InterviewSolution
Saved Bookmarks
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.
| 51. |
Sounds from two identical sources `S_(1)` and `S_(2)` reach a point `P`. When the sounds reach directly, and in the same phase, the intensity at `P` is `I_(0)`. The power of `S_(1)` is now reduced by `64%` and the phase difference between `S_(1)` and `S_(2)` is varied continuously. The maximum and minimum intensities recorded at `P` are now `I_("max")` and `I_("min")`A. `I_(max) = 0.64 I_(0)`B. `I_(min) = 0.36 I_(0)`C. `(I_(max))/(I_(min)) = 16`D. `(I_(max))/(I_(min)) = (1.64)/(0.36)` |
|
Answer» Correct Answer - A::C Let a= Initial amplitude due to `S_(1)` and `S_(2)` each `I_(0) = k(4a)^(2)`, where k is a constant After reduction of power of `S_(1)`, Due to superposition `a_(max)=a+0.6a=1.6a,` and `a_(min)=1-0.6a=0.4a` `I_(max)//I_(min)=(a_(max)//a_(min))^(2)=(1.6a//0.4a)=16` |
|
| 52. |
Transistor with `beta = 75` is connected to common-base configuration. What will be the maxima collector current for an emitter current of 5 mA ? |
|
Answer» `beta=75,I_(e)=5mA` `beta=(alpha)/(1-alpha)rArr 75=(alpha)/(1-alpha)rArr 75-75alpha=alpha or76 alpha=75 or alpha = (75)/(76)` `alpha=(I_(c))/(I_(e)):. I_(c)=alphaI_(e)=(75)/(76)xx5=4.93 mA`. |
|
| 53. |
In a transistor, the value of `beta` is 50. Calculate the value of `alpha` |
| Answer» `beta=50 ,beta=(alpha)/(1-alpha)rArr 50=(alpha)/(1-alpha)rArr 50-50alpha =alpha rArr alpha(50)/(51)=0.98` | |
| 54. |
For a common emitter amplifier, current gain = 50. If the emitter current is 6.6 mA, calculate the collector and base current. Also current gain. When emitter is working as common base amplifier, |
|
Answer» `beta=50,I_(e)=6.6mA becausebeta=(I_(c))/(I_(b)):. I_(c)=betaI_(e)=50 I_(b)`...(i) `I_(e)=I_(c)+I_(b)` using equation (i) we get `6.6 = 50 I_(c)+I_(b) = 51 I_(b) = (6.6)/(51)=0.129 mA` Hence `I_(c)=50 xx(6.6)/(51)=6.47mA and alpha=(beta)/(1+beta)=(50)/(51)=0.98` |
|
| 55. |
A detector at `x=0` receives waves from three sources each of amplitude `A` and frequencies `f + 2,f` and `f-2`. The equation of waves are : `y_(1)=A sin[2 pi(f+2),y_(2)=A sin 2 pift` and `y_(3)=A sin[2pi (f-2)t]`. The time at which intensity is minimum isA. `t=0,1//4,1//2,3//4,…sec`B. `t=1//6,1//3,2//3,5//6,…sec`C. `t=0,1//2,3//2,5//2,..sec`D. `t=1//2,1//4,1//6,1//8,…sec` |
|
Answer» Correct Answer - B `y=y_(1)+y_(2)+y_(3)=A sin 2 pift+A sin[2 pi(f-2)t]+Asin[2 pi(f+2)t]=A sin 2pi ft+2 A sin 2 pi ft cos 4 pi t` `= A[1+2 cos4 pi t]sin 3 pift=A_(0) sin 2pit` [where `A_(0)`= Amplitude of the resultant oscillation `=A[1+2 cos 4 pi t`] Intensity `prop A_(0)^(2) :. I prop(1+2 cos 4 pi t)^(2)` For maxima or minima of the intensity `(dI)/(dt)=0rArr 2(1+2 cos 4pit)(2)(-sin 4pit)4 pi=0 rArr 1+2 cos 4 pit =0` or `sin 4 pit =0` `:. cos 4pit=-(1)/(2)rArr 4pi t=2pin pm (2pi)/(3) :. t=(n)/(2)+(1)/(6)rArr t=(1)/(6),(1)/(3),(2)/(5),(5)/(6)....` (point of minimum intensity). |
|
| 56. |
Pure Si at 300K has equal electron `(n_(e))` and hole `(n_(h))` concentrations of `1.5xx10^(16)m^(-3)` Doping by indium increases `n_(h)` to `3xx10^(22)m^(-3)`. Calculate `n_(e)` in the doped Si. |
