Answer» a. Let the depth of the sea be d. the total distance travelled by the sigN/Al is 2d. By the question, `2d=(1500ms^-1)(0.8s)=1200m` or `d=600m` b. using the equation `v=vlamda` `lamda=v/v=(1500ms^1)/(50xx10^3s^-1)=3.0cm` |
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| 134. |
The displacement of a particle in a medium due to a wave travelling in the `x-`direction through the medium is given by `y = A sin(alphat - betax)`, where `t =` time, and `alpha` and `beta` are constants:A. The frequency of the wave is `alpha`B. The frequency of the wave is `alpha//(2pi)`C. The wavelength is `(2pi//beta)`D. The velocity of the wave is `alpha/beta` |
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Answer» Correct Answer - B::C::D |
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| 135. |
Assertion : Speed of sound in gass is independent of pressure of gas. Reason : With increase in temperature of gas speed of sound will increase.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 It is independent of pressure as long as temperature remains constant. |
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| 136. |
The equation of a sound wave in air is given by `rho = (0.01 N//m^(2)) [sin (1000 s^(-1)) t - (3.0 m^(-1))x]` (a) Find the frequency, wavelength and the speed of sound wave in air. (b) If the equilibrium pressure of air is `1.0 xx 10^(5) N//m^(2)`, what are the maximum and minimum pressure at a point as the wave passes through that point ? |
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Answer» `p = 0.01 sin (1000 t - 3 x)` (i) `p = p_(0) sin (omega t - k x)` (ii) Comparing (i) and (ii) `p_(0) = 0.01 N//m^(2), omega = 1000 rad//s, k = 3 m^(-1)` (a) `f = (omega)/(2 pi) = (1000)/(2 pi) = (500)/(pi) Hz` `k = (pi)/(lambda) rArr lambda = (2 pi)/(k) = (2 pi)/(3)m` `v = f lambda = (500)/(pi) xx (2 pi)/(3) = (1000)/(3) m//s` (b) `p_(max) = P_(0) + p_(0) = (1.01 xx 10^(5) + 0.01) N//m^(2)` `p_(min) = P_(0) - p_(0) = (1.01 xx 10^(5) - 0.01) N//m^(2)` `P_(0)`: atmospheric pressure |
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| 137. |
Speed of sound in gas is proportional toA. square root of isothermal elasticityB. square root pf adiabatic elasticityC. isothermal elasticityD. adiabatic elasticity |
| Answer» Correct Answer - B | |
| 138. |
It is possible to distinguish between the transverse and longitudinal waves by studying the property ofA. interferenceB. diffractionC. relflectionD. polarisation |
| Answer» Correct Answer - D | |
| 139. |
If copper has moduls of rigidity `12xx10^(10)N//m^(3)`and Bulk muodulus `12xx10^(11)`N/m and density 9 `g//cm^(3)`then find the velocity of longitudinal wave, when set-up in solid copper.A. 4389 m/sB. 5000 m/sC. 4000 m/sD. 4300 m/s |
| Answer» Correct Answer - A | |
| 140. |
The velocity of sound is not affected by change inA. temperatuerB. mediumC. pressureD. wavelength |
| Answer» Correct Answer - C | |
| 141. |
The velocity of sound in air is affected by change in the (i) atmospheric pressure (ii) moisture content of air (iii) temperature of air (iv) composition of air.A. (i), (ii), (iii)B. (i), (ii), (iv)C. (ii), (iii), (iv)D. (ii), (iii) |
| Answer» Correct Answer - C | |
| 142. |
In the above siltuation, bus is at rest blowing horm of frequency `f_(0).` A boy is at rest at some distance. What will be apparent frwuecy of sound, if the air star moving with the speed of 20 m/s from bus towards boy? A. `ltf_(0)`B. `gtf_(0)`C. `=f_(0)`D. None of these |
| Answer» Correct Answer - C | |
| 143. |
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 wel as the inertia propertyD. neither the elastic property nor the inertia property |
| Answer» Correct Answer - C | |
| 144. |
When we hear a sound, we can identify its source fromA. amplitude of soundB. intensity of soundC. wavelength of soundD. overtones present in the sound |
| Answer» Correct Answer - D | |
| 145. |
A place of cork is floating on water in a small tank. The cork oscillates up and down vertically when small ripples pass over the surface of water. The velocity of the ripples being `0.21 ms^(-1)`, wave length `15 mm`, and amplitude `5 mm`, the maximum velocity of the piece of cork is `(pi = (22)/(7))` A. `0.44 ms^(-1)`B. `0.24 ms^(-1)`C. `2.4 ms^(-1)`D. `4.4 ms^(-1)` |
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Answer» Correct Answer - A `n = (V)/(lambda) - (.21)/(15 xx 10^(-3)) = (210)/(15)` `V_(max) = Aomega = 5 xx 10^(-3) xx (210)/(15) xx 2pi` `= 70 xx 2 xx (22)/(7) xx 10^(-3) = .44m//sec`. |
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| 146. |
The apparent frequency is `f_(1)` when a souece of sound approches a stationary obsever with a speed u and `f_(2)` when the observe approaches the stationary source with same speed. If v is the velocity of sound, thenA. `f_(1)=f_(2)`B. `f_(1) gt f_(2) if u ltv`C. `f_(2) gt f_(1) if u ltv`D. `f_(2) gt f_(1) if u ltv` |
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Answer» Correct Answer - D `f_(1)=((v)/(v-v_(s)))f=((v)/(v-v))f=[((1)/(1-u/v))]f` When observer approaches towards the source `f_(2)=((v+v_(o))/(v))f=((v+u)/(v))f=(1-u/v)f` `therefore" "f_(2) gt f_(1)` |
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| 147. |
A sounding body emitting a frequency of `150 H_(Z)` is dropped from a height. During its fall under gravity it crosses a balloon moving upwards with a constant velocity of `2 m//s` one second after it started to fall . The difference in the frequency observer by the man in balloon just before and just afer crossing the body will be (velocity of sound `= 300 m//s`, `g = 10 m//s^(2))`A. 12B. 6C. 8D. 4 |
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Answer» Correct Answer - A `f=f_(0)((vpmv_(0))/(vpmv_(s)))` When approaching `f_(a)=150[(300+2)/(300-10)]` when receding `f_(r)=150[(300-2)/(300+10)]rArr f_(a)-f_(r)~=12` hence (A) |
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| 148. |
The temperature at which the velocity of sound in oxygen will be same as that of nitrogen at `15^(@) C` isA. `112^(@) C`B. `72^(@) C`C. `56^(@) C`D. `17 ^(@) C` |
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Answer» Correct Answer - C `nu = sqrt((gammaRT) /(M)) prop sqrt((T) / (M))` `(gamma = 1.4` for both) :. `T_(o_2)/M_(o_2) = T_(N_2)/M_(N_2)` :. `T_(o_2) = ((M_(o_2))/(M_(N_2))) T_(N_2)` = `((32) /(28)) (273 + 15)` = `329 K = 56^(@) C` |
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| 149. |
Calculate the speed of sound in oxygen at `273 K`. |
| Answer» Correct Answer - A::C | |
| 150. |
At what temperature will the speed of sound be double of its value at `0^@C ` ? |
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Answer» Correct Answer - A::C `nu prop sqrt(T)` `:. nu_(2)/(nu_(1)) = sqrt(T_(2)/(T_(1)))` or `T_(2) = (nu_(2)/(nu_(1)))^(2) T_(1) = (2)^(2) (273)` `= 1092 K` `= 819^(@)C` |
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