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.
| 1. |
A sinusoidal voltage of amplitude 25 volts and frequency 50 Hz is applied to a half wave rectifier using PN diode. No filter is used and the load resistor is `1000 Omega`. The forward resistance `R_(f)` ideal diode is `10 Omega`. Calculate. (i) Peak, average and rms values of load current (ii) d.c power output (ii) a.c power input (iv) % Rectifier efficiency (v) Ripple factor. |
|
Answer» (i) `I_("in")=(V_(m))/(R_(f)+R_(L))=(25)/((10+1000))=24.75A`, `I_(dc)=(I_(m))/(pi)=(24.75)/(3.14)=7.88mA`, `I_(max)=(I_(m))/(pi)=(24.75)/(2)=12.38mA` (ii) `P_(dc)=I_(dc)^(2)xxR_(L)=(7.88xx10^(-3))^(2)xx10^(3)~~63mW` (iii) `P_(dc)=I_("rms")^(2)(R_(f)+R_(L))=(12.38xx10^(-3))^(2)xx(10+1000)~~155mW` (iv) Rectifier efficiency `eta=(P_(dc))/(P_(ac))xx100=(62)/(155)xx100=40%` (v) Ripple factor `= [((I_("rms"))/(I_(ac)))^(2)-1]^(1//2)=[((12.38)/(7.88))^(2)-1]=1.21` |
|
| 2. |
A fullwave P.N diode rectifier used load resistor of `1500 Omega`. No filter is used. Assume each diode to have idealized characteristic with `R_(f) = 10 Omega` and `R_(r) = ac`. Since wave voltage applied to each diode has amplitude of 30 volts and frequency 50 Hz. Calculate. (i) Peak, d.c rms load current (ii) d.c power input (iii) A.C power input (iv) Rectifier efficiency. |
|
Answer» (i) Peak current `I_(m)=(V_(m))/(R_(f)+R_(L)):. I_(m)=(30"volts")/(10+1500)=19.9mA` d.c load current `I_(dc)=(2I_(m))/(pi)=0.636 I_(m)=0.636xx19.9mA=12.66mA` `I_("rms")=(I_(m))/(sqrt(2))=(19.9)/(sqrt(2))=14 mA` (ii) DC Power output `P_(dc)=I_(dc)^(2)xxR_(L)=(12.66xx10^(-3))^(2)xx1500 "Watt" = 240.41 mW` (iii) AC power input `P_("in")=I_("rms")^(2)(R_(f)+R_(2))=(14xx10^(-3))^(2)(10+1500)"Walt" = 295.96mW` |
|
| 3. |
Figure shows a string of linear mass density `1.0g cm^(-1)` on which a wave pulse is travelling. Find the time taken by pulse in travelling through a distance of `50 cm` on the string. Take `g = 10 ms^(-2)` |
|
Answer» Correct Answer - `0.05 s` `v=sqrt((1xx10)/(10^(-3)//10^(-2)))=10m//s,t=(L)/(v)=(0.5)/(10)=0.05 sec` |
|
| 4. |
Following are equations of four waves : (i) `y_(1) = a sin omega ( t - (x)/(v))` (ii) `y_(2) = a cos omega ( t + (x)/(v))` (iii) `z_(1) = a sin omega ( t - (x)/(v))` (iv) `z_(1) = a cos omega ( t + (x)/(v))` Which of the following statements are correct ?A. One superposition of waves (i) and (iii), a travelling wave having amplitude `a sqrt(2)` will be formed.B. Superposition of waves (ii) and (iii) is not possibleC. On superposition of waves (i) and (ii), a transverse stationary wave having maximum amplitude `a sqrt(2)` will be formed.D. On superposition of waves (iii) and (iv), a transverse stationary wave will be formed. |
|
Answer» Correct Answer - A::D Superposition of waves (i) & (iii) will give travelling wave having amplitude of `a sqrt(2)` [waves are along x-axis but particle displacements are along `y` & z-axis respectively] `z_(1)+z_(2)=a[sin omega (t-(x)/(v))+sin{omega(t+(x)/(v))+(pi)/(2)}]`. |
|
| 5. |