|
Answer» For a doped semi-conductor in thermal equilibrium `n_(e)n_(h) = n_(1)^(2)` (Law of mass action) `n_(e)=(n_(1)^(2))/(h_(h))=((1.5xx10^(16))^(2))/(3xx10^(22))=7.5 xx10^(9)m^(-3)` |
|
| 57. |
Three sound waves of equal amplitudes have frequencies `(v - 1), v, (v + 1)`. They superpose to give beats. The number of beats produced per second will be :A. 2B. 1C. 4D. 3 |
|
Answer» Correct Answer - A `{:(A,B,C),(v-1,v,v+1):}` Between `A & B 1 b//s 1` `B & C 1 b//s 1` `C & A 2 b//s (1)/(2) ((2)/(2))` `rArr 2 b//s` |
|
| 58. |
The transverse displacement `y(x, t)` of a wave on a string is given by `y(x, t)= e ^(-(ax^(2) + bt^(2) + 2sqrt((ab))xt)`. This represents a :A. Standing wave of frequency `sqrt(b)`B. standing wave of frequency `(1)/(sqrt(b))`C. wave moving in `+x` direction with speed `sqrt((a)/(b))`D. wave moving in -x direction with speed `sqrt((b)/(a))` |
|
Answer» Correct Answer - D `y(x_(1)t)=e^(-[sqrt(ax)+sqrt(bt)]^(2))` `v = omega//K=(sqrt(b))/(sqrt(a))` in `-ve` x direction |
|
| 59. |
A source of sound of frequency 256 Hz is moving towards a wall with a velocity of 5 m/s. How many beats per second will be heard by an observer O standing in such a positoin that the source S is between O and wall? `(c=330(m)/(s)`) |
|
Answer» Correct Answer - `7.87 Hz` Frequency reaching the wall `f_(1)=f_(0)((v)/(v-v_(s)))` Frequency received by the observer `f_(1)=f_(1),((v+v_(s))/(v))=f_(0)((v+v_(s))/(v-v_(s)))` `:.` Beat frequency `Delta f=f_(0)((v+v_(s))/(v-v_(s)))-f_(0)=256((330+5)/(330-5))-256` `=7.87 Hz` |
|
| 60. |
Statement-1 : Earthquakes cause vast devastation. Sometimes short and tall structures remain unaffected while the medium height structures fall Statement-2 : The natural frequency of the medium structures coincides with the frequency of the seismic wave.A. Statement-1 is true, Statement-2 is true , Statement-2 is correct explanation for Statement-12B. Statement-1 is true,Statement-2 is true , Statement-2 is NOT a correct explanation for Statement-12C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true |
| Answer» Correct Answer - A | |
| 61. |
Statement-1 : During thunderstrom, light is seen much earlier than the sound is heard Statement-2 : Light travels faster than sound.A. Statement-1 is true, Statement-2 is true , Statement-2 is correct explanation for Statement-10B. Statement-1 is true,Statement-2 is true , Statement-2 is NOT a correct explanation for Statement-10C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true |
| Answer» Correct Answer - A | |
| 62. |
A thunder tap is heard 5.5 s after the lightening flash. The distance of the flash is (velocity of sound in air is 330 m/s) :-A. 3560 mB. 300 mC. 1780 mD. 1815 m |
|
Answer» Correct Answer - D `d=vt = 330 xx 5.5 =1815 m` |
|
| 63. |
Statement-1 : The flash of lightening is seen before the sound of thunder is heard Statement-2 : The sound of thunder is produced after the flash of lightening.A. Statement-1 is true, Statement-2 is true , Statement-2 is correct explanation for Statement-11B. Statement-1 is true,Statement-2 is true , Statement-2 is NOT a correct explanation for Statement-11C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true |
