MCQ 511 Mark
What will be the wave velocity, if the radar gives 54 waves per min and wavelength of the given wave is $10 \mathrm{~m}$
- A
$4 \mathrm{~m} / \mathrm{sec}$
- B
$6 \mathrm{~m} / \mathrm{sec}$
- ✓
$9 \mathrm{~m} / \mathrm{sec}$
- D
$5 \mathrm{~m} / \mathrm{sec}$
AnswerCorrect option: C. $9 \mathrm{~m} / \mathrm{sec}$
(c) $n=\frac{54}{60} H z, \lambda=10 m \Rightarrow v=n \lambda=9 \mathrm{~m} / \mathrm{s}$.
View full question & answer→MCQ 521 Mark
A man standing on a cliff claps his hand hears its echo after $1 \mathrm{sec}$. If sound is reflected from another mountain and velocity of sound in air is $340 \mathrm{~m} / \mathrm{sec}$. Then the distance between the man and reflection point is
- A
$680 \mathrm{~m}$
- B
$340 \mathrm{~m}$
- C
$85 \mathrm{~m}$
- ✓
$170 \mathrm{~m}$
AnswerCorrect option: D. $170 \mathrm{~m}$
(d) If $d$ is the distance between man and reflecting surface of sound then for hearing echo$2 d=v \times t \Rightarrow d=\frac{340 \times 1}{2}=170 \mathrm{~m}$
View full question & answer→MCQ 531 Mark
A string is producing transverse vibration whose equation is $y=0.021 \sin (x+30 t)$, Where $x$ and $y$ are in meters and $t$ is in seconds. If the linear density of the string is $1.3 \times 10^{-4} \mathrm{~kg} / \mathrm{m}$, then the tension in the string in $N$ will be
AnswerCorrect option: D. $0.117$
$y=0.021 \sin (x+30 t) \Rightarrow v=\frac{\omega}{k}=\frac{30}{1}=30 \mathrm{~m} / \mathrm{s}$.
Using, $v=\sqrt{\frac{T}{m}} \Rightarrow 30=\sqrt{\frac{T}{1.3 \times 10^{-4}}} \Rightarrow T=0.117 \mathrm{~N}$
View full question & answer→MCQ 541 Mark
A stretched string of $1 \mathrm{~m}$ length and mass $5 \times 10^{-4} \mathrm{~kg}$ is having tension of $20 \mathrm{~N}$. If it is plucked at $25 \mathrm{~cm}$ from one end then it will vibrate with frequency
- A
$100 \mathrm{~Hz}$
- ✓
$200 \mathrm{~Hz}$
- C
$256 \mathrm{~Hz}$
- D
$400 \mathrm{~Hz}$
AnswerCorrect option: B. $200 \mathrm{~Hz}$
As we know plucking distance from one end
$=\frac{l}{2 p}$$\Rightarrow 25=\frac{100}{2 p} \Rightarrow p=2$.
Hence frequency of vibration$n=\frac{p}{2 l} \sqrt{\frac{T}{m}}=\frac{2}{2 \times 1} \sqrt{\frac{20}{5 \times 10^{-4}}}=200 \mathrm{~Hz} .$
View full question & answer→MCQ 551 Mark
A cylindrical tube, open at both ends, has a fundamental frequency $f_0$ in air. The tube is dipped vertically into water such that half of its length is inside water. The fundamental frequency of the air column now is
- A
$3 f_0 / 4$
- ✓
$f_0$
- C
$f_0 / 2$
- D
$2 f_0$
Answer(b)$n_{\text {open }}=\frac{v}{2 l_{\text {open }}}$
$n_{\text {closed }}=\frac{v}{4 l_{\text {closed }}}=\frac{v}{4 l_{\text {open }} / 2}=\frac{v}{2 l_{\text {open}}} $
$\left(\text { As } l_{\text {closed }}=\frac{l_{\text {open }}}{2}\right), $
$\text { i.e. frequency remains unchanged. }$
View full question & answer→MCQ 561 Mark
Two similar sonometer wires given fundamental frequencies of $500 \mathrm{~Hz}$. These have same tensions. By what amount the tension be increased in one wire so that the two wires produce $5$ beats/sec
- A
$1 \%$
- ✓
$2 \%$
- C
$3 \%$
- D
$4 \%$
AnswerCorrect option: B. $2 \%$
To produce $5$ beats $/ \mathrm{sec}$.