A parabolic pulse given by equation `y ("in cm") = 0.3 - 0.1(x-5t)^(2) (y ge 0)` travelling in a uniform string. The pulse passes through a boundary beyond which its velocity becomes `2.5 m//s`. What will be the amplitude of pulse in this medium after transmission ? |
| Answer» Correct Answer - `0.2 cm` | |
| 6. |
A man generates a symmetrical pulse in a string by moving his hand up and down. At `t=0` the point in his hand moves downward. The pulse travels with speed of `3 m//s` on the string & his hands passes `6` times in each second from the mean position. Then the point on the string at a distance `3m` will reach its upper extreme first time at time `t=`A. `1.25 sec`B. 1 secC. `(13)/(12) sec`D. none |
|
Answer» Correct Answer - A Frequency of wave `= (6)/(2)-3 rArr T=(1)/(3) s, lamda -vT=(3)((1)/(3))=1m` Total time taken `= (3)/(3)+(3T)/(4)=1.25 sec`. |
|
| 7. |
A person going away from a factory on his scooter at a speed of 36 km/hr listens to the siren of the factory. If the actual frequency of the siren is 700 Hz and a wind is blowing along the direction of the scooter at 36 km//hr, find the observed frequency heard by the person. (Given speed of sound `= 340 m//s`) |
|
Answer» Correct Answer - `680 Hz` `f_("observer")=f_(0)((v+v_(w)-v_(0))/(v+v_(w)))=700((340+10-10)/(340+10))` `=680Hz` |
|
| 8. |
A car has two horns having a difference in frequency of 180 Hz. The car is approaching a stationary observer with a speed of `60 ms^(-1)`. Calculate the difference in frequencies of the notes as heard by the observer, If velocity of sound in air is `330 ms^(-1)`. |
|
Answer» Correct Answer - `220 Hz` `Delta f=f_(0)((v)/(v-v_(S)))=180((300)/(330-60))=220Hz` |
|
| 9. |
Two engines pass each other moving in opposite directions with uniform speed of of them is blowing a whistle of frequency 540 Hz. Calculate the frequency heard by driver of second engine before they pass each other. Speed of sound is 300 m/sec :A. 648 HzB. 450 HzC. 270 HzD. 540 Hz |
| Answer» Correct Answer - A | |
| 10. |
A person observes a change of 2.5 % in frequency of sound of horn of a car. If the car is approaching forward the person & sound velocity is 320 m/s, then velocity of car in m/s will be approximaterly :-A. 8B. 800C. 7D. 6 |
|
Answer» Correct Answer - A `Deltaf=f_(app)-f_(o)=f_(o)((v)/(v-_(s)))-f_(o)` `(Deltaf)/(f_(o))=(v)/(v-v_(s))-1=(v_(s))/(v-v_(s))=(2.5)/(100)` `rArr (v_(s))/(v-v_(s))=(1)/(40)rArrv_(s)=(v)/(41)=8m//s` |
|
| 11. |
Two trains A and B are moving in the same direction with velocities 30 m/s and 10 m/s respectively, B is behind from A, blows a horn of frequency 450 Hz. Then the apparent frequency heard by B is (The velocity of sound is 330 m/s) :-A. 425 HzB. 300 HzC. 450 HzD. 350 Hz |
|
Answer» Correct Answer - A `f_(B)=f_(a)((v+v_(B))/(v+v_(A)))=450((330+10)/(330+30))=425 Hz` |
|
| 12. |
A toy-car, blowing its horn, is moving with a steady speed of 5 m//s, away from a wall. An observer, towards whom the toy car is moving, is able to heat 5 beats per second. If the velocity of sound in air is 340 m/s, the frequency of the horn of the toy car is close to :A. 340 HzB. 510 HzC. 170 HzD. 680 Hz |
| Answer» Correct Answer - C | |
| 13. |
