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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The equation of stationary wave along a stretched string is given by `y=5 sin ((pix)/(3)) cos 40 pi t `, where x and y are in cm and t in second. The separation between two adjacent nodes is |
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Answer» Correct Answer - 3 cm |
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| 2. |
In stationary waves, distance between a node and its nearest antinode is 20 cm . The phase difference between two particles having a separation of 60 cm will be |
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Answer» Correct Answer - `3pi//2` |
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| 3. |
The equation of a stationary wave is ` y = 0.8 cos ((pi x)/(20)) sin 200 pi t` where ` x` is in cm and t is in s. The separation between consecutive nodes will beA. `20 cm`B. `10 cm`C. `40 cm`D. `30 cm` |
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Answer» Correct Answer - A Standard equation ` y = A cos ( 2 pi x)/( lambda) sin ( 2 pi vt)/(lambda)` By comparing this equation with given equation. `( 2 pi x)/( lambda) = (pi x)/( 20) rArr lambda = 40 cm` Distance between two nodes ` = lambda//2 = 20 cm`. |
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| 4. |
Two waves are approaching each other with a velocity of `20m//s` and frequency `n`. The distance between two consecutive nudes isA. `(20)/(n)`B. `(10)/(n)`C. `(5)/(n)`D. `(n)/(10)` |
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Answer» Correct Answer - B |
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| 5. |
Spacing between two successive nodes in a standing wave on a string is `x`. If frequency of the standing wave is kept unchanged but tension in the string is doubled, then new sapcing between successive nodes will become:A. `x//sqrt(2)`B. `sqrt(2)x`C. `x//2`D. 2x |
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Answer» Correct Answer - B |
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| 6. |
If `v_(1) , v_(2) and v_(3)` are the fundamental frequencies of three segments of stretched string , then the fundamental frequency of the overall string isA. `v_(1) + v_(2) + v_(3)`B. `[(1)/(v_(1)) + (1)/(v_(2)) + (1)/( v_(3))]^(-1)`C. `v_(1) v_(2) v_(3)`D. `[v_(1) v_(2) v_(3)]^(1//3)` |
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Answer» Correct Answer - B `l = l_(1) + l_(2) + l_(3)` [ `because v prop (1)/(l)`] `(k)/(v) = (k)/(v_(1)) + (k)/(v_(2)) + (k) + (k)/( v_(3))` `(1)/(v) = (1)/(v_(1)) + (1)/( v_(2)) + (1)/(v_(3))` `v = [(1)/(v_(1)) + (1)/( v_(2)) + (1)/(v_(3))]^(-1)` |
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| 7. |
If the length of a stretched string is shortened by `40 %` and the tension is increased by `44 %`, then the ratio of the final and initial fundamental frequencies isA. `3 : 4`B. `4 : 3`C. `1 : 3`D. `2 :1` |
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Answer» Correct Answer - D `n = (1)/( 2l) sqrt((T)/(m))` `(n_(2))/(n_(1)) = (l_(1))/(l_(2)) sqrt((T_(2))/(T_(1)))` `= (l_(1))/([l_(1) - (40)/(100) l_(1)]) sqrt (((T_(1) + (44)/(100) T_(1))/(T_(1)))` `= (100)/(60) xx (12)/(10) = 2 :1` |
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| 8. |
To decrease the fundamental frequency of a stretched string fixed at both ends one mightA. increase its tensionB. increase its wave velocityC. increase its lengthD. decrease its linear mass density |
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Answer» Correct Answer - C `f_(0) = (1)/(2 l) sqrt((T)/(mu))` |
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| 9. |
A stretched string of length ` 1 m` fixed at both ends , having a mass of `5 xx 10^(-4) kg` is under a tension of `20 N`. It is plucked at a point situated at `25 cm` from one end . The stretched string would vibrate with a frequency ofA. `400 Hz`B. `100 Hz`C. `200 Hz`D. `256 Hz` |
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Answer» Correct Answer - C At `25 cm` , there will be antinode . So , wire will vibrate in two loops . ` v = (2)/( 2 l) sqrt (( T xx l)/( M)) or v = sqrt((T)/(Ml)) = sqrt((20)/(5 xx 10^(-4) xx 1))` ` = sqrt( 4 xx 10^(4)) Hz = 200 Hz` |