Frequency of one wire should be increase up to $505 \mathrm{~Hz}$.
i.e. increment of $1 \%$ in basic frequency.
$n \propto \sqrt{T}$ or $T \propto n^2 \Rightarrow \frac{\Delta T}{T}=2 \frac{\Delta n}{n}$$(\Rightarrow$percentage change in Tension $=2(1 \%)=2 \%)$
View full question & answer→MCQ 571 Mark
Which two of the given transverse waves will give stationary waves when get superimposed $z_1=a \cos (k x-\omega t) , z_2=a \cos (k x+\omega t) , z_3=a \cos (k y-\omega t)$
AnswerCorrect option: A. $A$ and $B$
Waves $A$ and $B$ satisfied the conditions required for a standing wave.
View full question & answer→MCQ 581 Mark
An observer standing at station observes frequency $219 \mathrm{~Hz}$ when a train approaches and $184 \mathrm{~Hz}$ when train goes away from him. If velocity of sound in air is $340 \mathrm{~m} / \mathrm{s}$, then velocity of train and actual frequency of whistle will be
- A
$15.5 \mathrm{~ms}^{-1}, 200 \mathrm{~Hz}$
- B
$19.5 \mathrm{~ms}^{-1}, 205 \mathrm{~Hz}$
- ✓
$29.5 \mathrm{~ms}^{-1}, 200 \mathrm{~Hz}$
- D
$32.5 \mathrm{~ms}^{-1}, 205 \mathrm{~Hz}$
AnswerCorrect option: C. $29.5 \mathrm{~ms}^{-1}, 200 \mathrm{~Hz}$
(c) When train is approaching frequency heard by the observer is$n_a=n\left(\frac{v}{v-v_S}\right) \Rightarrow 219=n\left(\frac{340}{340-v_S}\right)$when train is receding (goes away), frequency heard by the observer is$n_r=n\left(\frac{v}{v+v_S}\right) \Rightarrow 184=n\left(\frac{340}{340+v_S}\right)$On solving equation (i) and (ii) we get $n=200 \mathrm{~Hz}$ and $v_S=29.5 \mathrm{~m} / \mathrm{s}$.
View full question & answer→MCQ 591 Mark
Two cars are moving on two perpendicular roads towards a crossing with uniform speeds of $72 \mathrm{~km} / \mathrm{hr}$ and $36 \mathrm{~km} / \mathrm{hr}$. If first car blows horn of frequency $280 \mathrm{~Hz}$, then the frequency of horn heard by the driver of second car when line joining the cars make $45^{\circ}$ angle with the roads; will be
- A
$321 \mathrm{~Hz}$
- ✓
$298 \mathrm{~Hz}$
- C
$289 \mathrm{~Hz}$
- D
$280 \mathrm{~Hz}$
AnswerCorrect option: B. $298 \mathrm{~Hz}$
View full question & answer→MCQ 601 Mark
If the temperature increases, then what happens to the frequency of the sound produced by the organ pipe
Answer(a) Due to rise in temperature, the speed of sound increases. Since $n=\frac{v}{\lambda}$ and $\lambda$ remains unchanged, hence $n$ increases.
