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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. |
The minimum intensity of sound is zero at a point due to two sources of nearly equal frequencie4s whenA. two sources are vibration in the opposite phaseB. the amplitude of two sources are equalC. at the point of observation, the amplitudes of two S.H.M. produced by two sources are equal and both the S.H.M. are along the same straight lineD. both the sources are in the same phase |
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Answer» Correct Answer - C |
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| 102. |
A standing wave on a string is given by `y = ( 4 cm) cos [ x pi] sin [ 50 pi t]`, where `x` is in metres and t is in seconds. The velocity of the string section at `x = 1//3 m at t = 1//5 s`, isA. zeroB. ` pi m//s`C. `840 pi m//s`D. noneof these |
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Answer» Correct Answer - B velocity of the string section can be given as `v = ( delta y)/( delta t) = 4 cos pi x xx (50 pi) cos ( 50 pi t)` ` v = 4 cos [ pi xx (1)/(3)] xx 50 pi cos [ 50 pi xx (1)/( 5)]` `= 200 pi xx (1)/(2) xx 1 = 100 pi cm//s = pi m//s` |
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| 103. |
A wave is given by the equation ` y = 10 sin 2 pi (100 t - 0.02 x) + 10 sin 2 pi (100 t + 0.02 x)` Find the loop length , frequency , velocity and maximum amplitude of the stationary wave produced. |
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Answer» Correct Answer - maximum amplitude `20 units ` wave velocity is `5000 units`. Frequency `100 units` loop length is `25 units` We know the equation of a stationary wave is given by `y = A sin [ ( 2 pi)/( lambda) ( vt - x)] + A sin [ ( 2 pi)/(lambda) ( vt + x)]` `= 2 A cos ( 2pi x)/( lambda) sin ( 2pi v t)/( lambda)` ` = R sin ( 2pi)/( lambda) vt` Here , `R = 2 A cos (2 pi x// lambda)` is the amplitude of medium particle situated at a distance `x`. The given equation can be expressed as `y = 10 sin [( 2 pi)/( 50) (5000 t + x)]` ` = 2 xx 10 cos (( 2pi x)/( 50)) sin (( 2pi)/(50) 5000 t)` Comparing it with standard equation of stationary wave , we get wavelength `lambda = 50 units` Wave velocity ` v = 5000 units` Thus amplitude `R = 2 xx 10 cos ( 2 pi x)/( 50)` and maximum amplitude `R_(max) = 2 xx 10 = 20 units` Frequency `= (v)/( lambda) = ( 5000)/(50) = 100 units` And loop length is `(lambda)/(2) = (50)/(2) = 25 units` |
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| 104. |
For a certain organ pipe , three successive resonance observed are `425 , 595 and 765 Hz`. Taking the speed of sound to be `340 ms^(-1)` , find the length of the pipe , in meter . |
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Answer» Correct Answer - `1` Since frequencies are in odd number ratio , the pipe has to be a closed pipe. Ratio of `3` frequencies `= 425 : 595 : 765` `= 5 : 7 : 9` So fundamental frequency ` = f = (425)/(5) = 85 Hz` For fundamental frequency `l = (v)/( 4f) = ( 340)/(4 xx85) = 1m` |
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| 105. |
A string of length L, fixed at its both ends is vibrating in its `1^(st)` overtone mode. Consider two elements of the string of the same small length at positions `l_(1)=0.2L"and"l_(2)=0.45L` from one end. If `K_(1)"and" K_(2)` are their respective maximum kinetic energies,thenA. `K_(1)=K_(2)`B. `K_(1)gtK_(2)`C. `K_(1)ltK_(2)`D. it is not possible to decide the relation |
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Answer» Correct Answer - B |
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| 106. |
A longitudinal standing wave ` y = a cos kx cos omega t` is maintained in a homogeneious medium of density `rho`. Here `omega` is the angular speed and `k` , the wave number and `a` is the amplitude of the standing wave . This standing wave exists all over a given region of space. The space density of the potential energy `PE = E_(p)(x , t)` at a point `(x , t)` in this space isA. `E_(p) = ( rho a^(2) omega^(2))/(2)`B. `E_(p) = ( rho a^(2) omega^(2))/(2) cos^(2) kx sin^(2) omega t`C. `E_(p) = ( rho a^(2) omega^(2))/(2) sin^(2) kx cos^(2) omega t`D. `E_(p) = ( rho a^(2) omega^(2))/(2) sin^(2) kx sin^(2) omega t` |