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| 10. |
A sonometer wire , `100 cm` in length has fundamental frequency of `330 Hz`. The velocity of propagation of tranverse waves along the wire isA. `330 m//s`B. `660 m//s`C. `115 m//s`D. `990 m//s` |
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Answer» Correct Answer - B For fundamental mode `(lambda//2) = 100 cm or lambda = 200 cm` As `n = 330 Hz`, hence `V = n lambda = 330 xx (200)/(100) = 660 m//s` |
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| 11. |
Both neon `[ M_(Ne) = 20 xx 10^(-3) kg]` and helium `[M_(He) = 4 xx 10^(-3) kg]` are monoatomic gases and can be assumed to be ideal gases. The fundamental frequency of a tube (open at both ends ) of neon is `300 Hz at 270 K (R = ( 25//3) J//K mol)` The length of the tube isA. `(5)/(12) m`B. `(sqrt(3))/(12) m`C. `( 5 sqrt(3))/(12) m`D. ` ( 5 sqrt(3)) m` |
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Answer» Correct Answer - C Fundamental frequency `n_(Ne) = (1)/( 2L) sqrt((gamma RT)/( M_(Ne)))` `n_(Ne) = 300 Hz , M_(Ne) = 20 xx 10^(-3) kg` `gamma = (5)/(3) , R = (20)/(3) J//mol K` `T = 270 K` `L = (1)/( 2 xx 300) sqrt((5/3 xx 25/3 xx 270)/(20 xx 10^(-3)))` ` = ( 250 sqrt(3))/( 2 xx 300) = ( 5 sqrt(3))/(12) m` |
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| 12. |
Mark out the correct statement(s) regarding waves.A. standing waves appear to be stationary but transfer of energy from one particle to another continues to take place.B. A standing wave not only appears to be stationary but net transfer of energy from one particle to the other is also equal to zero.C. A standing wave does not appear to be stationary and net transfer of energy from one particle to the other is also non - zero.D. A standing wave does not appear to be stationary , but net transfer of energy from one particle to the other is zero. |
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Answer» Correct Answer - B Standing waves from two waves of equal amplitude , same frequency , same wavelength travelling in opposite directions superimose , as a result , the net transfer of energy through any cross - section is zero in standing waves. |
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| 13. |
Two speakers are placed as shown in Fig.7.98. Mark out the correct statement(s) A. If a person is moving along `AB` , he will hear the sound as loud , faint , loud and so onB. If a person moves along `CD`, he will hear loud , faint , loud and so onC. If a person moves along `AB` , he will hear uniform intense soundD. If a person moves along `CD` , he will hear uniform intense sound |
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Answer» Correct Answer - B::C At any point on line `AB` , the phase difference between two waves is zero and hence waves will interfere constructively . Along `CD` , the phase difference changes and waves interfere constructively and destructively and , hence sound will be loud , faint and so on . |
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| 14. |
Mark the correct statementsA. If all the particle of a string are oscillating in same phase , the string is resonating in its fundamental toneB. To observe interference , two sources of same frequency must be placed some distance apart from each otherC. To observe beats , two sources of same amplitude must be placed some distance apart from each otherD. None of the above |
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Answer» Correct Answer - A::B In case of a stationary wave , all the particles lying between two consecutive nodes , oscillate in the same phase . Since all the particles of given string are oscillating in the same phase , therefore , all the particles of the string lie between two consecutive nodes . Hence , the string is oscillating in the single loop . It means ,it is oscillating in its fundamental tone . Hence , option (a) is correct. Interference is a phenomenon of obtaining constant intensity at a fixed position but the intensity varies with position of the point to point . Hence , to observe interference , two sources having same frequency must be placed some distance apart . Hence option (b) is correct. Beats is a phenomenon of obtaining an intensity which varies with time . To obtain beats , two sources having different frequencies are required . Therefore , option ( c) is wrong. |