| Answer» Correct Answer - C | |
| 64. |
A steel wire of length `1m`, mass `0.1kg` and uniform cross-sectional area `10^(-6)m^(2)` is rigidly fixed at both ends. The temperature of the wire is lowered by `20^(@)C`. If transverse waves are set up by plucking the string in the middle.Calculate the frequency of the fundamental mode of vibration. Given for steel `Y = 2 xx 10^(11)N//m^(2)` `alpha = 1.21 xx 10^(-5) per ^(@)C` |
|
Answer» Correct Answer - 11 Hz For wire, `DeltaL=L alpha theta = (FL)/(AY)rArr D=AY alpha theta` `:. f=(1)/(2L)sqrt((F)/(mu))=(1)/(2xx1)sqrt((10^(-6)xx2xx10^(11)xx1.21 xx10^(5)xx20)/(0.1))` `=11Hz` |
|
| 65. |
A tuning fork of `512 H_(Z)` is used to produce resonance in a resonance tube expertiment. The level of water at first resonance is `30.7 cm` and at second resonance is `63.2 cm` . The error in calculating velocity of sound is (a) `204.1 cm//s` (b) `110 cm//s` (c) `58 cm//s` (d) `280 cm//s`A. 204.1 cm/sB. 110 cm/sC. 58 cm/sD. 280 cm/s |
|
Answer» Correct Answer - D The question is incomplete, as speed of sound is not given. Let us assume speed of sound as 330 m/s. Then, method will be as under. `(lamda)/(2)=(63.2-30.7)cmor lamda=0.65m` `:.` speed of sound observed `V_(0)=f lamda=512 xx0.65=332.8 m` `:.` Error is calculated velocity of sound `= 2.8 m//s = 280 cm//s` |
|
| 66. |
An open pipe is in resonance in `2nd` harmonic with frequency `f_(1)`. Now one end of the tube is closed and frequency is increased to `f_(2)` such that the resonance again ocuurs in `nth` harmonic. Choose the correct optionA. `n =3,f_(2) = (3)/(4) f_(1)`B. `n =3,f_(2) = (5)/(4) f_(1)`C. `n=5,f_(2) = (5)/(4) f_(1)`D. `n =5,f_(2) = (3)/(4) f_(1)` |
|
Answer» Correct Answer - C `f_(1)=(v)/(l)` (2nd harmonic of open pipe) `f_(2)=n((v)/(4l))(n^(th)` harmonic of closed pipe) Here, n is odd and `f_(2) gt f_(1)` It is possible when n = 5 because with n = 5 `rArr f_(2)=(5)/(4)((v)/(l))=(5)/(4)f_(1)` |
|
| 67. |
An open pipe is suddenly closed at one end with the result that the frequency of third harmonic of the closed pipe is found to be higher by `100 Hz` then the fundamental frequency of the open pipe. The fundamental frequency of the open pipe isA. 200 HzB. 300 HzC. 240 HzD. 480 Hz |
|
Answer» Correct Answer - A `f_(1)lamda_(1)=v` `f_(1)((4L)/(3))=vrArrf_(1)=(3v)/(4L),f_(2)(2L)=vrArrf_(2)=(v)/(2L)` Given that `f_(1)-f_(2)=100` `rArr (3v)/(2L)-(v)/(2L)=100 rArr(v)/(2L)=200Hz` |
|
| 68. |
In the arrangement shown in figure, the string has a mass of 4.5 g. How much time will it take for a transverse disturbance produced at the floor to reach the pulley ? Take `g=10ms^-2` . |
|
Answer» Correct Answer - `0.02 s` `v=sqrt((T)/(mu))=sqrt((2xx10)/((4.5xx10^(-3))/(2.25)))=100m//s` `t=(L)/(v)=(2)/(100)=0.02sec` |
|
| 69. |
A metallic rod of length 1m is rigidly clamped at its mid point. Longirudinal stationary wave are setup in the rod in such a way that there are two nodes on either side of the midpoint. The amplitude of an antinode is `2 xx 10^(-6) m`. Write the equation of motion of a point 2 cm from the midpoint and those of the constituent waves in the rod, (Young,s modulus of the material of the rod `= 2 xx 10^(11) Nm^(-2)` , density `= 8000 kg-m^(-3)`). Both ends are free. |