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Answer» Correct Answer - C The given longitudinal standing wave is `y = a cos kx cos omega t` `(i)` The nodes of this wave are located where `cos kx = 0 ` (i.e., ) at the values `x = (lambda)/( 4) , ( 3 lambda)/(4) ,…` and the antinodes are located where `cos kx = +- (i.e.,)` at the values `x = 0 , (lambda)/(2) ,...` At the nodes , the space density of kinetic energy ( kinetic energy per unit vanishes for the nodes , i.e., `x = (lambda)/(4) , ( 3 lambda)/(4) `etc. Also , `y` is maximum at `t = 0` , as we see from Eq.(i). Hence potential energy must be maximum at `t = 0` . Hence the time factor in potential energy density must enter as ` cos^(2) omega t`. Also , the sum of kinetic and potential energy densities must always be constant for a given `x` as it represents total energy at that point. Hence the potential energy density is `E_(p) = ( rho a^(2) omega^(2))/(2) sin^(2) kx cos^(2) omega t` `(ii)` and the kinetic energy density is `E_(k) = ( rho a^(2) omega^(2))/(2) cos^(2) kx sin^(2) omega t` `(iii)` |
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| 107. |
A longitudinal standing wave ` y = a cos kx cos omega t` is maintained in a homogeneious medium of density `rho`. Here `omega` is the angular speed and `k` , the wave number and `a` is the amplitude of the standing wave . This standing wave exists all over a given region of space. The space density of the kinetic energy . `KE = E_(k) ( x, t)` at the point `(x, t)` is given byA. `E_(k) = ( rho a^(2) omega^(2))/(2) cos^(2) kx cos^(2) omega t`B. `E_(k) = ( rho a^(2) omega^(2))/(2) sin^(2) kx cos^(2) omega t`C. `E_(k) = (rho a^(2) omega^(2))/( 2)`D. `E_(k) = ( rho a^(2) omega^(2))/(2) cos^(2) kx sin^(2) omega t` |
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Answer» Correct Answer - D The given longitudinal standing wave is `y = a cos kx cos omega t` `(i)` The nodes of this wave are located where `cos kx = 0 ` (i.e., ) at the values `x = (lambda)/( 4) , ( 3 lambda)/(4) ,…` and the antinodes are located where `cos kx = +- (i.e.,)` at the values `x = 0 , (lambda)/(2) ,...` At the nodes , the space density of kinetic energy ( kinetic energy per unit vanishes for the nodes , i.e., `x = (lambda)/(4) , ( 3 lambda)/(4) `etc. Also , `y` is maximum at `t = 0` , as we see from Eq.(i). Hence potential energy must be maximum at `t = 0` . Hence the time factor in potential energy density must enter as ` cos^(2) omega t`. Also , the sum of kinetic and potential energy densities must always be constant for a given `x` as it represents total energy at that point. Hence the potential energy density is `E_(p) = ( rho a^(2) omega^(2))/(2) sin^(2) kx cos^(2) omega t` `(ii)` and the kinetic energy density is `E_(k) = ( rho a^(2) omega^(2))/(2) cos^(2) kx sin^(2) omega t` `(iii)` |
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| 108. |
A longitudinal standing wave ` y = a cos kx cos omega t` is maintained in a homogeneious medium of density `rho`. Here `omega` is the angular speed and `k` , the wave number and `a` is the amplitude of the standing wave . This standing wave exists all over a given region of space. If a graph `E ( = E_(p) + E_(k))` versus `t` , i.e., total space energy density verus time were drawn at the instants of time `t = 0` and `t = T//4`, between two successive nodes separated by distance `lambda//2` which of the following graphs correctly shows the total energy `(E)` distribution at the two instants.A. B. C. D. |
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Answer» Correct Answer - A The graphs shown are between two successive nodes , say `x_(1) = lambda//4 at N_(1)` and `x_(2) = 3 lambda//4 at N_(2)`. Total energy density is `E = E_(p) + E_(k))` `= ( rho a^(2) omega^(2))/(2) [ sin^(2) kx cos^(2) omega t + cos^(2) kx sin^(2) omega t]` Putting ` x = lambda//2` , ( first antinode) `(E)_(lambda//2) = ( rho a^(2) omega^(2))/(2) [ sin^(2) omega t]` At ` t = 0 , (E)_(lambda//2) = 0` and at `t = (T)/(4)` `(E)_(lambda//2) = ( rho a^(2) omega^(2))/(2)` This is truly reflected in the graph `(a)`. |
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| 109. |
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: If `y = 10 sin 5 x cos 2 t m` represents a stationary wave then , the possible one of the travelling waves causing this isA. `y = 10 sin ( 5x - 2 t)`B. `y = 5 sin ( 2 t - 5x)`C. `y = 10 sin 2t`D. `y = 5 cos 5x` |
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Answer» Correct Answer - B A stationary wave is shown as `y = 2 A sin kx cos omega t` in general . So the amplitude of the original wave is `A`. |
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| 110. |
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: Speed of a wave in a string forming a stationary wave does not depend onA. TensionB. Mass of wire for a given lengthC. Length of the wire for a given massD. Harmonics of string |