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| 15. |
Which of the following statements are correct ?A. The decrease in the speed of sound at high altitudes is due to a fall in pressure .B. The standing wave on a string under a tension , fixed at its ends , does not have well - defined nodes .C. The phenomenon of beats is not observable in the case of visible light waves.D. The apparent frequency is `f_(1)` when a source of sound approached a stationary observer with a speed `u` and is `f_(2)` when the observer approaches the same stationary source with the same speed . Then `f_(2) lt f_(1) , if u lt v`, where `v` is the speed of sound. |
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Answer» Correct Answer - B::C::D Statement `(a)` is incorrect . A change in pressure has no effect on the speed of sound . The decrease in the speed at high altitudes is due to fall in temperature . Statement `(b)` is correct . Standing waves are produced due to superposition of the incident waves and the waves reflected from the fixed ends of the string . Since , the ends are never perfectly rigidly fixed , the amplitude of the reflected wave is always less than that of the incident wave. Consequently , the resultant amplitudes at nodes is not exactly zero . Thus , the nodes are not well defined . Statement `( c)` is also correct . To observe beats , the difference between the two interfering frequencies must be less than about `10 - 16 Hz`. Since , visible light waves have very high frequencies , beats are not observed due to persistence of vision. Statement `(d)` is also correct . We know that `f_(1) = (f)/( 1 - (u)/(v))` `(i)` And `f_(2) = f( 1 + (u)/(v))` `(ii)` Expression Eq. `(i)` may be written as `f_(1) = f ( 1 - (u)/(v))^(-1)` Extanding binomically and retaining terms up to order `u^(2)//v^(2)`, we have `f_(1) = v( 1 + (u)/(v) + (u^(2))/(v^(2)))` `(iii)` Comparing Eqs. `(ii)` and `(iii)` , we find that `f_(1) gt f_(2)`. |
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| 16. |
Which of the following functions represent a stationary wave ? Here `a , b` and `c` are constants:A. `y = a cos (bx) sin ( ct)`B. `y = a sin (bx) cos ( ct)`C. `y = a sin (bx + ct)`D. `y = a sin ( bx + ct ) + a sin ( bx - ct)` |
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Answer» Correct Answer - A::B::D A stationary wave is characterized by a function of type ` y = f(t) g(x)`. Hence , choices `(a)` and `(b)` represent a stationary wave . Choice `(d)` is superposition of two oppositely travelling waves of the same amplitude and same frequency , which gives rise to a stationary wave. Hence choice `(d)` also represents a stationary wave. |
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| 17. |
Two wires are fixed in a sonometer. Their tensions 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 360 Hz. The beat frequency when the two wires are sounded together isA. 5B. 8C. 6D. 10 |
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Answer» Correct Answer - D |
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| 18. |
In a standing transerse wave on a string :A. In one time period all the particles are simultaneously at rest twice.B. All the particles must be at their positive extremes simultaneously once in one time periodC. All the particles may be at their positive extremes simultaneously once in a time period.D. All the particles are never at rest simultaneously. |
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Answer» Correct Answer - A::C |
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| 19. |
Two plane harmonic sound waves are expressed by the equations. `y_(1)(x,t)-A cos (0.5 pi x-100 pit), y_(2)(x,t)=A cos(0.46 pix-92pi t)` (All parameters are in MKS) : How many times does an observer hear maximum intensity in one second :-A. `4`B. `6`C. `8`D. `10` |