View full question & answer→MCQ 611 Mark
Two whistles $A$ and $B$ produces notes of frequencies $660 \mathrm{~Hz}$ and $596 \mathrm{~Hz}$ respectively. There is a listener at the mid-point of the line joining them. Now the whistle $B$ and the listener start moving with speed $30 \mathrm{~m} / \mathrm{s}$ away from the whistle A. If speed of sound be 330 $\mathrm{m} / \mathrm{s}$, how many beats will be heard by the listener
Answer(b) For observer note of $B$ will not change due to zero relative motion.Observed frequency of sound produced by $A$$=660 \frac{(330-30)}{330}=600 \mathrm{~Hz}$$\therefore$ No. of beats $=600-596=4$
View full question & answer→MCQ 621 Mark
A closed pipe and an open pipe have their first overtones identical in frequency. Their lengths are in the ratio
- A
$1: 2$
- B
$2: 3$
- ✓
$3: 4$
- D
$4: 5$
AnswerCorrect option: C. $3: 4$
If is given that First over tone of closed pipe $=$ First over tone of open pipe
$\Rightarrow 3\left(\frac{v}{4 l_1}\right)=2\left(\frac{v}{2 l_2}\right) ;$ where $l$ and $l$ are the lengths of closed and open organ pipes hence $\frac{l_1}{l_2}=\frac{3}{4}$
View full question & answer→MCQ 631 Mark
A wave is represented by the equation $y=0.5 \sin (10 t-x) m$. It is a travelling wave propagating along the $+x$ direction with velocity
- ✓
$10 \mathrm{~m} / \mathrm{s}$
- B
$20 \mathrm{~m} / \mathrm{s}$
- C
$5 \mathrm{~m} / \mathrm{s}$
- D
AnswerCorrect option: A. $10 \mathrm{~m} / \mathrm{s}$
(a) The velocity of wave$v=\frac{\omega(\text { Co }- \text { efficient of } t)}{k(\text { Co }- \text { efficient of } x)}=\frac{10}{1}=10 \mathrm{~m} / \mathrm{s}$
View full question & answer→MCQ 641 Mark
The stationary wave $y=2 a \sin k x \cos \omega t$ in a closed organ pipe is the result of the superposition of $y=a \sin (\omega t-k x)$ and
- A
$y=-a \cos (\omega t+k x)$
- ✓
$y=-a \sin (\omega t+k x)$
- C
$y=a \sin (\omega t+k x)$
- D
$y=a \cos (\omega t+k x)$
AnswerCorrect option: B. $y=-a \sin (\omega t+k x)$
(b) In closed organ pipe. If $y_{\text {incident }}=a \sin (\omega t-k x)$ then $y_{\text {reflected }}=a \sin (\omega t+k x+\pi)=-a \sin (\omega t+k x)$Superimposition of these two waves $v$ give $h$ the $d$ required stationary wave.Go to Settings to activate Windo
View full question & answer→MCQ 651 Mark
In a large room, a person receives direct sound waves from a source 120 metres away from him. He also receives waves from the same source which reach him, being reflected from the 25 metre high ceiling at a point halfway between them. The two waves interfere constructively for wavelength of
- ✓
$20,20 / 3,20 / 5$ etc
- B
$10,5,2.5$ etc
- C
$10,20,30 \mathrm{etc}$
- D
$15,25,35$ etc
AnswerCorrect option: A. $20,20 / 3,20 / 5$ etc
View full question & answer→MCQ 661 Mark
Two open organ pipes of length $25 \mathrm{~cm}$ and $25.5 \mathrm{~cm}$ produce 10 beat/sec. The velocity of sound will be
- ✓
$255 \mathrm{~m} / \mathrm{s}$
- B
$250 \mathrm{~m} / \mathrm{s}$
- C
$350 \mathrm{~m} / \mathrm{s}$
- D
AnswerCorrect option: A. $255 \mathrm{~m} / \mathrm{s}$
(a)$\Delta n=n_1-n_2$
$ \Rightarrow 10=\frac{v}{2 l_1}-\frac{v}{2 l_2}=\frac{v}{2}\left[\frac{1}{l_1}-\frac{1}{l_2}\right] $
$\Rightarrow 10=\frac{v}{2}\left[\frac{1}{0.25}-\frac{1}{0.255}\right]$
$ \Rightarrow v=255 \mathrm{~m} / \mathrm{s}$
View full question & answer→MCQ 671 Mark
Two closed organ pipes, when sounded simultaneously gave 4 beats per sec. If longer pipe has a length of $1 \mathrm{~m}$. Then length of shorter pipe will be, $(v=300 \mathrm{~m} / \mathrm{s})$
- A
$185.5 \mathrm{~cm}$
- ✓
$94.9 \mathrm{~cm}$
- C
$90 \mathrm{~cm}$
- D
$80 \mathrm{~cm}$
AnswerCorrect option: B. $94.9 \mathrm{~cm}$
(b) For first pipe $n_1=\frac{v}{4 l_1}$ and for second pipe $n_2=\frac{v}{4 l_2}$
So, number of beats $=n_2-n_1=4$
$\Rightarrow 4=\frac{v}{4}\left(\frac{1}{l_2}-\frac{1}{l_1}\right) $
$\Rightarrow 16=300(\frac{1}{l_2}-\frac{1}{1})$
$\Rightarrow l_2=94.9\mathrm{~cm}$
View full question & answer→MCQ 681 Mark
Velocity of sound in air is
- A
Faster in dry air than in moist air
- B
Directly proportional to pressure
- C
Directly proportional to temperature
- ✓
Independent of pressure of air
AnswerCorrect option: D. Independent of pressure of air
(d) $v=\sqrt{\frac{\gamma P}{\rho}}$; as $P$ changes, $\rho$ also changes. Hence $\frac{P}{\rho}$ remains constant so speed remains constant.