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Answer» Correct Answer - D Speed , `V = sqrt((T)/((m//l)))` , So it is dependent on tension , mass of the string and its length. |
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| 111. |
The frequency of `B` is `3%` greater than that of `A` . The frequency of `C is 2%` less than that of `A` . If `B and C` produce `8 beats//s`, then the frequency of `A` isA. `136 Hz`B. `168 Hz`C. `164 Hz`D. `160 Hz` |
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Answer» Correct Answer - D `v_(B) = v_(A) + (3)/(100) v_(A)` ` v_(c ) = v_(A) - (2)/(100) v_(A)` `v_(B) - v_(c ) = 8` `(3)/(100) v_(A) + (2)/(100) v_(A) = 8` or `v_(A) xx (5)/(100) = 8 or v_(A) = 160 Hz` |
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| 112. |
An open and a closed pipe have same length . The ratio of frequency of their `n th` overtone isA. `( n + 1)/( 2n + 1)`B. ` 2 ( n + 1)/( 2n + 1)`C. `( n)/( 2n + 1)`D. `( n + 1)/( 2n)` |
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Answer» Correct Answer - B Let `l` be the length of the pipe and `v` the speed of sound . Then frequency of open organ pipe of `n th` overtone is `f_(1) = ( n + 1) (v)/( 2 l)` and frequency of closed organ pipe of `n th` overtone `f_(2) = ( 2 n + 1) (v)/( 4 l)` Therefore , the desired ratio is `(f_(1))/(f_(2)) = ( 2 (n + 1))/(( 2 n + 1))` |
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| 113. |
`n th` harmonic of a closed organ is equal to `m th` harmonic of an pipe . First overtone frequency of the closed organ pipe is also equal to first overtone frequaency of an organ pipe . Find the value of `n`, if `m = 6`. |
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Answer» Correct Answer - `9` `n((V)/( 4 L_(c ))) = m ((V)/( 2L_(0)))` `(i)` Also `3((V)/( 4L_( c))) = 2 ((V)/( 2 L_(0)))` `(ii)` FromEq. (ii) `(L_(c))/(L_(0)) = (3)/(4)` From Eq.(i) ` (n)/(m) = 2((L_(c ))/(L_(0))) = (6)/(4) = (3)/(2) = (9)/(6)` `n = 9` if `m = 6` |
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| 114. |
Which of the following statements is correct for stationary wavesA. Nodes and antinodes are formed in case of stationary transverse wave onlyB. In case of longitudinal stationary wave , compressions and rarefactions are obtained in place of nodes and antinodes respectivelyC. Suppose two plane wave , one longitudinal and the other transverse having same frequency and amplitude are travelling in a medium in opposite directions with the same period , by superposition of these waves , stationary waves cannot be obtainedD. None of the above |
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Answer» Correct Answer - C Node means a point at which medium particles do not displace from its mean position and antinode mean a point at which particles oscillate with maximum possible amplitude . Nodes and antinodes are obtained for both types of stationary waves , transverse and longitudinal . Hence , options (a) and (b) both are wrong . To obtain a stationary wave , two waves travelling in opposite directions , having same amplitude , same frequency are required . They must have same nature , means either both of the waves should be longitudinal or both of them should be transverse . Hence , option ( c) is correct . |
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| 115. |
A sound wave starting from source `S`, follows two paths `AOB and ACB` to reach the detector `D`. If `ABC` is an equilateral traingle , of side `l` and there is silence at point `D`, the maximum wavelength `(lambda)` of sound wave must be A. `l`B. ` 2 l`C. ` 3 l`D. `4 l` |
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Answer» Correct Answer - B Path difference ` = ( 2l - l) = lambda//2` (for minimum) ` lambda = 2 l` |
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| 116. |
A sound wave of wavelength ` lambda` travels towards the right horizontally with a velocity `V`. It strikes and reflects from a vertical plane surface , travelling at a speed `v` towards the left . The number of positive crests striking in a time interval of `3 s` on the wall isA. `3(V + v)//lambda`B. `3(V - v)//lambda`C. `(V + v)// 3 lambda`D. `(V - v)// 3 lambda` |
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Answer» Correct Answer - A The relative velocity of sound waves w.r.t. the wall is `V + x`. Hence , the apparent frequency of the waves striking the surface of the wall is `(V + v)//lambda`. The number of positive crests striking per second is same frequency . In ` 3 s` , the number is ` [ 3 (V + v)]//lambda`. |