|
Answer» Correct Answer - `y=10^(-6)sin(0.1pit)sin(25000pit),y_(1)=10^(-6)sin(25000pit-5pit),y_(2)=10^(-6)sin(25000 pit+5 pi x)`] `v=sqrt((Y)/(rho))=sqrt((2xx10^(11))/(8000))=5000m//s` `(5)/(2)lamda =LrArr lamda=(2L)/(5)=(2xx1)/(5)=0.4m`. |
|
| 70. |
The figure shows a snap photograph of a vibrating string at `t = 0`. The particle `P` is observed moving up with velocity `20sqrt(3) cm//s`. The tangent at `P` makes an angle `60^(@)` with x-axis. (a) Find the direction in which the wave is moving. (b) Write the equation of the wave. (c) The total energy carries by the wave per cycle of the string. Assuming that the mass per unit length of the string is `50g//m`. |
|
Answer» Correct Answer - (i) Negative x (ii) `y = 0.4 sin (10 pi t + (pi)/(2) x + (pi)/(4))` (x,y are in cm) (iii) `1.6 xx 10^(-5) J` (i) For the particle P `(dely)/(delt)=-v((dely)/(delx))rArr +20sqrt(3)=v(sqrt(3))` `rArrv=-20cm//s` (along negative x-axis) (ii) Equation of wave `y=Asin(omegat+kx+phi)` at`t =0,x=0,y=2sqrt(2),A=4` `rArr4=2sqrt(2)sinphirArrphi=(pi)/(4),lamda=5.5-1.5=4cm` `f=(v)/(lamda)=(20 cm//s)/(4cm)=5Hz` `:.y=4sin(10pit+(pix)/(2)+(pi)/(4))` (iii) Energy carried in one wavelength `E=(1)/(2)muA^(2)omega^(2)lamda` `=(1)/(2)xx(50)/(1000)xx(4xx10^(-2))^(2)xx(10pi)^(2)xx(4)/(100)` `=16 pi^(2) xx 10^(-5) J` |
|
| 71. |
A sound absorber attenuates the sound level by `20 dB`. The intensity decreases by a factor ofA. 1000B. 10000C. 10D. 100 |
|
Answer» Correct Answer - D Intensity change in decibel `=10 "log"(I_(2))/(I_(1))=20rArr "log"(I_(2))/(I_(1))=2rArr(I_(2))/(I_(1))=10^(2)=100` |
|
| 72. |
A uniform rope having some mass hinges vertically from a rigied support. A transverse wave pulse is produced at the lower end. The speed (v) of the wave pulse varies with height (h) from the lower end as :-A. B. C. D. |
|
Answer» Correct Answer - C `v=sqrt((T)/(mu))=sqrt((muxg)/(mu))=sqrt(gx)rArr v^(2)=gx`. |
|
| 73. |
A copper wire is held at the two ends by rigid supports. At `30^(@)C`, the wire is just taut, with negligible tension. Find the speed of transverse waves in this wire at `10^(@)C`. Given : Young modulus of copper `= 1.3 xx 10^(11) N//m^(2)`. Coefficient of linear expansion of copper ` = 1.7 xx 10^(-5) .^(@)C^(-1)`. Density of copper `= 9 xx 10^(3) kg //m^(3)`.A. 210 m/sB. 110 m/sC. 90 m/sD. 70 m/s |
|
Answer» Correct Answer - D `DeltaL=(TL)/(AY)rArr T=(DeltaLAY)/(L)=(LalphathetaAY)/(L)=alpha thetaAy` `mu =dA :. v=sqrt((T)/(mu))=sqrt((alpha theta AY)/(dA))=70 m//s` |
|
| 74. |
One end of a taut string of length `3m` along the x-axis is fixed at `x = 0`. The speed of the waves in the string is `100ms^(-1)`. The other end of the string is vibrating in the y-direction so that stationary waves are set up in the string. The possible wavelength`(s)` of these sationary waves is (are)A. `y(t) = A sin (pi x)/(6) cos (50 pi t)/(3)`B. `y(t) = A sin (pi x)/(3) cos (100 pi t)/(3)`C. `y (t) = A sin (5 pi x)/(6) cos (250 pi t)/(3)`D. `y(t) = A sin (5 pi x)/(2) cos 250 pi t` |