View full question & answer→MCQ 691 Mark
A plane progressive wave is represented by the equation $y=0.1 \sin \left(200 \pi t-\frac{20 \pi x}{17}\right)$ where $y$ is displacement in $m, t$ in second and $x$ is distance from a fixed origin in meter. The frequency, wavelength and speed of the wave respectively are
- ✓
$100 \mathrm{~Hz}, 1.7 \mathrm{~m}, 170 \mathrm{~m} / \mathrm{s}$
- B
$150 \mathrm{~Hz}, 2.4 \mathrm{~m}, 200 \mathrm{~m} / \mathrm{s}$
- C
$80 \mathrm{~Hz}, 1.1 \mathrm{~m}, 90 \mathrm{~m} / \mathrm{s}$
- D
$120 \mathrm{~Hz}, 1.25 \mathrm{~m}, 207 \mathrm{~m} / \mathrm{s}$
AnswerCorrect option: A. $100 \mathrm{~Hz}, 1.7 \mathrm{~m}, 170 \mathrm{~m} / \mathrm{s}$
Comparing the given equation with standard equation We get $\omega=2 \pi n=200 \pi$
$ \Rightarrow n=100 \mathrm{~Hz}$
$k=\frac{20 \pi}{17} $
$\Rightarrow \lambda=\frac{2 \pi}{k}=\frac{2 \pi}{20 \pi / 17}=1.7 \mathrm{~m}$ and $v=\frac{\omega}{k}=\frac{200 \pi}{20 \pi / 17}=170 \mathrm{~m} / \mathrm{s}$.
View full question & answer→MCQ 701 Mark
Two open organ pipes give 4 beats/sec when sounded together in their fundamental nodes. If the length of the pipe are $100 \mathrm{~cm}$ and $102.5 \mathrm{~cm}$ respectively, then the velocity of sound is :
- A
$496 \mathrm{~m} / \mathrm{s}$
- ✓
$328 \mathrm{~m} / \mathrm{s}$
- C
$240 \mathrm{~m} / \mathrm{s}$
- D
$160 \mathrm{~m} / \mathrm{s}$
AnswerCorrect option: B. $328 \mathrm{~m} / \mathrm{s}$
(b)}$\Delta n=n_1-n_2 $
$\Rightarrow 4=\frac{v}{2 l_1}-\frac{v}{2 l_2}=\frac{v}{2}\left[\frac{1}{1.00}-\frac{1}{1.025}\right] $
$\Rightarrow 8=[1-0.975]$
$ \Rightarrow v=\frac{8}{0.025} \approx 328 \mathrm{~m} / \mathrm{s} .$
View full question & answer→MCQ 711 Mark
The phase difference between two points separated by $0.8 \mathrm{~m}$ in a wave of frequency is $120 \mathrm{~Hz}$ is $\frac{\pi}{2}$. The velocity of wave is
- A
$720 \mathrm{~m} / \mathrm{s}$
- ✓
$384 \mathrm{~m} / \mathrm{s}$
- C
$250 \mathrm{~m} / \mathrm{s}$
- D
$1 \mathrm{~m} / \mathrm{s}$
AnswerCorrect option: B. $384 \mathrm{~m} / \mathrm{s}$
(b) Phase difference $=\frac{2 \pi}{\lambda} \times$ pathdifference
$\Rightarrow \frac{\pi}{2}=\frac{2 \pi}{\lambda} \times 0.8 $
$\Rightarrow \lambda=4\times0.8=3.2 \mathrm{~m} $
$\text { Velocity } v=n \lambda=120 \times 3.2=384 \mathrm{~m}\mathrm{s} .$
View full question & answer→MCQ 721 Mark
Sound velocity is maximum in
- ✓
$\mathrm{H}_2$
- B
$N_2$
- C
$\mathrm{He}$
- D
$\mathrm{O}_2$
AnswerCorrect option: A. $\mathrm{H}_2$
(a) $v=\sqrt{\frac{\gamma R T}{M}} \Rightarrow v \propto \frac{1}{\sqrt{M}}$. Since $M$ is minimum for $H_2$ so sound velocity is maximum in $H$.