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Answer» Correct Answer - A `y_(1) = A cos (0.5 pi x - 100 pi t)` `y_(2) = A cos (0.46 pi x - 92 pi t)` For the first wave angular frequency is `omega_(1) = 100 pi , f_(1) = 50 Hz`. For the second wave angular frequency is `omega_(2) = 92 pi , f_(2) = 46 Hz`. Frequency at which the amplitude of resultant wave varies `f_(A) = (f_(1) - f_(2))/(2) = ( 50 - 46)/(2) = 2` Time interval between this is maximum. `Delta t = (1)/( 2 f_(A))` `Delta t = (1)/(4)` Therefore , the number of time intensity is maximum in time `1 s is 4`. |
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| 20. |
Two tuning forks `A and B` are sounded together and `8 beats//s` are heard . `A` is in resonance with a column of air `32 cm` long in a pipe closed at one end and `B` is increased by one `cm`. Calculate the frequency of fork . |
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Answer» Correct Answer - `264 Hz , 256 Hz` `n_(A) - n_(B) = 8` Also `n_(A) = (v)/( 4 (0.32)) , n_(B) = (v)/( 4 (0.33))` `(v)/( 4 (0.32)) - (v)/( 4(0.33) = 8` `v = 338 m//s` `:. n_(A) = (v)/( 4 xx 0.32) = (338)/(4 xx 0.32) = 264 Hz`. `n_(B) = n_(A) - 8 = 256 Hz`. |
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| 21. |
The vibrations of string of length `60 cm` fixed at both ends are presented by the equations ` y = 4 sin ( pi x//15) cos ( 96 pi t)` where `x and y` are in cm and t in `s`. The maximum displacement at `x = 5 cm` isA. `2 sqrt(3) cm`B. `4 cm`C. zeroD. `4 sqrt(2) cm` |
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Answer» Correct Answer - A For ` x = 5 , y = 4 sin ((5 pi)/( 15)) cos (96 pi t)` ` = 2 sqrt(3) cos (96 pi t)` So , `y` will be minimum when `cos (96 pi t ) = max = 1` `y_(max) = 2 sqrt(3) "cm at" x = 5` |
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| 22. |
The length of two open organ pipes are `l` and `(l+deltal)` respectively. Neglecting end correction, the frequency of beats between them will b approximately.A. `v/(2l)`B. `v/(4l)`C. `(vDeltal)/(2l^(2))`D. `(vDeltal)/(l)` |
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Answer» Correct Answer - C |
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| 23. |
A stiff wire is bent into a cylinder loop of diameter `D`. It is clamped by knife edges at two points opposite to each other . A transverse wave is sent around the loop by means resonance frequency (fundamental mode) of the loop in terms of wave speed `v` and diameter `D` isA. `(v)/(D)`B. `( 2v)/( pi D)`C. `(v)/( pi D)`D. `(v)/( 2 pi D)` |
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Answer» Correct Answer - C Supports for the loop are resonable for nodes at two points . `pi r = n ((lambda)/(2))` `pi (D)/(2) = n (lambda)/(2)` `f = n((v)/(pi D))` Fundamental frequency ` = v// pi D` |
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| 24. |
Two waves of slightly different frequencies `f_(1) and f_(2) (f_(1) gt f_(2))` with zero phase difference , same amplitudes ,travelling in the same direction superimpose .A. Phenomenon of beats is always observed by human ear.B. Intensity of resultant wave is a constant.C. Intensity of resultant wave varies periodically with time with maximum intensity `4 a^(2)` and minimum intensity zero.D. A maxima appears at a time `1//[2 (f_(1) - f_(2))]` later ( or earlier ) than a minima appears . |
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Answer» Correct Answer - C::D `t = 0 , (1)/((f_(1) - f_(2))) , (2)/((f_(1) - f_(2))) , (3)/((f_(1) - f_(2))) ,…..` are times at which maxima are obtained `t = (1/2)/((f_(1) - f_(2))) , (3/2)/((f_(1) - f_(2))) , (5/2)/((f_(1) - f_(2))) ,…` are times at whhich minima are obtained. |
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| 25. |
Two waves of equal frequency `f` and velocity `v` travel in opposite directions along the same path. The waves have amplitudes `A` and `3 A` . Then:A. the amplitude of the resulting wave varies with position between maxima of amplitude `4 A` and minima of zero amplitude.B. the distance between a maxima and adjacent minima of amplitudes is `v//2f`C. maximum amplitude is `4 A` and minimum amplitude is ` 2A`D. The position of a maxima or minima of amplitude does not change with time |