|
Answer» Correct Answer - A::C::D At `{:(x=0 y=0),(x=3 y=0),((omega)/(k)=100 m//s):}]` The equation satisfying all three conditions is correct. |
|
| 75. |
What will be conductivity of pure sillicon crystal at 300 K temp? if electron hole pairs per `cm^(3)` is `1.072xx10^(10)` at this temp. `mu_(n)=1350 cm^(2)//"volt"` sec and `mu_(p)=480 cm^(2)//"volt"` sec- |
| Answer» `sigma=n_(1)emu_(e)+n_(1)emu_(h)=n_(1)e(mu_(e)+mu_(h))=3.14xx10^(-6)"mho/cm"` | |
| 76. |
Pure `Si` at `500K` has equal number of electron `(n_(e))` and hole `(n_(h))` concentration of `1.5xx10^(16)m^(-3)`. Dopping by indium. Increases `n_(h)` to `4.5xx10^(22) m^(-3)`. The doped semiconductor is of |
| Answer» `n_(e)=(n_(1)^(2))/(n_(h))=((1.5xx10^(16))^(2))/((4.5xx10^(22)))=5xx10^(9)m^(-3)` | |
| 77. |
A police car moving at 22 m/s, chases a motorcylist. The police man sounds his horn at 176 Hz, while both of them move towards a stationary siren of frequency 165 Hz. Calculate the speed of the motorcycle, if it is given that he does not observes any beats A. 33 m/sB. 22 m/sC. zeroD. 11 m/s |
|
Answer» Correct Answer - B The motorcyclist observes no beats, So, the apparent frequency observed by him from the two sources must be equal. ` f_(1)=f_(2):.176((330-v)/(330-22))=165((330+v)/(330))` Solving this equation, we get `v = 22 m//s`. |
|
| 78. |
In the experiment for the determination of the speed of sound in air using the resonance column method, the length of the air column that resonates in the fundamental mode, with a tuning fork is `0.1m`. When this length is changed to `0.35m`, the same tuning fork resonates with the first overtone. Calculate the end correction.A. 0.012 mB. 0.025 mC. 0.05 mD. 0.024 m |
|
Answer» Correct Answer - B Let `Delta l` be the end correction Given that, fundamental tone for a length `0.1 m` = first overtone for the length 0.35 m `(v)/(4(0.1+Deltal))=(3v)/(4(0.35+Deltal))` we get `Deltal=0.025m=2.5 xx 10^(-2)m`. |
|
| 79. |
A detector is released from rest over height h a source of sound of frequency `f_(0) = 10^(3) Hz`. The frequency observed by the detector at time t is plotted in the graph. The speed of sound in air is `(g = 10 m//s^(2))` A. 330 m/sB. 350 m/sC. 300 m/sD. 310 m/s |
|
Answer» Correct Answer - C `f=f_(o)((v+g t)/(v))rArr (df)/(g t)=f_(o)(0+(g)/(v))` `rArr (1000)/(30)=(1000xx10)/(v)rArrv=300 m//s` |
|
| 80. |
A SONAR system fixed in a submarine operates at a frequency 40.0kHz. An enemy submarine moves towards the SONAR with a speed of 360 km`h^(-1)`. What is the frequency of sound reflected by the submarine ? Take the speed of sound in water to be 1450 `ms^(-1)`. |
|
Answer» Correct Answer - `45.93 kHz` Frequency received by subnarine `f_(1)=f_(1) =f_(0)((v)/(v-v_(s)))` `f_(2)=f_(1)((v+v_(s))/(v))=f_(0)((v+v_(s))/(v-v_(s)))=140((1450+100)/(1450+100))` `=45.93kHz` |
|
| 81. |
The resultant amplitude, when two waves of two waves of same frequency but with amplitudes `a_(1)` and `a_(2)` superimpose at phase difference of `pi//2` will be :-A. `a_(1) + a_(2)`B. `a_(1) - a_(2)`C. `sqrt(a_(1)^(2) + a_(2)^(2))`D. `a_(1)^(2) + a_(2)^(2)` |