View full question & answer→MCQ 731 Mark
If the equation of transverse wave is $Y=2 \sin (k x-2 t)$, then the maximum particle velocity is
- ✓
$4$ units
- B
$2$ units
- C
$8$ units
- D
$6$ units
AnswerCorrect option: A. $4$ units
Maximum particle velocity $=a \omega=2 \times 2=4$ units.
View full question & answer→MCQ 741 Mark
A source emits a sound of frequency of $400 \mathrm{~Hz}$, but the listener hears it to be $390 \mathrm{~Hz}$. Then
- A
The listener is moving towards the source
- B
The source is moving towards the listener
- ✓
The listener is moving away from the source
- D
The listener has a defective ear
AnswerCorrect option: C. The listener is moving away from the source
(c) Since apparent frequency is lesser than the actual frequency, hencetherelativeseparation between source and listener should be increasing.
View full question & answer→MCQ 751 Mark
It takes 2.0 seconds for a sound wave to travel between two fixed points when the day temperature is $10^{\circ} \mathrm{C}$. If the temperature rise to $30^{\circ} \mathrm{C}$ the sound wave travels between the same fixed parts in
- ✓
$1.9 \mathrm{sec}$
- B
$2.0 \mathrm{sec}$
- C
$2.1 \mathrm{sec}$
- D
$2.2 \mathrm{sec}$
AnswerCorrect option: A. $1.9 \mathrm{sec}$
(a) Suppose the distance between two fixed points is $d$ then
$t=\frac{d}{v} \text { also } v \propto \sqrt{T} $
$\Rightarrow \frac{t_1}{t_2}=\frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}} $
$\Rightarrow \frac{t_1}{t_2}=\sqrt{\frac{303}{283}}$
$\Rightarrow t_2=1.9\mathrm{sec}$
View full question & answer→MCQ 761 Mark
A source is moving towards an observer with a speed of $20 \mathrm{~m} / \mathrm{s}$ and having frequency of $240 \mathrm{~Hz}$. The observer is now moving towards the source with a speed of $20 \mathrm{~m} / \mathrm{s}$. Apparent frequency heard by observer, if velocity of sound is $340 \mathrm{~m} / \mathrm{s}$, is
- A
$240 \mathrm{~Hz}$
- B
$280 \mathrm{~Hz}$
- ✓
$270 \mathrm{~Hz}$
- D
$360 \mathrm{~Hz}$
AnswerCorrect option: C. $270 \mathrm{~Hz}$
(c) $n^{\prime}=n\left(\frac{v+v_O}{v-v_S}\right)=240\left(\frac{340+20}{340-20}\right)=270 \mathrm{~Hz}$.