A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, `y(x,t)=(0.01 m) sin[(62.8 m^(-1)x] cos[(628 s^(-1))t]`. Assuming `p = 3.14`, the correct statement(s) is (are)A. The number of node is 5B. The length of the string is 0.25 mC. The maximum displacement of the midpoint of the string. From its equilibrium position is 0.01 mD. The fundamental frequency is 100 Hz |
| Answer» Correct Answer - B::C | |
| 14. |
The ends of a stretched wire of length `L` are fixed at `x = 0 and x = L`. In one experiment, the displacement of the wire is `y_(1) = A sin(pi//L) sin omegat` and energy is `E_(1)` and in another experiment its displacement is `y_(2) = A sin (2pix//L ) sin 2omegat` and energy is `E_(2)`. ThenA. `E_(2) = E_(1)`B. `E_(2) = 2 E_(1)`C. `E_(2) = 4 E_(1)`D. `E_(2) = 16 E_(1)` |
|
Answer» Correct Answer - C Energy `prop ("amplitude")^(2) ("frequency")^(2)` Amplitude (A) is same in both the cases, but frequency `2omega`, in the second case is two times the frequency `(omega)` in the first case, `E_(2) = 4E_(1)` |
|
| 15. |
A massless rod BD is suspended by two identical massless strings AB and CD of equal lengths. A block of mass m is suspended at point P such that BP is equal to x, If the fundamental frequency of the left wire is twice the fundamental frequency of right wire, then the value of x is :- A. `(l)/(5)`B. `(l)/(4)`C. `(4l)/(5)`D. `(3 l)/(4)` |
|
Answer» Correct Answer - A `f propv propsqrt(T)` `f_(AB)=2f_(CD)` `:. T_(AB)=4T_(CD)` Further `Sigma tau_(p)=0` `:. F_(AB)(X)=T_(CD)(l-x)` `rArr 4X=l-X ` (as `T_(AB)=4T_(CD)`) `rArr X = l//5` |
|
| 16. |
A string of length 1 m fixed one end and on the other end a block of mass `M = 4 kg` is suspended. The string is set into vibration and represented by equation `y =6 sin ((pi x)/(10)) cos (100 pi t)` where x and y are in cm and t is in seconds. (i) Find the number of loops formed in the string. (ii) Find the maximum displacement of a point at `x = 5//3 cm` (iii) Calculate the maximum kinetic energy of the string. (iv) Write down the equations of the component waves whose superposition given the wave. . |
|
Answer» Correct Answer - (i) 10 (ii) 3 cm (iii) 36 J (iv) `y_(1)=3sin((pix)/(10)-100pit),y_(2)=3sin((pix)/(10)+100pit)` `(2pi)/(lamda)=(pix)/(10)rArr lamda=20cm` (i) Total no of wavelength `=(L)/(lamda)=(100)/(20)=5` `:.` Number of loops formed `=2xx 5=10` (ii) Maximum displacement at `x = (5)/(3)` `A=6sin((pi)/(10)xx(5)/(3))=3cm` (iii) `y=6sin((pix)/(10))cos(100pit)` `=6sin(10pix)cos(100pit)` `v=6pisin(10pix)sin (100 pit)m//s` `:.KE_(max)=underset(0)overset(1)(int)(1)/(2)muv^(2)dx=(1)/(2)mu underset(0)overset(4)(int)6pisin(10pix)]^(2) dx` where `sqrt((T)/(mu))=(100pi)/(10pi)=10=mu=0.4:. KE_(max)=36J` (iv) `y=6 sin((pix)/(10))cos(100pit)=y_(1)+y_(2)` `rArr y_(1)=3sin((pix)/(10)-100t),y_(2)=3sin((pix)/(10)+100 pit)`. |
|
| 17. |
A sonometer wire of length `1.5m` is made of steel. The tension in it produces an elastic strain of `1%`. What is the fundamental frequency of steel if density and elasticity of steel are `7.7 xx 10^(3) kg//m^(3)` and `2.2 xx 10^(11) N//m^(2)` respectively ?A. 188.5 HzB. 178.2 HzC. 200.5 HzD. 770 Hz |