|
Answer» Correct Answer - C `a=sqrt(a_(1)^(2)+a_(2)^(2)+2a_(1)a_(2)"cos"(pi)/(2))=sqrt(a_(1)^(2)+a_(2)^(2))` |
|
| 82. |
A string of length `0.4 m` and mass `10^(-2) kg` is clamped at one end . The tension in the string is `1.6 N`. The identical wave pulses are generated at the free end after regular interval of time , `Delta t` . The minimum value of `Delta t` , so that a constructive interference takes place between successive pulses isA. 0.05 sB. 0.10 sC. 0.20 sD. 0.40 s |
|
Answer» Correct Answer - B Mass per unit length of the string `m =(10^(-2))/(0.4)=2.5 xx10^(-2)kg//m` `:.` Velocity of wave in the string `v=sqrt((T)/(m))=sqrt((1.6)/(2.5xx10^(-3)))rArr v=8m//s` For constructive interference between successive pulses `Delta t_(min)=(2l)/(v)=((2)(0.4))/(8)=0.10s` After two reflections, the wave pulse is in same phase as it was produced, since in one reflection its phase changes by `pi`, and if at this moment next identical pulse is produced, then constructive interference will be obtained. |
|
| 83. |
A flute which we treat as a pipe open at both ends is `60 cm` long. (a) What is the fundamental frequency when all the holes are covered? (b) How far from the mouthpieces should a hole be uncovered for the fundamental frequency to be `330 H_(Z)` ? Take speed of sound in air as `340 m//s`. |
| Answer» Correct Answer - `283.33 Hz` | |
| 84. |
A narrow tube is bend in the form of a circle of radius R, as shown in the figure. Two small holes S and D are made in the tube at the positions right angle to each other. A source placed at S generated a wave of intensity `I_(0)` which is equally divided into two parts : One part travels along the longer path, while the other travels along the shorter path. Both the parts wave meet at the point D where a detector is placed. The maximum intensity produced at D is given by :-A. `4 I_(0)`B. `2 I_(0)`C. `I_(0)`D. `3 I_(0)` |
|
Answer» Correct Answer - B `I_(max)=(sqrt((I_(0))/(2))+sqrt((I_(0))/(2)))^(2)=2I_(0)` |
|
| 85. |
A narrow tube is bend in the form of a circle of radius R, as shown in the figure. Two small holes S and D are made in the tube at the positions right angle to each other. A source placed at S generated a wave of intensity `I_(0)` which is equally divided into two parts : One part travels along the longer path, while the other travels along the shorter path. Both the parts wave meet at the point D where a detector is placed. The maximum value of `lamda` to produce a maxima at D is given by :-A. `pi R`B. `2 pi R`C. `(pi R)/(2)`D. `(3 pi R)/(2)` |
|
Answer» Correct Answer - A `lamda_(max)` to produce maxima at `D = piR` |
|
| 86. |
A narrow tube is bend in the form of a circle of radius R, as shown in the figure. Two small holes S and D are made in the tube at the positions right angle to each other. A source placed at S generated a wave of intensity `I_(0)` which is equally divided into two parts : One part travels along the longer path, while the other travels along the shorter path. Both the parts wave meet at the point D where a detector is placed. The maximum value of `lamda` to produce a minima at D is given by :-A. `pi R`B. `2 pi R`C. `(pi R)/(2)`D. `(3 pi R)/(2)` |