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| 117. |
A sound wave starting from source `S`, follows two paths `SEFD and SEABFD`. If `AB = 1 , AE = BF = 0.6 l ` and wavelength of wave is `lambda = 11 m`. If maximum sound is heard at `D` , then maximum value of length `l` is A. `11 m`B. `6 m`C. `2.5 m`D. `5 m` |
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Answer» Correct Answer - D For maximum path difference `Delta = n lambda` `2 xx 0.6 l = lambda` `l = (lambda)/( 1.2) = (6)/(1.2) = 5 m` |
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| 118. |
Two uniform strings `A and B` made of steel are made to vibrate under the same tension. If the first overtone of `A` is equal to the second overtone of `B` and if the radius of `A` is twice that of `B`, the ratio of the lengths of the strings is |
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Answer» Correct Answer - `1:3` |
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| 119. |
Two uniform strings `A and B` made of steel are made to vibrate under the same tension. If the first overtone of `A` is equal to the second overtone of `B` and if the radius of `A` is twice that of `B`, the ratio of the lengths of the strings isA. `2 : 1`B. `3 : 2`C. `3 : 4`D. `1 : 3` |
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Answer» Correct Answer - D According to equation `2 n_(1) = 3n_(2)` or `(2)/( 2l_(1)) sqrt((T)/(m_(1))) = (3)/( 2 l_(2)) sqrt((T)/(m_(2)))` or `(l_(1))/(l_(2)) = (2)/(3) sqrt((m_(2))/(m_(1))) = (2)/(3) sqrt(( a_(2) rho)/( a_(1) rho))` or `(l_(1))/(l_(2)) = (2)/(3) sqrt((r_(r)^(2))/(r_(1)^(2))) = (2)/(3) sqrt(((1)/(2))^(2)` or `(l_(1))/(l_(2)) = (1)/(3)` |
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| 120. |
Two closed - end pipes , when sounded together produce ` 5 beats//s`. If their lengths are in the ratio `100 : 101` , then fundamental notes ( in Hz) produced by them areA. `245 , 250`B. `250 , 255`C. `495 ,500`D. `500 , 505`. |
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Answer» Correct Answer - D `(f_(1))/(f_(2)) = (101)/(100)` `f_(1) - f_(2) = 5` `(101)/(100) f_(2) - f_(2) = 5 or f_(2) = 500 Hz` and `f_(1) = f_(2) + 5 = 505 Hz` |
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| 121. |
Two tunig forks of frequency `250Hz` and `256Hz` produce beats. If a maximum is observed just now, after how much time the next maximum is observed at the same place?A. `(1)/(18) s`B. `(1)/(6) s`C. `(1)/(12) s`D. `(1)/(24) s` |
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Answer» Correct Answer - C Beat frequency , `Delta f = 6 Hz` Time interval between two consecutive maxima is `1//6 s`. So , the required time `1//2 s`. |
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| 122. |
A chord attached about an end to a vibrating fork divides it into 6 loops when its tension is 36 N. The tension at which it will vibrate 4 loops is:A. ` 24 N`B. `36 N`C. `64 N`D. `81 N` |
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Answer» Correct Answer - D `f = (p)/( 2 L) sqrt((T)/(mu))` `rArr p_(1) sqrt(T_(1)) = p_(2) sqrt(T_(2))` `rArr 6 sqrt(36) = 4 sqrt(T_(2)) rArr T_(2) = 81 N` |
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| 123. |
A sonometer wire resonates with a given tuning fork forming ` 5` antinodes when a mass of `9 kg` is suspended from the wire . When this resonates with the same tuning fork forming three antinodes for the same positions of the bridges . The value of `M` isA. `25 kg`B. `5 kg`C. `12.5 kg`D. `(1//25) kg` |
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Answer» Correct Answer - A `f_(0) = (5)/( 2l) sqrt((9 g)/(mu)) = (3)/(2 l) sqrt((M g)/(mu))` `:. M = 25 kg` |
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| 124. |
A sonometer wire supports a 4 kg load and vibrates in fundamental mode with a tunig fork of frequency 416 Hz. The length of the wire between the bridges is now doubled. In order to maintain fundamental mode, the load should be changed toA. `1 kg`B. `2 kg`C. `8 kg`D. `16 kg` |
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Answer» Correct Answer - D For the given problem , `( sqrt(T))/(l) = constant ` or `T prop l^(2)` If `l` is to be doubled , `T` would be quadrupled. |
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| 125. |
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 `330 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 = (330)/(500) = 0.66 m` The resonance occurs at `(lambda)/(4) , ( 3lambda)/(4) , ( 5 lambda)/(4) , ( 7 lambda)/(4) ,….` i.e ., at `0.165 m , 0.495 m , 0.825 m , 1.115 m`. As the length of the tube is only `1.0 m, hence `3` resonances will be observed. |