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Answer» Correct Answer - C::D `y = y_(1) + y_(2)` ` = A sin (omega t - kx) + 3 A sin(omega t + kx)` `= A sin omega t cos kx - A cos omega t sin kx + 3 A sin omega t cos kx + 3 A cos omega t sin kx` ` = 4 A sin omega t cos kx + 2 A cos omega t sin kx` ` = 2 A sin omega t cos k x + 2 A sin (omega t + kx)` It is combination of a stationary and travelling wave. Maximum amplitude `= 4 A` Minimum amplitude ` = 2 A` Distance between points having amplitude ` 4 A` and `2 A` will be ` = lambda//4 = v//4 f` |
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| 26. |
Statement I : The principle of superpositions states that amplitudes , velocities , and , accelerations of the particles of the medium due to the simultaneous operation of two or more progressive simple harmonic waves are the vector sum of the separate amplitude , velocity and acceleration of those particles under the effect of each such wave acting alone in the medium Statement II : Amplitudes , velocities and accelerations are linear functions of the displacement of the particle and its time derivates.A. Statement I is true , Statement II is true , Statement II is a correct explanation for Statement I.B. Statement I is true , Statement II is true , Statement II is NOT a correct explanation for Statement I.C. Statement I is true , Statement II is false.D. Statement I : is false , Statement II is true. |
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Answer» Correct Answer - C Principle of superposition holds true only when the vectors are linear functions of variable and its derivatives. |
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| 27. |
The equation for the fundamental standing sound wave in a tube that is closed at both ends if the tube is `80 cm` long and speed of the wave is `330 m//s` is (assume that amplitude of wave at antinode to be `s_(0)`)A. `y = s_(0) cos (3.93 t) sin (1295 x)`B. `y = s_(0) sin (7.86 t) cos (1295 x)`C. `y = s_(0) cos (7.86 t) sin (1295 x)`D. `y = s_(0) cos (1295 t) sin (3.93 x)` |
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Answer» Correct Answer - D Since `f_(n) = n ((v)/( 2 L)) = n((330)/( 1.6)) = 206 n` `lambda_(n) = ( 2 L)/(n) = (1.6)/(n)` `{ L = (n lambda_(n))/(2)}` And the standing wave equation with nodes at both ends is `s = s_(0) sin(3.93 n x) cos (1295 n t)` For fundamental mode//frequency ` n = 1` `s = s_(0) = sin (3.93 x) cos (1295 t)` |
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| 28. |
A cylindrical tube open at both ends, has a fundamental frequency `f` in air. The tube is dipped vertically in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is nowA. `v//2`B. `v`C. `3 v//4`D. ` 2 v` |
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Answer» Correct Answer - B `f_(open) = (v)/( 2l) = v` `f_(closed) = (v)/(4 ((l)/(2))) = (v)/( 2l) = v` |
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| 29. |
A wave representing by the equation `y = a cos(kx - omegat)` is suerposed with another wave to form a stationary wave such that point `x = 0` is a node. The equation for the other wave isA. ` a sin ( kx + omega t)`B. ` - a cos ( kx - omega t)`C. ` - a cos (k x + omega t)`D. `- a sin ( kx - omega t)` |
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Answer» Correct Answer - C Since the point ` x = 0` is a node and reflectiion is taking place from point ` x = 0` . This means that reflection must be taking place from the fixed end and hence the reflected ray must suffer an additional phase change of `pi` or a path change of `lambda//2`. So , If `y_("incident") = a cos (kx - omega t)` , then `y _("reflected") = -a cos (omega t + kx)` |
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| 30. |
Vibrating tuning fork of frequency `n` is placed near the open end of a long cylindrical tube. The tube has a side opening and is fitted with a movable reflecting piston. As the piston is moved through `8.75 cm`, the intensity of sound changes from a maximum to minimum. If the speed of sound is `350 m//s`. Then `n` is |