|
Answer» Correct Answer - B `lamda_(max)` to produce minima at `D = 2 pi R` |
|
| 87. |
Plane sound waves of wavelength 0.12 m are incident on two narrow slits in a box nonreflecting walls, as shown. At a distance of 5.0 m from the center of the slits, a first-order maximum occurs at point P, which is 3.0 m from the central maximum. The distance between the slits is most nearly. .A. 0.07 mB. 0.09 mC. 0.16 mD. 0.20 m |
|
Answer» Correct Answer - D `d sin theta =n lamda` `n=1` `d sin 37^(@)=lamda` `d=0.2` |
|
| 88. |
Two vibrating tuning fork produce progressive waves given by `y_(1) = 4 sin 500 pi t` and `y_(2) = 2 sin 506 pi t`. Number of beats produced per minute is :-A. 3B. 360C. 180D. 60 |
|
Answer» Correct Answer - C `f_(1)=(500pi)/(2pi)=250 ,f_(2)=(506)/(2pi)=253` `:. Delta f=3s^(-1)=3xx60 "min"^(-1)=180 "min"^(-1)` |
|
| 89. |
Two open pipes of length L are vibrated simultaneously. If length of one of the pipes is reduced by y. then the number of beats heats per second will be if the velocity of sound is v and `y lt lt L` :-A. `(vy)/(2L^(2))`B. `(vy)/(L^(2))`C. `(vy)/(2L)`D. `(2L^(2))/(vy)` |
|
Answer» Correct Answer - A `lamda_(1)=2L,lamda_(2)=2(L-y)` `Delta f=f_(2)-f_(1)=(v)/(lamda_(2))-(v)/(lamda_(1))=(v)/(2)[(1)/(L-y)-(1)/(L)]` `rArr (vy)/(2(L-y)L)=(vy)/(2L^(2))` |
|
| 90. |
Two open pipes of length 25 cm and 25.5 cm produced 0.1 beat/second. The velocity of sound will be :-A. 255 cm/sB. 250 cm/sC. 350 cm/sD. None of these |
|
Answer» Correct Answer - A `v=f_(1)xx50=f_(2)xx51` `rArr f_(1)-f_(2)=(v)/(50)-(v)/(51)=0.1 rArrv=255 m//s` |
|
| 91. |
A sonometer wire under tension of 64 N vibrating in its fundamental mode is in resonance with a vibrating tuning fork. The vibrating portion of the sonometer wire has a length of 10 cm and mass of 1 kg. The vibrating tuning fork is now moved away from the vibrating wire with a constant speed and an observer standing near the sonometer hears one beat per second. Calculate the speed with which the tuning fork is moved, if the speed of sound in air is 300 m/s. |
|
Answer» Correct Answer - `0.073 m//s` `mu=(1gm)/(10cm)=(10^(-3)kg)/(0.1m)=10^(-4)kg//m,T=64N` `v = sqrt((T)/(mu))rArrf_(0)=(v)/(2L)=(1)/(2xx0.1)sqrt((64)/(10^(-4)))` `=5xx8xx100` `=4000m//s` `rArr v_("string")=v_("fork")=1` `=5xx8xx100` `=4000m//s` `rArr v_("string")=v_("fork")=1` `rArr 4000-4000((300)/(300-v))=1rArr v=0.073m//s`. |
|
| 92. |
Length of a sonometer wire is either 95 cm or 100 cm. In both the cases a tuning fork produces 4 beats then the frequency of tuning fork is :-A. 152B. 156C. 160D. 164 |
|
Answer» Correct Answer - B For sonometer wire `nxx100=(n+1)xx95` n=no of harmonics `rArr n =19` `:. f=19((L)/(2))+4=20((L)/(2))-4rArr L=16` `rArrf=20((L)/(2))-4=156 Hz` |
|
| 93. |
The distance between two consecutive crests in a wave train produced in string is 5 m. If two complete waves pass through any point per second, the velocity of wave is :-A. 2.5 m/sB. 5 m/sC. 10 m/sD. 15 m/s |