|
Answer» Correct Answer - B Fundamental frequency `f=(V)/(2l)=(1)/(2xx1.5)sqrt((T)/(eq))=(1)/(3)sqrt((yxx"strain" xxS)/(rho S))` (S `rarr` cross - section Area) `= =(1)/(3)sqrt((2.2xx10^(11)xx(1)/(100))/(7.7xx10^(3)))=178.2Hz` |
|
| 18. |
A whistle giving out `450 H_(Z)` approaches a stationary observer at a speed of `33 m//s`. The frequency heard the observer (in `H_(Z)`) is (speed of sound `= 330 m//s`)A. 409B. 429C. 517D. 500 |
|
Answer» Correct Answer - D `f=f_(o)((v)/(v_(obs)))` `=450((330)/(330-33))=500Hz` |
|
| 19. |
A progressive wave on a string having linear mass density `rho` is represented by `y=A sin((2 pi)/(lamda)x-omegat)` where `y` is in mm. Find the total energy (in `mu J`) passing through origin from `t=0` to `t=(pi)/(2 omega)`. [Take : `rho = 3 xx 10^(-2) kg//m , A = 1mm , omega = 100 rad..sec , lamda = 16 cm`].A. 6B. 7C. 8D. 9 |
|
Answer» Correct Answer - A Total energy `(1)/(2) rho A^(2) omega^(2)xx(lamda)/(4)`. |
|
| 20. |
The figure represents the instantaneous picture of a transverse wave travelling along the negative x-axis. Choose the correct alternative (s) related to the movement of the 9 points shown in the figure. (Instantaneous velocity) The points moving upward is/are :-A. aB. cC. fD. g |
|
Answer» Correct Answer - A::D `v_(uarr)` = Point a |
|
| 21. |
The figure represents the instantaneous picture of a transverse wave travelling along the negative x-axis. Choose the correct alternative (s) related to the movement of the 9 points shown in the figure. (Instantaneous velocity) A perfectly elastic uniform string is suspended vertically with its upper end fixed to the ceiling and the lower end loaded with the weight. If a transverse wave is imparted to the lower end of the string, the pulse willA. not travel along the length of the stringB. travel upwards with increasing speedC. travel upwards with decreasing speedD. travelled upwards with constant acceleration |
|
Answer» Correct Answer - B::D `v=sqrt((T)/(mu))=sqrt((w+muxg)/(mu))=sqrt((w)/(mu)+xg)` `rArr v^(2)=(w)/(mu)+xgrArr2v(dv)/(dx)=g rArr a=(g)/(2)` |
|
| 22. |
The figure represents the instantaneous picture of a transverse wave travelling along the negative x-axis. Choose the correct alternative (s) related to the movement of the 9 points shown in the figure. (Instantaneous velocity) The points moving with maximum velocity is/are :-A. bB. cC. dD. h |
|
Answer» Correct Answer - C::D `v_(max)`= points 0,d,h |
|
| 23. |
A transverse wave travelling along the positive x-axis, given by `y=A sin (kx - omega t)` is superposed with another wave travelling along the negative x-axis given by `y=A sin (kx + omega t)`. The point `x=0` isA. a nodeB. an antinodeC. neither a node nor an antinodeD. a node or antinode depending on `t` |
|
Answer» Correct Answer - B At `x=0,y_(1) = A sin(-omega t)` and `y_(2) = -A sin omega t , y_(1)+y_(2) = -2A sin omega t` (antinode). |
|
| 24. |
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) : What is the speed of the sound :-A. 200 m/sB. 180 m/sC. 192 m/sD. 96 m/s |
|