View full question & answer→MCQ 771 Mark
Stationary waves are formed when
- A
Two waves of equal amplitude and equal frequency travel along the same path in opposite directions
- ✓
Two waves of equal wavelength and equal amplitude travel along the same path with equal speeds in opposite directions
- C
Two waves of equal wavelength and equal phase travel along the same path with equal speed
- D
Two waves of equal amplitude and equal speed travel along the same path in opposite direction
AnswerCorrect option: B. Two waves of equal wavelength and equal amplitude travel along the same path with equal speeds in opposite directions
View full question & answer→MCQ 781 Mark
The following phenomenon cannot be observed for sound waves
Answer(d) Sound waves are longitudinal in nature so they can not be polarised
View full question & answer→MCQ 791 Mark
$A$ is singing a note and at the same time $B$ is singing a note with exactly one $-$ eighth the frequency of the note of $A$. The energies of two sounds are equal, the amplitude of the note of $B$ is
AnswerCorrect option: D. Eight times as that of $A$
$\frac{t_1}{t_2}$ Energy $ \propto a^2 n^2 $
$\Rightarrow \frac{a_B}{a_A}=\frac{n_A}{n_B} \quad(\because $ energy is same$)$
$ \Rightarrow\frac{a_B}{a_A}=\frac{8}{1}$
View full question & answer→MCQ 801 Mark
A wave of frequency $500 \mathrm{~Hz}$ has velocity $360 \mathrm{~m} / \mathrm{sec}$. The distance between two nearest points $60^{\circ}$ out of phase, is
- A
$0.6 \mathrm{~cm}$
- ✓
$12 \mathrm{~cm}$
- C
$60 \mathrm{~cm}$
- D
$120 \mathrm{~cm}$
AnswerCorrect option: B. $12 \mathrm{~cm}$
The distance between two points i.e. path difference between
$\text {them } \Delta=\frac{\lambda}{2 \pi} \times \phi $
$=\frac{\lambda}{2\pi} \times\frac{\pi}{3}$
$=\frac{\lambda}{6}=\frac{v}{6 n}(\because v=n \lambda) $
$\Rightarrow \Delta=\frac{360}{6 \times 500}=0.12 \mathrm{~m}=12 \mathrm{~cm}$
View full question & answer→MCQ 811 Mark
The walls of the halls built for music concerts should
MCQ 821 Mark
Which of the following statements is wrong
- A
Sound travels in straight line
- B
Sound is a form of energy
- C
Sound travels in the form of waves
- ✓
Sound travels faster in vacuum than in air
AnswerCorrect option: D. Sound travels faster in vacuum than in air
(d) Air is more rarer for sound to travel as compared to vacuum.
View full question & answer→MCQ 831 Mark
A tuning fork sounded together with a tuning fork of frequency $256$ emits two beats. On loading the tuning fork of frequency $256,$ the number of beats heard are $1$ per second. The frequency of tuning fork is
Answer$n_1=$ Known frequency $=256, n_{=}=?$$x=2$ bps, which is decreasing after loading (i.e. $x \downarrow$ ) known tuning fork is loaded so $n \downarrow$
Hence $n\downarrow-n=x \downarrow$... (i)$n-n \downarrow=x \downarrow$Gorrect
${cnn}\downarrow=x \downarrow \quad \text {... (ii) }$
$\Rightarrow n_x=n_{-}x=256-2=254\mathrm{~Hz}$
View full question & answer→MCQ 841 Mark
When an aeroplane attains a speed higher than the velocity of sound in air, a loud bang is heard. This is because
- A
- ✓
It produces a shock wave which is received as the bang
- C
Its wings vibrate so violently that the bang is heard
- D
The normal engine noises undergo a Doppler shift to generate the bang
AnswerCorrect option: B. It produces a shock wave which is received as the bang
View full question & answer→MCQ 851 Mark
Sound travels in rocks in the form of
- A
Longitudinal elastic waves only
- B
Transverse elastic waves only
- ✓
Both longitudinal and transverse elastic waves
- D
AnswerCorrect option: C. Both longitudinal and transverse elastic waves
View full question & answer→MCQ 861 Mark
Two tuning forks have frequencies $380$ and $384 \mathrm{~Hz}$ respectively. When they are sounded together, they produce $4$ beats. After hearing the maximum sound, how long will it take to hear the minimum sound
- A
$\frac{1}{2} \mathrm{sec}$
- B
$\frac{1}{4} \mathrm{sec}$
- ✓
$\frac{1}{8} \mathrm{sec}$
- D
$\frac{1}{16} \mathrm{sec}$
AnswerCorrect option: C. $\frac{1}{8} \mathrm{sec}$
Beat period $T=\frac{1}{n_1 \sim n_2}=\frac{1}{384-380}=\frac{1}{4} \mathrm{sec}$.
Hence minimum time interval between maxima and minima $t=\frac{T}{2}=\frac{1}{8}\mathrm{sec}$.