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| 126. |
A tuning fork vibrating with a sonometer having 20 cm wire produces 5 beats per second. The beat frequency does not change if the length of the wire is changed to 21 cm. The frequency of the tuning fork (in Hertz) must beA. 200B. 210C. 205D. 215 |
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Answer» Correct Answer - C |
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| 127. |
Two identical sonometer wires have a fundamental frequency of `500 Hz` when kept under the same tension . The percentage change in tension of one of the wires that would cause an occurrence of `5 beats//s` , when both wires vibrate together isA. `0.5 %`B. `1 %`C. `2 %`D. `4 %` |
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Answer» Correct Answer - C `f = (1)/( 2l) sqrt((T)/(m))` `(Delta f)/( f) = (1)/(2) (Delta T)/(T)` `(Delta T)/(T) = 2 (( Delta f))` `( Delta T)/(T) xx 100 = ` percentage change in tension `( Delta T)/( T) = (1)/( 50)` `(Delta T)/( T) = (1)/(50)` `(Delta T)/(T) xx 100 = 2 %` |
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| 128. |
A long tube open at the top is fixed verticallly and water level inside the tube can be moved up or down . A vibrating tuning fork is held above the open end and the water level is pushed down gradually so as to get first and second resonance at `24.1 cm` and `74.1 cm` , respectively below the open end . The diameter of the tube isA. `5 cm`B. `4 cm`C. `3 cm`D. `2 cm` |
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Answer» Correct Answer - C `f_(1) = (v)/( 4 ( 24.1 + 0.3 D))` `f_(3) = ( 3 v)/( 4 (74.1 + 0.3 D))` `f_(1) = f_(3)` `rArr = (v)/( 4 ( 24.1 + 0.3 D)) = 3 (v)/( 4 (74.1 + 0.3 D))` `3 ( 24.1 + 0.3 D) = 74.1 + 0.3 D` `72.3 + 0.6 D = 74.1 + 0.3 D` `0.6 D = 74.1 - 72.3` `0.6 D = 1.8` `D = (1.8)/(0.6) = 3 cm` |
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| 129. |
An open organ pipe of length `l` is sounded together with another open organ pipe of length ` l + x` in their fundamental tones. Speed of sound in air is `v` . The beat frequency heard will be `(x lt lt l)`:A. `( vx)/( 4 l^(2))`B. `(vl^(2))/( 2x)`C. `( v x)/( 2 l^(2))`D. `( v x^(2))/( 2l)` |
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Answer» Correct Answer - C Beat frequency `= f_(1) - f_(2) = (v)/(2 l) - (v)/( 2( l + x))` ` = (v)/( 2 l) [ 1 - ( 1 + (x)/(l))^(-1)]` ` = (v)/( 2l) [ 1 - 1 + (x)/(l)]` `= ( vx)/( 2l^(2))` |
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| 130. |
In an organ pipe ( may be closed or open of `99 cm` length standing wave is setup , whose equation is given by longitudinal displacement `xi = (0.1 mm) cos ( 2pi)/( 0.8) ( y + 1 cm) cos 2 pi (400) t` where `y` is measured from the top of the tube in metres and `t` in second. Here `1 cm` is th end correction. Equation of the standing wave in terms of excess pressure is - (Bulk modulus of air `B = 5 xx 10^(5) N//m^(2)`)A. `P_(ex) = (125 pN//m^(2)) sin ( 2pi)/(0.8) ( y + 1 cm) cos 2 pi (400 t)`B. `P_(ex) = (125 pN//m^(2)) cos ( 2pi)/(0.8) ( y + 1 cm) sin 2 pi (400 t)`C. `P_(ex) = (225 pN//m^(2)) sin ( 2pi)/(0.8) ( y + 1 cm) cos 2 pi (200 t)`D. `P_(ex) = (225 pN//m^(2)) cos ( 2pi)/(0.8) ( y + 1 cm) sin 2 pi (200 t)` |
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Answer» Correct Answer - A `P_(ex) = -B( d xi)/( d x)` `= ( 5 xx 10^(5)) xx (0.1 xx 10^(-3)) ( 2pi)/(0.8) sin ( 2pi)/( 0.8)` `(y + 1 cm) cos 2 p(400) t` `= (125 pi N//m^(2)) sin ( 2pi)/(0.8) ( y + 1 cm) cos 2 pi (400 t)` |
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| 131. |
In an organ pipe ( may be closed or open of `99 cm` length standing wave is setup , whose equation is given by longitudinal displacement `xi = (0.1 mm) cos ( 2pi)/( 0.8) ( y + 1 cm) cos 2 pi (400) t` where `y` is measured from the top of the tube in `metres and t "in second"`. Here `1 cm` is th end correction. The upper end and the lower end of the tube are respectively :A. open - closedB. closed - openC. open - openD. closed - closed |
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Answer» Correct Answer - A `(0.1 mm) cos ( 2pi)/(0.8) ( y + 1 cm ) cos 2 pi 7 (400) t` end correction is `1 cm`. So at `y = - 1 cm`. `(0.1 mm) cos ( 2 pi)/(0.8) ( - 1 cm + 1 cm) = ( 0.1 mm) cos (0) =` Antinode So upper end is open. At lower end `y = 99 cm = 0.99 m` `= (0.1 mm) cos ( 2 pi)/( 0.8) (0.99 + 0.01)` ` = 0.01 cos ( 5 pi)/(2) = 0 rArr` Node So tube is open closed. |
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| 132. |