Answer» Correct Answer - A Speed of wave `v=(omega)/(R)=v=(100pi)/(0.5 pi)rArr(92pi)/(0.46pi)=200 m//s` |
|
| 25. |
Statement-1 : A balloon filled with `CO_(2)` gas acts as a coverging lens for a sound wave Statement-2 : Sound waves travel faster in air than in `CO_(2)`A. Statement-1 is true, Statement-2 is true , Statement-2 is correct explanation for Statement-16B. Statement-1 is true,Statement-2 is true , Statement-2 is NOT a correct explanation for Statement-16C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true |
| Answer» Correct Answer - A | |
| 26. |
A boat at anchor is rocked by waves whose crests are `100 m` apart and whose speed is `25 m//s`. These waves reach the boat once every :A. 2500 sB. 75 sC. 4 sD. 0.25 s |
|
Answer» Correct Answer - C `lamda=100 m,v=25m//s,T=(1)/(f)=(lamda)/(v)=(100)/(25)=4s` |
|
| 27. |
The equation of a wave is `y=(x,t)=0.05 sin [(pi)/(2)(10x-40t)-(pi)/(4)]m` find: (a) the wavelength, the frequency and the wave velocity (b) the participle velocity and acceleration at `x=0.5m`and `t = 0.05s`. |
|
Answer» (a) The equation may be rewritten as, `y(x,t) = 0.05 sin (5 pi x - 20 pi t - (pi)/(4))m` Comparing this with equation of plane progressive harmonic wave, `y(x,t)=A sin(kx-omegat+phi)` we have number `k = (2pi)/(lamda) = 5 pi rad//m :. Lamda = 0.4 m` The angular frequency is, `omega=2pif=20 pirad//s :. f =1 0 Hz` The wave velocity is, `v = f lamda = (omega)/(k) = 4 ms^(-1)` in -x direction (b) The particle velocity and acceleration are, `V_(p)=(dely)/(delt)=-(20pi)(0.05)cos((5pi)/(2)-pi-(pi)/(4))=2.22 m//s` `a_(p)=(del^(2)y)/(delt^(2))==(20pi)^(2)(0.05)sin((5pi)/(2)-pi-(pi)/(4))=140 m//s^(2)` |
|
| 28. |
A piezoelectric quartz plate of thickness 5mm is vibrating is resonance. Calculate its fundamental frequency if for quartz. `Y=8xx10^(10)N//m^(2)` and `rho=2.65xx10^(3)kg//m^(3)`. |
|
Answer» We known that for longitudinal waves in solids `v =sqrt((Y)/(rho))` so `v=sqrt((8xx10^(10))/(265xx10^(3)))=5.5 xx10^(3)m//s` Further more for fundamental mode of plate `-(lamda//2)=L` So `lamda=2xx5xx10^(-3)=10^(-2)m` But as `v = f lamda`, i.e.,`f=(v//lamda)` So `f=[5.5xx10^(3)//10^(-2)]=5.5xx10^(5)Hz=550kHz`. |
|
| 29. |
(a) Speed of sound in air 332 m/s at NTP. What will the speed of sound in hydrogen at NTP if the density of hydrogen at NTP is (1/16) that of air. (b) Calculate the ratio of the speed of sound in neon to that in water vapour any temperature. [Molecular weight if neon `= 2,02 xx 10^(-2) kg//mol` and for water vapours `= 1.8 xx 10^(-2) kg//mol` ] |
|
Answer» The velocity of sound in air is given by `= sqrt((E)/(rho))=sqrt((gamma P)/(rho))=sqrt((gamma RT)/(M))` (a) In terms density and pressure `(V_(H))/(V_("air"))=sqrt((P_(H))/(rho_(H))xx(rho_("air"))/(P_("air")))=sqrt((rho_("air"))/(rho_(H)))["as" P_("air")=P_(H)] rArr V_(H)=V_("air")xxsqrt((rho_("air"))/(rho_(H)))=332xxsqrt((16)/(1))=1328 m//s` (b) In terms of temperature and molecular weight `(V_(Ne))/(V_(W))=sqrt((gamma_(Ne))/(M_(Ne))xx(M_(W))/(gamma_(W)))["as" T_(N)=T_(W)]` Now as neon is mono atomic `(gamma = 5//3)` while water vapours poly atomic `(gamma = 4//3)` so `(V_(Ne))/(V_(W))=sqrt(((5//3)xx1.8xx10^(-2))/((4//3)xx2.02xx10^(-2)))=sqrt((5)/(4)xx(1.8)/(2.02))=1.055` |