View full question & answer→MCQ 871 Mark
The frequency of fundamental tone in an open organ pipe of length $0.48 \mathrm{~m}$ is $320 \mathrm{~Hz}$. Speed of sound is $320 \mathrm{~m} / \mathrm{sec}$. Frequency of fundamental tone in closed organ pipe will be
- A
$153.8 \mathrm{~Hz}$
- ✓
$160.0 \mathrm{~Hz}$
- C
$320.0 \mathrm{~Hz}$
- D
$143.2 \mathrm{~Hz}$
AnswerCorrect option: B. $160.0 \mathrm{~Hz}$
(b) $n_{\text {Closed }}=\frac{1}{2}\left(n_{\text {Open }}\right)=\frac{1}{2} \times 320=160 \mathrm{~Hz}$
View full question & answer→MCQ 881 Mark
When an engine passes near to a stationary observer then its apparent frequencies occurs in the ratio $5/3.$ If the velocity of engine is
- A
$540 \mathrm{~m} / \mathrm{s}$
- B
$270 \mathrm{~m} / \mathrm{s}$
- ✓
$85 \mathrm{~m} / \mathrm{s}$
- D
$52.5 \mathrm{~m} / \mathrm{s}$
AnswerCorrect option: C. $85 \mathrm{~m} / \mathrm{s}$
When engine approaches towards observer $n^{\prime}=n\left(\frac{v}{v-v_S}\right)$
when engine going away from observer $n^{\prime \prime}=\left(\frac{v}{v+v_S}\right) n$
$[\therefore \frac{n^{\prime}}{n^{\prime \prime}}=\frac{v+v_S}{v-v_S} \Rightarrow \frac{5}{3}=\frac{340+v_S}{340-v_S} \Rightarrow v_S=85 \mathrm{~m} / \mathrm{s} .$
View full question & answer→MCQ 891 Mark
Equation of a stationary wave is $y=10 \sin \frac{\pi x}{4} \cos 20 \pi t$. Distance between two consecutive nodes is
AnswerOn comparing the given equation with standard equation$\frac{2 \pi}{\lambda}=\frac{\pi}{4} \Rightarrow \lambda=8$
Hence distance between two consecutive nodes $\frac{\lambda}{2}=4$
View full question & answer→MCQ 901 Mark
What should be the velocity of a sound source moving towards a stationary observer so that apparent frequency is double the actual frequency (Velocity of sound is $v$ )
- A
$v$
- B
$2 v$
- ✓
$\frac{v}{2}$
- D
$\frac{v}{4}$
AnswerCorrect option: C. $\frac{v}{2}$
(c) By using $n^{\prime}=\left(\frac{v}{v-v_S}\right) \Rightarrow 2 n=n\left(\frac{v}{v-v_S}\right) \Rightarrow v_S=\frac{v}{2}$
View full question & answer→MCQ 911 Mark
A transverse wave is given by $y=A \sin 2 \pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)$. The maximum particle velocity is equal to 4 times the wave velocity when
- A
$100 \mathrm{~m} / \mathrm{s}$
- B
$250 \mathrm{~m} / \mathrm{s}$
- C
$750 \mathrm{~m} / \mathrm{s}$
- ✓
$1000 \mathrm{~m} / \mathrm{s}$
AnswerCorrect option: D. $1000 \mathrm{~m} / \mathrm{s}$
(d) $\quad v=\frac{\omega}{k}=\frac{100}{1 / 10}=1000 \mathrm{~m} / \mathrm{s}$
View full question & answer→MCQ 921 Mark
"Stationary waves" are so called because in them
- A
The particles of the medium are not disturbed at all
- B
The particles of the medium do not execute SHM
- ✓
There occurs no flow of energy along the wave
- D
The interference effect can't be observed
AnswerCorrect option: C. There occurs no flow of energy along the wave
View full question & answer→MCQ 931 Mark
A simple harmonic progressive wave is represented by the equation : $y=8 \sin 2 \pi(0.1 x-2 t)$ where $x$ and $y$ are in $\mathrm{cm}$ and $t$ is in seconds. At any instant the phase difference between two particles separated by $2.0 \mathrm{~cm}$ in the $x-$ direction is
Answer$(d)$ From the given equation $k=0.2 \pi$
