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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. |
On blowing a pipe harder it produces a frequency double the fundamental frequency , the pipe isA. Open at both endsB. Both open and closedC. Open at one end and closed to other endD. Can not be predicted |
| Answer» Correct Answer - A | |
| 102. |
When an air column in a tube open at both the ends is set into vibration withA. Odd harmonics are presentB. Some harmonics are presentC. All harmonic are presentD. Even harmonics are present |
| Answer» Correct Answer - C | |
| 103. |
When temperature of air column bounded by the tube increases , then the frequency of the vibration of air columnA. DecreasesB. Remains sameC. IncreasesD. becomes zero |
| Answer» Correct Answer - C | |
| 104. |
An empty vessel is partially filled with water, then the frequency of vibration of air column in the vesselA. IncreaseB. DecreaseC. Remain constantD. None of these |
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Answer» Correct Answer - A As n `prop` (1/l) there force l decreases hence n increases. |
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| 105. |
The distance between a node the next antinode isA. 0.01 mB. 0.02 mC. 0.0141 mD. 0.1 m |
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Answer» Correct Answer - D `lamda/4 = (0.4)/(4) = 0.1m` |
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| 106. |
In the above problem, end correction isA. 1cmB. 0.5cmC. 4cmD. 2cm |
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Answer» Correct Answer - A `e=(l_2-3l_1)/(2) = (77-3xx25)/(2) = 2/2 = 1cm` |
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| 107. |
With the data of the above problem, the frequency with which the string vibrates isA. 50HzB. 12.5 HzC. 62.5 HzD. 2500 Hz |
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Answer» Correct Answer - B `v=nlamda therefore n=v/lamda = 50/4 = 25/2 = 12.5 Hz` |
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| 108. |
The maximum velocity of the particle in the above problem at a distance of 10 m isA. 1.5 m/sB. 4.713 m/sC. Zero m/sD. 3 m/s |
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Answer» Correct Answer - B `v_("max") = R_omega = 0.01xxomega=0.01xx150pi` `=1.5pi=1.5xx3.142=4.7130` |
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| 109. |
In a stationary waves, the pressure variation is maximum atA. AntinodesB. Midway between node and antinodeC. NodeD. Can not be predicted |
| Answer» Correct Answer - C | |
| 110. |
A pipe open at both produces a note of fundamental frequency `v_(1)` When the pipe is kept with `3/4` th of its length in water, it produces a note of fundamental frequency `v_(2)` The ratio of `v_(1)/v_(2)` isA. `4/3`B. `3/4`C. 2D. `1//2` |
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Answer» Correct Answer - d Accordingly `v_(1)=v/(2l)` ...(i) and `v_(2)=v/(4l//4) = v/l` ...(ii) Hence, `v_(1)/v_(2)=1/2` |
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| 111. |
The frequency of tuning fork is 256 Hz. It will not resonate with a fork of frequencyA. 768 HzB. 738 HzC. 512 HzD. 256 Hz |
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Answer» Correct Answer - b The fork of frequency 256 Hz will resonate only with those tuning forks which have frequency in integral multiple of 256 Hz. It means 512 Hz and 768 Hz. |
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| 112. |
The maximum number of overtones emitted by an open organ pipe of length 15cm that can be heard by a person with normal hearing (velocity of sound in air is 330m/s ) will beA. 16B. 17C. 18D. 19 |
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Answer» Correct Answer - B `n=(v)/(2l)=(330)/(2xx15xx10^(-2)) = (330xx100)/(30)` n=1100 ltbr. Maximum audiable frequency = 20000 Number of harmonics = `20000/1100=18` Number of overtones = 17. |
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| 113. |
When an open pipe is vertically dipped in water 8cm is inside the water, then the fundamental frequency of air column is 400Hz. If it is 6cm inside the water then fundamental frequency of the air column is 300Hz. Then the length of the tube will beA. 22cmB. 14cmC. 20cmD. 16cm |