|
| 30. |
A wave disturbance in a medium is described by `y(x, t) = 0.02 cos(50pit + (pi)/(2))cos(10pix)` where `x and y` are in meter and `t` is in second`A. A node occurs at x = 0.15 mB. An antinode occurs at `x -0.3 m`C. The speed of wave is `5 ms^(-1)`D. The wavelength is `0.2 m` |
|
Answer» Correct Answer - A::B::C::D `(2pi)/(lamda)=10pi rArrlamda=0.2m` Node occurs at `x=(lamda)/(4),(3lamda)/(4), (5lamda)/(4)=0.05m` `0.15 m` Antinode occurs at `x=(lamda)/(2),lamda,3(lamda)/(2)`....= `0.1 m,0.2,0.3m` `v = (50 pi)/(10 pi) = 5 m//s` |
|
| 31. |
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 :-A. 10 HzB. 5 HzC. 4 HzD. 2 Hz |
|
Answer» Correct Answer - B `L=5(lamda)/(2)rArr 10=5(lamda)/(2)rArr lamda=4m:. f=(v)/(lamda)=(20)/(4)`. |
|
| 32. |
Calculate the ratio of intensity of wave train A to wave train B. |
|
Answer» Correct Answer - `1` `I prop ((A)/(T))^(2)rArr (I_(1))/(I_(2))=((A_(1))/(A_(2)))^(2)((T_(2))/(T_(1)))^(2)=((2)/(1))^(2)xx((1)/(2))^(2)=1` |
|
| 33. |
In an experiment to measure the focal length of a concave mirror. If was found that for an object distance of 0.30 m, the image distance come out to be 0.60 m. Let us determine the focal length. |
|
Answer» By mirror formula, `{:(,,u=-0.30m),((1)/(v)+(1)/(u)=(1)/(f), rArr,),(,,v=-0.60m):}` `rArr (1)/(f)=(-1)/(0.30)-(1)/(0.60)rArr (1)/(f)=(-3.0)/(0.60)rArrf =0.20m` As `(1)/(f)=(1)/(v)+(1)/(u)rArr(-df)/(f^(2))=(-dv)/(v^(2))-(du)/(u^(2))rArrdf=(0.20)^(2)[(0.01)/((0.60)^(2))+(0.01)/((0.30)^(2))]rArrdf=0.0055~~ 0.01 m` `:.` Focal length `f=(0.20 pm0.01)m`. |
|
| 34. |
Calculate the change in intensity level when the intensity of sound increases by `10^(6)` times its original intensity. |
|
Answer» Here `= (I)/(I_(0))=("Final intensity")/("Initial intensity")=10^(6)` Increase in intensity level, `L=10 log_(10)((I)/(I_(0)))=10 log_(10)(10^(6))=10xxlog_(10)10=10xx6xx1=60` decibels. |
|
| 35. |
The length , breadth , and thickness of a metal sheet are `4.234 m , 1.005 m , and 2.01 cm`, respectively. Give the area and volume of the sheet to the correct number of significant figures. |
|
Answer» Length `(l) = 4.324 m` Breadth `(b) = 1.005 m` Thickness `(t) = 2.01 cm = 2.01 xx 10^(4) m` Therefore are of the sheet `= 2(lxxb+bxxt+txxl)` `=(4.25517+0.0202005+0.0851034)` `=2(4.255+0.0202+0.0851)` `=2(4.360)=8.7206 = 8.73)` Since area can contain a maximum of 3SF (Rule II article 2) therefore, rounding off. we get : Area `= 8.72 m^(2)` Like wise volume `= lxxbxxt=2.234xx1.005xx0.0201 m^(3)=0.0855289 m^(2)` Since volume can contan 3SF, therefore, rounding off, we get : volume `= 0.0855 m^(3)` |
|
| 36. |
The waves produced by a motor boat sailing in water are :A. TransverseB. LongitudinalC. Longitudinal and transverseD. Stationary |
|
Answer» Correct Answer - C Waves on surface of water are combination of longitudinal and transverse waves. |
|
| 37. |
Find the thickness of the wire. |
|