$\Rightarrow \frac{2 \pi}{\lambda}=0.2 \pi \Rightarrow \lambda=10 \mathrm{~cm} $
$\Delta \phi=\frac{2 \pi}{\lambda} \Delta x=\frac{2 \pi}{10} \times 2=\frac{2 \pi}{5}=72^{\circ} \mathrm{a}, \mathrm{b},\mathrm{c}) $
$I=2 \pi n^2 a^2 \rho v \Rightarrow I \propto n^2 a^2 v$
View full question & answer→MCQ 941 Mark
A source of sound of frequency 90 vibrations/ sec is approaching a stationary observer with a speed equal to $1 / 10$ the speed of sound. What will be the frequency heard by the observer
Answer(c)$n^{\prime}=n\left(\frac{v}{v-v_S}\right)=90\left(\frac{v}{v-\frac{v}{10}}\right)=100 \frac{\text { Vibration }}{\text { sec }}$
View full question & answer→MCQ 951 Mark
On sounding tuning fork $A$ with another tuning fork $B$ of frequency $384 \mathrm{~Hz}, 6$ beats are produced per second. After loading the prongs of $A$ with some wax and then sounding it again with $B, 4$ beats are produced per second. What is the frequency of the tuning fork $A$
- A
$388 \mathrm{~Hz}$
- B
$380 \mathrm{~Hz}$
- C
$378 \mathrm{~Hz}$
- ✓
$390 \mathrm{~Hz}$
AnswerCorrect option: D. $390 \mathrm{~Hz}$
(d)$n_u=?, n_s=384 \mathrm{~Hz}$
$x=6$ bps, which is decreasing (from 6 to 4) i.e. $x \downarrow$ Tuning fork $A$ is loaded so $n \downarrow$
$\text { Hence } n \downarrow-n_v=x \downarrow \quad \longrightarrow \text { Correct } $
$n_{-}n\downarrow=x \downarrow \quad \text { Wrong } $
$\Rightarrow n_x=n_{+}+x=384+6=390 \mathrm{~Hz} $
View full question & answer→MCQ 961 Mark
The speed of a wave in a certain medium is $960 \mathrm{~m} / \mathrm{s}$. If $3600$ waves pass over a certain point of the medium in $1$ minute, the wavelength is
- A
$2$ metres
- B
$4$ metres
- C
$8$ metres
- ✓
$16$ metres
AnswerCorrect option: D. $16$ metres
$n=\frac{3600}{60}=60 \mathrm{~Hz}$
$\Rightarrow \lambda=\frac{v}{n}=\frac{960}{60}=16 \mathrm{~m}$
View full question & answer→MCQ 971 Mark
In stationary waves all particles between two nodes pass through the mean position
- A
At different times with different velocities
- B
At different times with the same velocity
- C
At the same time with equal velocity
- ✓
At the same time with different velocities
AnswerCorrect option: D. At the same time with different velocities
View full question & answer→MCQ 981 Mark
The displacement of a particle is given by $ x=3 \sin (5 \pi t)+4 \cos (5 \pi t)$. The amplitude of the particle is
AnswerFor the given super imposing waves
$a_1=3, a_2=4 \text{and phase difference } \phi=\frac{\pi}{2} $
$\Rightarrow A=\sqrt{a_1^2+a_2^2+2 a_1 a_2 \cos \pi / 2}=\sqrt{(3)^2+(4)^2}=5$
View full question & answer→MCQ 991 Mark
Beats are produced with the help of two sound waves of amplitudes $3$ and $5$ units. The ratio of maximum to minimum intensity in the beats is
- A
$2: 1$
- B
$5: 3$
- C
$4: 1$
- ✓
$16: 1$
AnswerCorrect option: D. $16: 1$
$\frac{I_{\max }}{I_{\min }}=\left(\frac{a_1+a_2}{a_1a_2}\right)^2=\frac{(5+3)^2}{(5-3)^2}=\frac{16}{1}$
View full question & answer→MCQ 1001 Mark
Quality of a musical note depends on
- ✓
- B
- C
- D
Velocity of sound in the medium
Answer(a) The quality of sound depends upon the number of harmonics present. Due to different number of harmonics present in two sounds, the shape of the resultant wave is alsodifferent.
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