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Answer» Correct Answer - B `n_1(L-8) =n_2(L-6)` 400(L-8)=300(L-6) 4L - 32 = 3L - 18 4L - 3L = 32 - 18 L=14cm. |
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| 114. |
If the length of the vibrating string is kept constant , then frequency of the string will be directly proportional toA. `sqrtT`B. TC. `1/T`D. `sqrt(1/T)` |
| Answer» Correct Answer - A | |
| 115. |
The harmonics which are present in a pipe open at one end areA. odd harmonicsB. even harmonicsC. even as well as odd harmonicsD. None of the above |
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Answer» Correct Answer - a Wavelenght of closed prgan pipe is ,`lambda = (4l)/((2n-1))` Putting `n=1,2,3...` we find that `lambda _(1) =4L, (4L)/3, (4L)/5, ...` So, frequency of vibration corresopending to modes `n=1,2,3,... `is `v_(1) = v/lambda_(1) = v/(4L)=v_(1) ` `v_(2)= v/lambda_(2)=v/(4L//3) = (3v)/(3L) = 3 v_(1)` `v_(3) = v/lambda_(3)=v/(4L//5) = (5v)/(4L)= 5 v_(1)` `therefore v_(1): v_(2) : v_(3) ....... = 1 : 3 : 5 : ...` So, only odd harmonics are present. |
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| 116. |
The fundamental frequency of an open pipe of length 34cm and how many harmonics may be heard by a person having normal hearing ?A. 100Hz, 10B. 300 Hz, 30C. 200 Hz, 20D. 500 Hz, 40 |
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Answer» Correct Answer - D `n_0 = (v)/(2l) = (340)/(2xx34xx10^(-2)) = 500Hz` Number of harmonics `=20000/500=40` |
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| 117. |
An open pipe emits harmonics in the ratio ofA. `1:2:3`B. `1:5:9:`C. `1:3:5`D. `2:4:6` |
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Answer» Correct Answer - A In a open pipe , vibration of air column vibrate with all harmonics are present. n=3600/60 =60 |
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| 118. |
Which of the following statement is not correct in stationary waves ?A. At a given instant, all particles in a segment are in same phase.B. All the particles cross the mean position simultaneously.C. The amplitude of vibration is different for different particles .D. The time period of vibration is different for different particles |
| Answer» Correct Answer - D | |
| 119. |
The harmonics which are present in a pipe open at one end areA. Odd harmonicsB. Even harmonicsC. Even as well as odd harmonicsD. None of these |
| Answer» Correct Answer - A | |
| 120. |
The true statement about stationary waves isA. Dispalcement at nodes is maximum B. Displacement at antinodes is minimumC. Displacement at nodes is zero and that at antinodes is maximumD. None of these |
| Answer» Correct Answer - C | |
| 121. |
In a stationary waves, the maximum strees is atA. Half the distance form nodeB. Displacement antinodeC. Pressure nodeD. Node |
| Answer» Correct Answer - D | |
| 122. |
Stationary waves can not be produced in aA. Thin wireB. Thick wireC. Short wireD. Infinitely long wire |
| Answer» Correct Answer - D | |
| 123. |
The velocity of the stationary waves isA. ZeroB. IncreasesC. ConstantD. Decreases |
| Answer» Correct Answer - A | |
| 124. |
All the particles in a loop of a stationary waves areA. In opposite phase B. With phase difference of `pi/2` radC. In phaseD. Can not be predicted |
| Answer» Correct Answer - C | |
| 125. |
The true statement about stationary waves isA. Dispalcement at nodes is maximumB. Displacement at antinodes is minimumC. Displacement at nodes is zero and that at antinodes is maximumD. None of these |
| Answer» Correct Answer - C | |
| 126. |
Stationary waves are called, "stationary" because in itA. Everything remains at restB. The particles of the medium are not disturbed at allC. The particles of the medium are only slightly disturbedD. There is no flow of energy along the waves |
| Answer» Correct Answer - D | |
| 127. |
Standing stationary waves can be obtained in an air column even if the interfering waves areA. Different frequenciesB. Different amplitudesC. Different velocitiesD. Different wavelength |
| Answer» Correct Answer - B | |
| 128. |