Answer» Excess reading (zero error) = 0.03 mm It is giving 7.67 mm in which there is 0.03 mm excess reading, which has to be removed (subracted) so actual reading = 7.67 - 0.03 = 7.64 mm. |
|
| 38. |
A wave of frequency 500 HZ travels between X and Y, a distance of 600 m in 2 sec. How many wavelength are there in distance XY :-A. 1000B. 300C. 180D. 2000 |
|
Answer» Correct Answer - A `lamda =(v)/(f)=(600//2)/(500)=(3)/(5)m` `:.` Number of waves `=(600)/(lamda)=(600xx5)/(3)=1000`. |
|
| 39. |
A source of sound of frequency 600 Hz is placed inside water. The speed of sound in water is `1500 m//s` and in air it is `300 m//s`. The frequency of sounds recorded by an observer who is standing in air is :-A. 200 HzB. 3000 HzC. 120 HzD. 600 Hz |
|
Answer» Correct Answer - D The frequency us a characteristic of source. It is independent of the medium. |
|
| 40. |
Statement-1 : The fundamental frequency of an organ pipe increases as the temperature increases Statement-1 : If transverse waves are produced in a very long string fixed at one end. Near the free end only progressive wave is observed, in parctice Statement-2 : As the temperature increases, the velocity of sound increases more rapidly than length of the pipe.A. Statement-1 is true, Statement-2 is true , Statement-2 is correct explanation for Statement-20B. Statement-1 is true,Statement-2 is true , Statement-2 is NOT a correct explanation for Statement-20C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true |
| Answer» Correct Answer - A | |
| 41. |
Statement-1 : Partially transwave waves are possible on a liquid surface Statement-2 : Surface tension provide some rigidity on a liquid surfaceA. Statement-1 is true, Statement-2 is true , Statement-2 is correct explanation for Statement-6B. Statement-1 is true,Statement-2 is true , Statement-2 is NOT a correct explanation for Statement-6C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true |
| Answer» Correct Answer - A | |
| 42. |
Statement-1 : When two vibrating tuning forks have `f_(1) = 300 Hz` and `f_(2) = 350 Hz` and held close to each other , beats cannot be heard Statement-1 : If transverse waves are produced in a very long string fixed at one end. Near the free end only progressive wave is observed, in parctice Statement-2 : The principle of superposition is valid only when `f_(1) - f_(2) lt 10 Hz`A. Statement-1 is true, Statement-2 is true , Statement-2 is correct explanation for Statement-19B. Statement-1 is true,Statement-2 is true , Statement-2 is NOT a correct explanation for Statement-19C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true |
| Answer» Correct Answer - C | |
| 43. |
Statement-1 : In frasonic waves are generally produced by large vibrating bodies Statement-2 : Infrasonic waves have frequency range lies below 20 HzA. Statement-1 is true, Statement-2 is true , Statement-2 is correct explanation for Statement-5B. Statement-1 is true,Statement-2 is true , Statement-2 is NOT a correct explanation for Statement-5C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true |
| Answer» Correct Answer - A | |
| 44. |