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Answer» Correct Answer - 1000 Hz |
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| 31. |
A string fixed at both ends whose fundamental frequency is `240 Hz` is vibrated with the help of a tuning fork having frequency `480 Hz`, thenA. The string will vibrate with a frequency of `240 Hz`B. The string will vibrate in resonance with the tuning forkC. The string will vibrate in resonance with a frequency of `480 Hz`, but is not a resonance with the tuning forkD. The string is in resonance with the tuning fork and hence vibrate with a frequency of `240 Hz` |
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Answer» Correct Answer - B If one of the natural frequencies of the string matches with the source frequency , then resonance condition will arise and the string will vibrate with source frequency . |
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| 32. |
A glass tube `1.5 m` long and open at both ends, is immersed vertically in a water tank completely. A tuning fork of 660 Hz is vibrated and kept at the upper end of the tube and the tube is gradually raised out of water the total number of resonances heard before the tube comes out of water taking velocity of sound air `330m//s` isA. 12B. 6C. 8D. 4 |
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Answer» Correct Answer - B |
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| 33. |
A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode. Amplitude at the centre of the string is 4 mm. Minimum distance between the two points having amplitude 2 mm is:A. ` 1 m`B. `75 cm`C. `60 cm`D. `50 cm` |
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Answer» Correct Answer - A `lambda = 2l = 3m` Equation of standing wave ( As `x = 0` is taken as a node) `y = 2 A kx cos omega t`, given ` 2 A = 4 mm` To find value of `x` for which amplitude is `2 mm` , we have ` 2 mm = ( 4mm) sin kx` `( 2pi)/( lambda) x = ( pi)/(6) rArr x_(1) = (1)/(4) m` `( 2pi)/( lambda) x = (pi)/( 2) + (pi)/( 3) rArr x_(2) = 1.25 m` `x_(2) - x_(1) = 1 m` |
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| 34. |
Stationary waves are produced in 10 m long stretched string. If the string vibrates in 5 segments and wave velocity 20 m/s then the frequency is :- |
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Answer» Correct Answer - 5 Hz. |
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| 35. |
A `40 cm` long brass rod is dropped one end first onto a hard floor but is caught before it topples over . With an `3 kHz` tone . The speed of sound in brass isA. `600 m//s`B. `1200 m//s`C. `2400 m//s`D. `4800 m//s` |
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Answer» Correct Answer - C The brass rod is open at both ends . So the longitudinal waves will have a fundamental frequency `f_(0) = (v)/( 2l)`. `v = (3000) (2) ((40)/(100))` `v = 2400 m//s` |
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| 36. |
If the phase difference between the two wave is `2 pi` during superposition, then the resultant amplitude is |
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Answer» Correct Answer - Maximum |
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| 37. |
The fundamental frequency of a sonometre wire is n . If its radius is doubled and its tension becomes half, the material of the wire remains same, the new fundamental frequency will be |
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Answer» Correct Answer - `(n)/(2sqrt2)` |
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| 38. |
Velocity of sound in air is `320 m//s`. The resonant pipe shown in Fig. 7.81 cannot vibrate with a sound of frequency . A. `80 Hz`B. `240 Hz`C. `320 Hz`D. `400 Hz` |
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Answer» Correct Answer - C ` f = (v)/( 4 l) = (320)/( 4) Hz = 80 Hz` Since even harmonics cannot be present therefore `320 Hz` `(= 4 xx 80)` is ruled out. |
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| 39. |
The length , radius , tension and density of string`A` are twice the same parameters of string `B`. Find the ratio of fundamental frequency of `B` to the fundamental frequency of `A`. |