|
Answer» Correct Answer - C `f=(2 "waves")/(1 sec)=2Hz,lamda=5m :. v =f lamda=10m//s`. |
|
| 94. |
An organ pipe `P_(1)` open at one end vibrating in its first harmonic and another pipe `P_(2)` open at ends vibrating in its third harmonic are in resonance with a given tuning fork. The ratio of the length of `P_(1)` to that `P_(2)` isA. `(8)/(3)`B. `(3)/(8)`C. `(1)/(6)`D. `(1)/(3)` |
|
Answer» Correct Answer - C `f_(1)=f_(2),(v)/(lamda_(1))=(v)/(lamda_(2))rArr lamda_(1)=4L_(1)=lamda_(2)=(2)/(3)L_(2)` `rArr (L_(1))/(L_(2))=(2)/(3xx4)=(1)/(6)` |
|
| 95. |
A travelling wave represented by `y=Asin (omegat-kx)` is superimposed on another wave represented by `y=Asin(omegat+kx).` The resultant isA. A standing wave having nodes at `x=(n+(1)/(2))(lamda)/(2),n=0,1,2`B. A wave travelling along + x directionC. A wave travelling along -x directionD. A standing wave having nodes at `x = (n lamda)/(2) , n=0,1,2` |
|
Answer» Correct Answer - A `y_(1)=Asin(wt-kx)&y_(2)=Asin(wt-kx)` By superposition principle `y = y_(1) +y_(2)` `=A sin (wt-kx)+A sin (wt +kx)` `=2A sin wt cos kx` Amplitude = 2A cos kx At nodes displacement is minimum `2A coc kx =0 rArr cos kx =0` `kx = (2n+1) (pi)/(2) rArr (2pi)/(2)x=(2n+1)(pi)/(2)` `x =(2n+1)(pi)/(4)` where `n =0,1,2` |
|
| 96. |
A wave `y = a sin (omegat - kx)` on a string meets with another wave producing a node at `x = 0`. Then the equation of the unknown wave isA. `y = a sin(omega t + kx)`B. `y = -a sin(omega t +kx)`C. `y = a sin (omega t - kx)`D. `y = -a sin (omega t - kx)` |
| Answer» Correct Answer - B | |
| 97. |
Statement-1`:` Two longitudinal waves given by equation `y_(1)(x,t)=2asin(omegat-kx)` and `y_(2)(x,t)=a sin (2 omegat-2kx)` will have equal intensity. Statement-2 `:` Intensity of waves of given frequency in same medium is proportional to square of amplitude only.A. Statement-1 is flase, statement-2 is trueB. Statement-1 is true, Statement-2 is falseC. Statement-1 is true, Statement-2 true , Statement-2 is the correct axplanation of statement-1D. Statement-1 is true, Statement-2 is true , Statement-2 is not correct explanation of Statement-1. |
|
Answer» Correct Answer - B `y_(1)(x,t)=2a sin (wt-kx)` `y_(2)(x,t)=a sin(2wt-2kx)` `I=2pi^(2)n^(2)a^(2)rhovrArr(I_(1))/(I_(2))=((2a)/(a)xx(n)/(2n))^(2)=(1)/(1)` Intensity depends on frequency and amplitude So statememt-1 is true statement-2 is false. |
|
| 98. |
A tuning fork having n = 300 Hz produces 5 beats/s with another tuning fork. If impurity (wax) is added on the arm of known tuning fork, the number of beats decreases then calculate the frequency of unknown tuning fork. |
|
Answer» The frequency of unknown tuning fork should be `300 pm 5 = 295 Hz or 305 Hz` When wax is added, it is would be 305 Hz, beats would have increases but with 295 Hz beats is decreases so frequency of unknown tuning fork 295 Hz. |
|
| 99. |
Calculate the collector and emitter current for which `I_(b) = 20 mA, beta = 100` |
|
Answer» `beta=100,I_(b)=20muA` `I_(c)=betaI_(b)=100xx20xx10^(-6)=2000 muA` `I_(e)=I_(b)+I_(c)=20+2000=2020muA=2.02xx10^(-3)A=2.02mA` |
|