Two waves travelling in opposite directions produce a standing wave . The individual wave functions are given by `y_(1) = 4 sin ( 3x - 2 t)` and `y_(2) = 4 sin ( 3x + 2 t) cm` , where `x` and `y` are in cmA. The maximum displacement of the motion at `x = 2.3 cm is 4.63 cm`.B. The maximum displacement of the motion at `t = 2.3 s is 4.63 cm`.C. Nodes are formed at `x` values given by ` 0 , pi//3 , 2 pi//3 , 4 pi//3 , ….`D. Antinodes are formed at `x` values given by `pi//6 , pi//2 , 5 pi//6 , 7 pi//6,…` |
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Answer» Correct Answer - A::C::D `y = y_(1) + y_(2)` y = 4[ sin ( 3x - 2t) + sin( 3x + 2t)]` `y = 4[ 2 sin ( 3x) cos(2 t)]` `y = 8 sin ( 3 x) cos (2t)` `y = R cos ( 2t)` `R = Resultant Amplitude ` = 8 sin ( 3x)` `R = 8 sin [ 3 (2.3)]` `R = 8 sin (6.9)` `R = 4.63 cm` Nodes are formed at points of zero intensity , i.e., `I_(R) = R^(2) = 0` `sin^(2) ( 3x) = 0` `sin ( 3 x) = 0` `3x = 0 , pi , 2 pi , 3 pi , 4 pi ,...` `x = 0 , (pi)/(3) , ( 2pi)/(3) , pi , (4 pi)/(3) ,...` Antinodes are formed in between. |
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| 133. |
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. The upper end and the lower end of the tube are respectively .A. open - closedB. closed - openC. open - openD. closed - closed |
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Answer» Correct Answer - A `xi = (0.1 mm) cos ( 2pi)/( 0.8) ( y + 1 cm) cos (400 t)` End correction is `1 cm` , so at `y = - 1 cm` `xi = (0.1 mm) cos ( 2 pi)/(0.8) ( - 1 cm + 1 cm)` `= (0.1 mm) cos (0) = "Antinode"` So upper end is open . At lower end `y = 99 cm = 0.99 m` `xi = (0.1 mm) cos ( 2pi)/( 0.8) (0.99 + 0.01)` `= 0.01 cos ( 5 pi)/(2) = 0` `rArr` Node So tube is open closed. |
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| 134. |
In an organ pipe ( may be closed or open of `99 cm` length standing wave is setup , whose equation is given by longitudinal displacement `xi = (0.1 mm) cos ( 2pi)/( 0.8) ( y + 1 cm) cos 2 pi (400) t` where `y` is measured from the top of the tube in metres and `t` in second. Here `1 cm` is th end correction. Assume end correction approximately equals to `(0.3) xx`(diameter of tube) , estimate the moles of air pressure inside the tube (Assume tube is at `NTP` , and at `NTP , 22.4 litre` contain `1 mole`)A. `(10 pi)/(36 xx 22.4)`B. `(10 pi)/(18 xx 22.4)`C. `(10 pi)/(72 xx 22.4)`D. `(10 pi)/(60 xx22.4)` |
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Answer» Correct Answer - A End correction ` = (0.3)d = 1cm rArr d = (10)/(3) cm` Vol. of tube `= (pi (d^(2))/(4))l = (pi)/(4) ((10)/(3))^(2) xx 100 cm^(3)` (take `1 = 0.99 m ~~ 1m) = (10 pi)/(36) L` Moles `= ( 10 pi)/( 36 xx 22.4)` moles ( `22.4` 1t contains `1` mole `( 10 pi)/(36)` lt contains `( 10 pi)/(36 xx 22.4)` mole) |
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| 135. |
Let the two waves `y_(1) = A sin ( kx - omega t)` and `y_(2) = A sin ( kx + omega t)` from a standing wave on a string . Now if an additional phase difference of `phi` is created between two waves , thenA. the standing wave will have a different frequencyB. the standing wave will have a different amplitude for a given pointC. the spacing between two consecutives nodes will changeD. None of the above |
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Answer» Correct Answer - B Initially the standing wave equation is ` y = 2 A sin kx cos omega t` If phase difference `phi` is added to one of waves , then resulting standing wave equation is ` y = 2 A ( kx + (phi)/(2)) cos ( omega t - (phi)/( 2))` Here, frequency does not change and also spacing between two successive nodes not change as its value for both is `pi//k`. But for a particle , in standing wave , amplitude changes. |
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| 136. |
Microwaves from a transmitter are directed normally toward a plane reflector. A detector moves along the normal to the reflector. Between positions of 14 successive maxima the detector travels a distance 0.14 m. The frequency of the transmitter is `(c = 3 xx 10^(8) m s^(-1))`.A. ` 1.5 xx 10^(10) Hz`B. `10^(10) Hz`C. `3 xx 10^(10) Hz`D. `6 xx 10^(10) Hz` |
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Answer» Correct Answer - A If detector moves ` x` distance , distance from direct sound increases by `x` and distance from reflected sound decreases by `x` so path difference created ` = 2 x` `2 (0.14) = 14 lambda = 14 c//f` `f = ( 14 xx 3 xx 10^(8))/( 0.14 xx 2) = 1.5 xx 10^(10) Hz` |