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Answer» Correct Answer - `4` `f prop ((T//mu)^(1//2))/( L)` When `mu = mass per unit length = rho a = rho ( pi r^(2))` So , `f prop ((T//rho)^(1//2))/( r L)` `(f_(2))/(f_(1)) = ((T_(2))/(T_(1)))^(1//2) ((rho_(1))/(rho_(2)))^(1//2) ((r_(1) L_(1))/(r_(2) L_(2)))` `= ((1)/(sqrt(2))) ( sqrt(2)) (4) = 4` |
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| 40. |
When a sound wave is reflected from a wall the phase difference between the reflected and incident pressure wave is:A. `0`B. `pi`C. `pi//2`D. `pi//4` |
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Answer» Correct Answer - A When a sound wave gets reflected from a rigid boundary , the particles at the boundary are unable to vibrate . Thus , a ref,ected wave is generated which interferes with the oncoming wave to produce zero displacement at the rigid boundary . At these points ( zero displacement) , the pressure variation is maximum . Thus , a reflected pressure wave has the same phase as the incident wave . |
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| 41. |
Two tuning forks `A and B` give `18 beats "in" 2 s`. A resonates with one end closed air column of `15 cm` long and `B` with both ends open column of `30.5` long. Calculate their frequencies. |
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Answer» Correct Answer - `549 Hz , 540 Hz`. `n_(A) = (v)/( 4 xx 0.15) , n_(B) = (v)/( 2 xx 0.305)` Clearly `n_(A) gt n_(B) = 9` `(v)/( 0.60) - (v)/( 0.61) = 9` ` v = 329.4 m//s` `n_(A) = (329.4)/(0.60) = 549 Hz` `n_(B) = n_(A) - 9 = 549 - 9 = 540 Hz`. |
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| 42. |
The superposition takes place between two waves of frequency f and amplitude a . The total intensity is directly proportional to |
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Answer» Correct Answer - `4a^(2)` |
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| 43. |
A sound wave of frequency `440 Hz` is passing through in air. An `O_(2)` molecule `( mass = 5.3 xx 10^(- 26) kg)` is set in oscillation with an amplitude of `10 ^(-6) m`. Its speed at the centre of its oscillation isA. `1.70 xx 10^(-5) m//s`B. `17.0 xx 10^(-5) m//s`C. `2.76 xx 10^(-3) m//s`D. `2.77 xx 10^(-5) m//s` |
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Answer» Correct Answer - C `v_(max) = omega_(n) A = ( 2 pi f) A = ( 2 pi) = (2 pi) (440) (10^(-6))` ` = 2.76 xx 10^(-3) m//s` |
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| 44. |
Estimation of frequency of a wave forming a standing wave represented by `y = A sin kx cos t` can be done if the speed and wavelength are known using `speed = "Frequency" xx "wavelength"` . Speed of motion depends on the medium properties namely tension in string and mass per unit length of string . A string may vibrate with different frequencies . The corresponding wavelength should be related to the length of the string by a whole number for a string fixed at both ends . Answer the following questions: A string fixed at both ends having a third overtone frequency of `200 Hz` while carrying a wave at a speed of `30 ms^(-1)` has a length ofA. `30 m`B. `22.5 cm`C. `30 cm`D. `10.25 cm` |
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Answer» Correct Answer - C Third overtone ( fourth harmonic) frequency in string is `( 4 v)/( 2 l) = 200 Hz` ` l = ( 4v)/(400) = (v)/(100) = (30)/(100) 0.30 m = 30 cm` |
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| 45. |
A long tube contains air at a pressure of `1 atm` and a temperature of `107^(@) C`. The tube is open at one end and closed at the other by a movable piston. A tuning fork near the open end is vibrating with a frequency of `500 Hz`. Resonance is produced when the piston is at distance `19 , 58.5 and 98 cm` from the open end. The molar mass of air is `28.8 g//mol`. The ratio of molar heat capacities at constant pressure and constant volume for air at this temperature is nearlyA. `1.66`B. `1.4`C. `1.33`D. `1.5` |