All the particles in a loop of a stationary waves areA. In opposite phaseB. With phase difference of `pi/2` radC. In phaseD. Can not be predicted |
| Answer» Correct Answer - C | |
| 129. |
In a stationary waves, the period of each oscillating particles isA. SameB. DecreasesC. IncreasesD. Can not be predicted |
| Answer» Correct Answer - A | |
| 130. |
In stationary waves , every particles performsA. S.H.M.B. S.H.M. except antinode pointC. S.H.M. except node pointD. S.H.M. of all points of medium |
| Answer» Correct Answer - C | |
| 131. |
The adjacent particles in the loop of stationary waves areA. In opposite phaseB. Phase difference is `pi/2` radC. Phase difference is `pi/3` radD. In phase |
| Answer» Correct Answer - D | |
| 132. |
Stationary waves is called standing waves becauseA. The particles of the medium remains stationaryB. Particles of the medium do not oscillates at allC. The wave remains localised to a part of the mediumD. Medium through which it travels is stationary |
| Answer» Correct Answer - C | |
| 133. |
In a stationary waves, the pressure variation is minimum atA. AntinodesB. Midway between node and antinodeC. NodeD. Can not be predicted |
| Answer» Correct Answer - A | |
| 134. |
The antinode isA. A position of particle which oscillates with maximum amplitude and maximum average speedB. The mean position of a particle which oscillated with highest amplitude and highest average speedC. A particle which does not oscillates at allD. A particle which oscillates with maximum amplitude and lowest average speed |
| Answer» Correct Answer - A | |
| 135. |
The pressure antinode in a stationary waves is aA. Displacement nodeB. Pressure nodeC. Displacement antinodeD. Node |
| Answer» Correct Answer - A | |
| 136. |
A string is taken and stretched so that it elongates by 2% . Then its fundamental frequencyA. Increase by 2%B. Increase by 1%C. Decrease by 2%D. Decrease by 1% |
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Answer» Correct Answer - C `n_2/n_1 = l_1/l_2` `therefore n_2 = 100/102 n_1 = 98% n_1` = decreases by 2% |
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| 137. |
The linear density of a stretched string is increased by 1%. Then the change in fundamental frequency of string isA. Increase by 0.5%B. Increase by 1%C. Decrease by 0.5%D. Decrease by 1% |
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Answer» Correct Answer - C `(Deltan)/(n) = -1/2 (Deltam)/(m) = -1/2 %` = decrease by 0.5% |
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| 138. |
The radius of stretched sring is decreased by 1% . Then the change in fundamental frequencyA. Decrease by 1%B. Increase by 2%C. Increase by 1%D. Decrease by 1% |
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Answer» Correct Answer - C `(Deltam)/(n) = (Deltar)/( R) = ` + 1% = increases by 1% |
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| 139. |
A sonometer wire is in unison with a tuning fork. When the length of the wire is increased by 2% , the number of beats heard per second is 5, then the frequency of the fork will beA. 245HzB. 250HzC. 255HzD. 260Hz |
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Answer» Correct Answer - C `l_2=1.02l_1 " and " n_1lamda_1 = n_2lamda_2` `n_1/n_2 = 1.02 therefore n_1-n_2=5` `1.02n_2-n_2= 5` `0.02n_2=5` `therefore n_2 = 250Hz` `n_1=n_2+5 = 250+5=255Hz` |
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| 140. |
The simplest mode of a vibration of the string isA. First harmonicB. Fundamental mode a vibrationsC. First overtoneD. Both a and b |
| Answer» Correct Answer - D | |
| 141. |
The speed of a transverse wave (v) along a stretched string is given byA. `sqrt(T/m)`B. `sqrt(m/T)`C. `sqrt(P/E)`D. `sqrt(1.m)` |
| Answer» Correct Answer - A | |
| 142. |
If a wave is propagated along the stretched string in the formA. Longitudinal waveB. Electromagnetic waveC. Transverse waveD. All of the above |
| Answer» Correct Answer - C | |
| 143. |
The equation of a stationary wave along a stretched string is given by `y = 4 sin frac{2pi"x"}{3} cos 4Opit` where, x and y are in cms and t is in sec. The separation between two adjacent nodes isA. 3 cmB. `1.5` cmC. 6 cmD. 4 cm |