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| 137. |
`S_1 and S_2` are two coherent sources of radiations separated by distance `100.25 lambda`, where `lambda` is the wave length of radiation. `S_1` leads `S_2` in phase by `pi//2`.A and B are two points on the line joining `S_1 and S_2` as shown in figure.The ratio of amplitudes of component waves from source `S_1 and S_2` at A and B are in ratio 1:2. The ratio of intensity at A to that of B `(I_A/I_B)` is A. `oo`B. `(1)/(9)`C. `0`D. `9` |
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Answer» Correct Answer - B For interference at `A : S_(2)` is behind of `S_(1)` by a distance of `100 lambda = lambda//2`. Hence the waves from `S_(1)` and `S_(2)` interfere at `A` with phase difference of `200.5 pi + 0.5 pi = 201 pi = pi`. Hence the net amplitude at `A is 2a - a = a` . For interference at `B : S_(2)` is ahead of `S_(1)` by a distance of `100 lambda + lambda//4` (equal to phase difference `pi//2`) .Further `S_(2)` lags `S_(1)` by `pi//2`. Hence the waves from `S_(1)` and `S_(2)` interfere at `B` with a phase difference of `200.5 pi - 0.5 pi = 200 pi = 0 pi` . Hence the net amplitude at `A is 2 a + a = 3 a`. Hence , `((I_(A))/(I_(B))) = ((a)/( 3 a))^(2) = (1)/(9)` |
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| 138. |
Tuning fork `F_1` has a frequency of 256 Hz and it is observed to produce `6 beats//second` with another tuning fork `F_2`. When `F_2` is loaded with wax, it still produces `6 beats//sec` with `F_1`. The frequency of `F_2` before loading wasA. 253 HzB. 262 HzC. 250 HzD. 259 Hz |
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Answer» Correct Answer - B |
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| 139. |
A string of 7m length has a mass of `0.035 kg.` If tension in the string is `60.N,` then speed of a wave on the string isA. `77m//s`B. `102 m//s`C. `110 m//s`D. `165 m//s` |
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Answer» Correct Answer - C |
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| 140. |
A strain of sound waves is propagated along an organ pipe and gets reflected from an open end . If the displacement amplitude of the waves (incident and reflected) are `0.002 cm` , the frequency is `1000 Hz` and wavelength is `40 cm`. Then , the displacement amplitude of vibration at a point at distance `10 cm` from the open end , inside the pipe isA. `0.002 cm`B. `0.003 cm`C. `0.001 cm`D. `0.000 cm` |
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Answer» Correct Answer - D The equation of stationary wave foe open organ pipe can be written as ` y = 2 A cos (( 2 pi x)/(lambda)) sin (( 2 pi f t)/(v))` where `x = 0` is the open end from where the wave gets reflected. Amplitude of stationary wave is `A_(s) = 2 A cos (( 2pi x)/(lambda))` For ` x = 0.1 m`, `A_(s) = 2 xx 0.02 cos [( 2pi xx 0.1)/(0.4)] = 0` |
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| 141. |
Two waves `y_(1) = A cos (0.5 pi x - 100 pi t)` and `y_(2) = A cos (0.46 pi x - 92 pi t)` are travelling in a pipe placed along the `x ` - axis. Find wave velocity of louder soundA. `100`B. `46`C. `192`D. `96` |
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Answer» Correct Answer - A At `x = 0` `y = y_(1) + y_(2) = A cos (100 pi t) + A cos ( 92 pi t)` `rArr y = 2 A cos ( 96 pi t) sin ( 4 pi t)` for `y = 0` , we have `sin ( 4 pi t) = 0` `rArr t = (n)/(4)` `rArr t = 0 , (1)/(4) , (2)/(4) , (3)/(4) ,(4)/(4) "in" 1 s` This is ` 5` times also for `y = 0` , we have `cos ( 96 pi t) = 0` `rArr 96 pi t = (( 2m - 1) pi)/( 2)` `rArr t = ( 2m - 1)/( 192)` `rArr t = (1)/(192) , (3)/(192) ,....,(191)/(192)` This is `95` times. So the total number of times , the value of `y = zero is 100`. |
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| 142. |
Two waves `y_(1) = A cos (0.5 pi x - 100 pi t)` and `y_(2) = A cos (0.46 pi x - 92 pi t)` are travelling in a pipe placed along the `x ` - axis. Find wave velocity of louder soundA. `100 m//s`B. `192 m//s`C. `200 m//s`D. `96 m//s` |
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Answer» Correct Answer - C Wave velocity of louder sound `v = (omega)/(k) = 200 m//s` |
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| 143. |
Two waves are passing through a region in the same direction at the same time . If the equation of these waves are `y_(1) = a sin ( 2pi)/(lambda)( v t - x)` and `y_(2) = b sin ( 2pi)/( lambda) [( vt - x) + x_(0) ]` then the amplitude of the resulting wave for `x_(0) = (lambda//2)` isA. `| a - b|`B. ` a + b`C. `sqrt(a^(2) + b^(2))`D. `sqrt( a^(2) + b^(2) + 2 ab cos x)` |