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Answer» Correct Answer - B `v = sqrt(( gamma Rt)/(M)) rArr v = 395 m//s` `M = 28.8 g//mol = 28.8 xx 10^(-3) kg//mol` `T = 107^(@) C = 380 K` `R = 8.3 J//mol K` `gamma = ( v^(2) M)/(RT) = ((395)^(2) xx 28.8 xx 10^(-3))/(8.3 xx 380) = 1.4` |
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| 46. |
In a standing wave experiment , a `1.2 - kg` horizontal rope is fixed in place at its two ends `( x = 0 and x = 2.0 m)` and made to oscillate up and down in the fundamental mode , at frequency of `5.0 Hz`. At `t = 0` , the point at `x = 1.0 m` has zero displacement and is moving upward in the positive direction of `y - axis` with a transverse velocity `3.14 m//s`. Tension in the rope isA. `60 N`B. `100 N`C. `120 N`D. `240 N` |
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Answer» Correct Answer - D `mu = (1.2)/(2) = 0.6 kg//m` `n = 5 Hz` `lambda = 2 l = 4 m` `V = n lambda = 5 xx 4 = 20 m//s` Using `v = sqrt((T)/(mu))` `T = 20^(2) xx 0.6 = 240 N` `((delta y)/( delta t))_(max) = 3.14 m//s` `( 2 A)omega = 3.14` Amplitude `2A = (3.14)/(2 xx (3.14) xx 5) = 0.1 m` Equation of standing wave is `y = (0.1) sin (pi)/(2) s sin (10 pi) t` |
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| 47. |
In a standing wave experiment , a `1.2 - kg` horizontal rope is fixed in place at its two ends `( x = 0 and x = 2.0 m)` and made to oscillate up and down in the fundamental mode , at frequency of `5.0 Hz`. At `t = 0` , the point at `x = 1.0 m` has zero displacement and is moving upward in the positive direction of `y - axis` with a transverse velocity `3.14 m//s`. Speed of the participating travelling wave on the rope isA. `6 m//s`B. `15 m//s`C. `20 m//s`D. `24 m//s` |
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Answer» Correct Answer - C `mu = (1.2)/(2) = 0.6 kg//m` `n = 5 Hz` `lambda = 2 l = 4 m` `V = n lambda = 5 xx 4 = 20 m//s` Using `v = sqrt((T)/(mu))` `T = 20^(2) xx 0.6 = 240 N` `((delta y)/( delta t))_(max) = 3.14 m//s` `( 2 A)omega = 3.14` Amplitude `2A = (3.14)/(2 xx (3.14) xx 5) = 0.1 m` Equation of standing wave is `y = (0.1) sin (pi)/(2) s sin (10 pi) t` |
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| 48. |
In a standing wave experiment , a `1.2 - kg` horizontal rope is fixed in place at its two ends `( x = 0 and x = 2.0 m)` and made to oscillate up and down in the fundamental mode , at frequency of `5.0 Hz`. At `t = 0` , the point at `x = 1.0 m` has zero displacement and is moving upward in the positive direction of `y - axis` with a transverse velocity `3.14 m//s`. What is the correct expression of the standing wave equation ?A. `(0.1) sin ( pi//2) x sin (10 pi) t`B. `(0.1) sin (pi) x sin (10 pi) t`C. `(0.05) sin (pi//2) x cos (10 pi) t`D. `(0.04) sin (pi//2) x sin (10 pi) t` |
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Answer» Correct Answer - A `mu = (1.2)/(2) = 0.6 kg//m` `n = 5 Hz` `lambda = 2 l = 4 m` `V = n lambda = 5 xx 4 = 20 m//s` Using `v = sqrt((T)/(mu))` `T = 20^(2) xx 0.6 = 240 N` `((delta y)/( delta t))_(max) = 3.14 m//s` `( 2 A)omega = 3.14` Amplitude `2A = (3.14)/(2 xx (3.14) xx 5) = 0.1 m` Equation of standing wave is `y = (0.1) sin (pi)/(2) s sin (10 pi) t` |
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| 49. |
The frequencies of two sound sources are 256 Hz and 260 Hz, At `t=0` the intesinty of sound is maximum. Then the phase difference at the time `t=1//16` sec will beA. ZeroB. `pi`C. `pi//2`D. `pi//4` |
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Answer» Correct Answer - C |
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| 50. |
A sounding fork whose frequency is `256 Hz` is held over an empty measuring cylinder. The sound is faint , but if just the right amount of water is poured into the cyclinder , it becomes loud. If the optimal amount of water produce an air column of length `0.31 m`, then the speed of sound in air to a first approximation isA. `317 m//s`B. `371 m//s`C. `340 m//s`D. `332 m//s` |
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Answer» Correct Answer - A `f_(closed) = (v)/(4 l)` `256 = (v)/( 4 (0.31))` `v = 317.44 m//s` |
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