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Answer» Correct Answer - b Compare the giben equation with the standerd form of stationary wave equation `y=2r sin frac{2pix}{lambda} cos frac{2rpivt}{lambda}` we get, ` frac{2pix}{lambda}= frac{2rpivt}{3}` `rArr lambda = 3` cm `therefore` Separation between two adjacent nodes `=lambda/2=1.5` cm |
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| 144. |
The equation of a standing wave is `y=0.2` `sinfrac{2pi}{0.3}xcdot cos frac{2pit}{0.01}` where x and y are in metres and t is in seconds. The velocity of propagation of the wave isA. 30 `ms^(-1)`B. 40 `ms^(-1)`C. 300 `ms^(-1)`D. 400 `ms^(-1)` |
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Answer» Correct Answer - a The given wave equation is `Y=0.2 sin frac(2pi)(0.3)xcos frac(2pit)(0.01)` ...(i) Comparing Eq. (i) with standard wave equation `y=a sin kx cos omega t,` we get `omega = (2pi)/(0.01) rads^(-1) rArr k = (2k)/(0.3)rad//m` `rArr` Wave velocity `omega/k=(2pi//0.01)/(2pi//0.3)=(0.3)/(0.01)=30 ms^(-1)` |
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| 145. |
If the frequency of first harmonic of a closed pipe is in unison with the third harmonic of an open pipe. Then, the ratio of lengths of the pipe closed at one end to the open at both the ends isA. `1//12`B. `3//4`C. `1//6`D. `6//7` |
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Answer» Correct Answer - C `n_c = 3n_0` `(v)/(4l_c) = 3xx(v)/(2l_0)` `therefore l_c/l_0 = 1/6` |
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| 146. |
Second overtone of a closed pipe of length , 1m is in unison with third overtone of an open pipe , then the length of the open pipe will beA. 1.6 mB. 0.625 mC. 0.8 mD. 3.2 m |
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Answer» Correct Answer - A `3n_c = 4n_0` `5(v)/(4l_c) = 4xx(v)/(2l_0)` `l_0 = 5/8 = 0.625m` |
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| 147. |
The first obertone of an open organ pipe beats with the first overtone of a closed organ pipe with a beat frequency of 2.2 Hz. The fudamental frequency of clos3ed organ pipe is 110 Hz. Find length of the open pipe. (Given, sound in air =330 m/s )A. 1.0067m onlyB. 0.9937m onlyC. 1.0067 or 0.9937mD. 0.75m |
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Answer» Correct Answer - C Let `l_c " and " l_0` represent the length of closed and open end organ pipes `n_c=110 therefore n_1=3n_c=3xx110=330` First overtone of closed pipe `n_10 - n_(1c) = pm2.2` `n_10-3n_c = pm 2.2` gt `n_10-330 = pm 2.2` `therefore n_10 = 322.2 " or " 327.8` `l_(01) = (330)/(330pm2.2) = (330)/(332.2) " or " (330)/(327.8)` ` l_(01) = 0,9937 " or " 1.0067` |
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| 148. |
The length of a sonometer wire between two fixed ends is 110cm. Where should the two bridges the placed so as to divide the wire into three segments, whose fundamental frequencies are in the ration` 1:2:3?`A. 60cm, 30cm and 20cmB. 20cm, 30cm and 60cmC. 30cm, 20cm and 60cmD. 60cm, 20cm and 30cm |
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Answer» Correct Answer - A `l_1+l_2+l_3=110` `n_1/n_2=l_2/l_1 therefore l_2/l_1 = 1/2 therefore l_2=l_1/2` and `n_1/n_3 = l_3/l_1 therefore l_3=(l_1)/(3)` `l_1+(l_1)/(2) +(l_1)/(3) = 110` `6l_1+3l_1+2l_1=110xx6` `11l_1 = 110xx6` `l_1=60cm` Similarly `l_2=30 " and " l_3 =20cm` |
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| 149. |
The equation of a stationary wave is `y=-4sin((pix)/(5)) cos(100pit)`. Amplitude the progressive wave which produced the stationary wave isA. 1mB. 4mC. 2mD. 8m |
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Answer» Correct Answer - C `y=-4sin((pix)/(5)) cos 100pit` Comparing it with stand equation of stationary wave, `y=A sin kx.cosomegat` A=4m |
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| 150. |
For the stationary wave `y=4sin((pix)/(15))cos(96pit)`, the distance between a node and the next antinode isA. `7.5`B. 15C. `22.5`D. 30 |
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Answer» Correct Answer - a On comparing giben equation with standard ewuation, we get `(2pi)/lambda = pi/15 rArr lambda = 30 ` Distance between the node and antinode `lambda/4=30/4=7.5` |
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