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Answer» Correct Answer - A Let `phi_(1) and phi_(2)` represent angles of the first and second waves , then `phi_(2) = ( 2 phi)/( lambda ) [( vt - x) + x_(0)]` and `phi_(1) = ( 2pi)/(lambda) ( v t - x)` But `x_(0) = (lambda)/(2)`, ` phi_(2) - phi_(1) = pi` Hence , phase difference , `phi = pi `. So , amplitude of resultant wave `R sqrt(a^(2) + b^(2) + 2 ab cos phi)` ` = sqrt(a^(2) + b^(2) + 2 ab cos pi) = sqrt(( a - b)^(2)) = a - b` or `R = | a - b |` |
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| 144. |
In a large room , a person receives waves from the same source which reach , being reflected from the `25 m` high ceiling at a point halfway between them . The two waves interfere constructively for a wavelength ofA. `20 , 20/3 , 20/5` , etc.B. `10 , 5 , 2.5 `, etc.C. `10 , 20 , 30 `, etc.D. ` 15 , 25 , 35 `, etc. |
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Answer» Correct Answer - B Length of the path for direct sound ` = 120 m` Length of the path for reflected sound ` = 2 sqrt((60)^(2) + (25)^(2)) = 130 m` Geometrical path difference ` = 130 - 120 = 10 m` Two waves interfere constructively when `10 = n lambda` Putting , `n = 1 , 2, 3,..., lambda = 10 , 5 , 2.5 , .....` |
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| 145. |
A long tube contains air pressure of `1 atm` and a temperature of `59^(@) 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 distances `16 cm , 49.2 cm and 82.4 cm` from open end. Molar mass of air is `28.8 g//mol`. Ratio of heat capacities at constant pressure and constant volume for air at `59^(@) C` isA. `1.4`B. `1.152`C. `1.60`D. `2` |
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Answer» Correct Answer - B `v = sqrt((gamma RT)/(M))` `M = 28.8 g//mol = 28.8 xx 10^(-3) kg//mol` `T = 332 K` `R = 8.3 J//K mol` `gamma = (v^(2) M)/(RT) = ((332)^(2) xx 28.8 xx 10^(-3))/(8.3 xx 332) = 1.152` |
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| 146. |
A long tube contains air pressure of `1 atm` and a temperature of `59^(@) 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 distances `16 cm , 49.2 cm and 82.4 cm` from open end. Molar mass of air is `28.8 g//mol`. The speed of sound in air at `59^(@) C` isA. `332 m//s`B. `342 m//s`C. `352 m//s`D. `362 m//s` |
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Answer» Correct Answer - A Speed sound in air at `59^(@) C = 2n (l_(2) - l_(1))` `= 2 xx 500 (49.2 - 16) xx 10^(-2)` `= 332 m//s` |
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| 147. |
A long tube contains air pressure of `1 atm` and a temperature of `59^(@) 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 distances `16 cm , 49.2 cm and 82.4 cm` from open end. Molar mass of air is `28.8 g//mol`. Radius of tube isA. `1.1 cm`B. `1 cm`C. `1.2 cm`D. `2 cm` |
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Answer» Correct Answer - B `(lambda)/(4) = l_(1) + e` `( 3lambda)/( 4) = l_(2) + e` `e = (l_(2) - 3l_(1))/(2) = (49.2 - 3(16))/(2) = 0.6` `rArr 0.6 r = 0.6` ` r = 1 cm` |
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| 148. |
A turning fork vibrating at `500 Hz` falls from rest accelerates at `10 m//s^(2)`. How far below the point of release is the tuning fork when wave with a frequency of `475 Hz` reach the release point ?A. `16.9 m`B. `16 m`C. `1.69 m`D. `1.6 m` |
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Answer» Correct Answer - A Distance travelled by the tuning fork `= (1)/(2) "gt"^(2) = (1)/(2) xx 10 xx 1.837^(2) = 16.875 m` `= 16.9 m` |
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| 149. |
The two parts of a sonometer wire divided by a movable knife edge , differ in length by `2 mm` and produce `1 beat//s` , when sounded together . Find their frequencies if the whole length of wire is `1.00 m`. |
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Answer» Correct Answer - `249.5 Hz , 250.5 Hz` `l_(1) + l_(2) = 100 cm , l_(1) - l_(2) = 0.2 cm` `:. l_(1) = 50.1 cm and l_(2) = 49.9 cm` Also `nl = constant ` ` n xx 50.1 = ( n + 1) xx 49.9 ` `n = 249.5 Hz` `:. n_(1) = 249.5 Hz and n_(2) = n + 1 = 250.5 Hz` |
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| 150. |
The figure shown at time `t = 0` second, a rectangular and triangular pulse on a uniform wire are approcaching other. The pulse speed is `0.5 cm//s` . The resultant pulse at `t = 2` second is A. B. C. D. |
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Answer» Correct Answer - D |
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