A tube, closed at one end and containing air, produces, when excited, the fundamental note of frequency `512 Hz`. If the tube is open at both ands the fundamental frequency that can be excited is (in Hz)A. 1024B. 512C. 256D. 128 |
|
Answer» Correct Answer - A `v=f_(1) lamda_(1)=f_(2)lamda_(2)rArr 512xx4L=f_(2)xx2L` `rArr f_(2)=1024Hz` |
|
| 45. |
A progressive wave having amplitude `5 m` and wavelength `3m`. If the maximum average velocity of particle in half time period is `5 m//s` and waves is moving in the positive x-direction then find which may be the correct equation (s) of the wave ? [wave `x` in meter].A. `5 sin ((2pi)/(5)t-(2pi)/(3)x)`B. `4 sin((pit)/(2)-(2pi)/(3))+3 cos ((pi t)/(2)-(2pi)/(3)x)`C. `5 sin ((pi t)/(2)-(2pi)/(3)x)`D. `3 cos((2pi)/(5)t-(2pi)/(3)x)-4 sin ((2pi)/(5)t-(2pi)/(3)x)` |
|
Answer» Correct Answer - B::C `because lamda = 3 m :. k=(2pi)/(lamda)=(2pi)/(3)` Maximum displacement in half time period `= 2a = 10 m` So maximum average velocity `(10)/(.^(T)//_(2))=5 rArr T=4s rArr omega = (2pi)/(T)=(2pi)/(4)=(pi)/(2)`. |
|
| 46. |
Three progressive waves `A,B` and `C` are shown in figure. With respect to wave `A` .A. The wave `C` lags behind in phase by `pi//2` and `B` leads by `pi//2`B. The wave `C` leads in phase by `pi` and `B` lags behind by `pi`C. The wave `C` leads in phase by `pi//2` and `B` lags behind by `pi//2`D. The wave `C` lags behind in phase by `pi` and `B` leads by `pi` |
| Answer» Correct Answer - A | |
| 47. |
A source of sound emits sound waves at frequency `f_(0)`. It is moving towards an observer with fixed speed `v_(s)(v_(s) lt v` where v is the speed in air). If the observer were to move towards the source with speed `v_(0)` one of the following two graphs (A and B) will give the correct variation of the frequency f heard by the observer as `v_(0)` is changed, The variation of f with `v_(0)` is given correctly by :-A. graph A with slope `= (f_(0))/((v+v_(s))`B. graph B with slope `= (f_(0))/((v-v_(s))`C. graph B with slope `= (f_(0))/((v+v_(s))`D. graph A with slope `= (f_(0))/((v-v_(s))` |
|
Answer» Correct Answer - D Apprent frequency `f=f_(0)((V+V_(0))/(V-V_(S)))=(f_(0)V)/(V-V_(S))+(f_(0)V_(0))/((V-V_(S)))` |
|
| 48. |
For a sine wave passing through a medium, let y be the displacement of a particle, v be its velocity and a be its acceleration :-A. y,v and a are always in the same phaseB. y and a are always in opposite phaseC. Phase difference between y and v is `(pi)/(2)`D. Phase difference between v and a is `(pi)/(2)` |
| Answer» Correct Answer - B::C::D | |
| 49. |
Two wave are represented by equation `y_(1) = a sin omega t` and `y_(2) = a cos omega t` the first wave :-A. leads the second by `pi`B. lags the second by `pi`C. leads the second by `(pi)/(2)`D. lags the second by `(pi)/(2)` |
|
Answer» Correct Answer - D `y_(1) = a sin omega t` `y_(2) = a cos omega t = a sin(omega t + pi//2)` `y_(1)` lags `y_(2)` behind by phase `(pi)/(2)` |
|
| 50. |
A 100 MHz xarrier wave frequency is modulated by 10 kHz sinusoidal modulating signal. If the maximum frequency deviation is 50 kHz, find the modulation index. |
| Answer» `m_(f)=(Deltaf)/(f_(m))=(50 kHz)/(10kHz)=5` | |