MCQ 1511 Mark
The following equations represent progressive transverse waves $ Z_1=A \cos (\omega t-k x), Z_2=A \cos (\omega t+k x) \text {, } Z_3=A \cos (\omega t+k y)$ and $Z_4=A \cos (2 \omega t-2 k y)$. A stationary wave will be formed by superposing
- ✓
$Z_1$ and $Z_2$
- B
$Z_1$ and $Z_4$
- C
$Z_2$ and $Z_3$
- D
$Z_3$ and $Z_4$
AnswerCorrect option: A. $Z_1$ and $Z_2$
When two waves of equal frequency and travelling in opposite direction superimpose, then the stationary wave is produced. Hence $Z$ and $Z$ produces stationary wave.
View full question & answer→MCQ 1521 Mark
If two waves having amplitudes $2 A$ and $A$ and same frequency and velocity, propagate in the same direction in the same phase, the resulting amplitude will be
- ✓
$3 A$
- B
$\sqrt{5} \mathrm{~A}$
- C
$\sqrt{2} A$
- D
$A$
Answer(a) In the same phase $\phi=0$ so resultant amplitude =$a_1+a_2=2 A+A=3 A$
View full question & answer→MCQ 1531 Mark
A plane wave is represented by $x=1.2 \sin (314\ t+12.56\ y)$
Where $x$ and $y$ are distances measured along in $x$ and $y$ direction in meters and $t$ is time in seconds. This wave has
- A
A wavelength of $0.25 \mathrm{~m}$ and travels in + ve $x$ direction
- B
A wavelength of $0.25 \mathrm{~m}$ and travels in + ve $y$ direction
- ✓
A wavelength of $0.5 \mathrm{~m}$ and travels in - ve $y$ direction
- D
A wavelength of $0.5 \mathrm{~m}$ and travels in - ve $x$ direction
AnswerCorrect option: C. A wavelength of $0.5 \mathrm{~m}$ and travels in - ve $y$ direction
(c) The given equation representing a wave travelling along $-y$ direction (because ' + ' sign is given between $t$ term and $x$ term).
On comparing it with $x=A \sin (\omega t+k y)$
We get $k=\frac{2 \pi}{\lambda}=12.56 \Rightarrow \lambda=\frac{2 \times 3.14}{12.56}=0.5 \mathrm{~m}$
View full question & answer→MCQ 1541 Mark
Two sound waves (expressed in CGS units) given by $y_1=0.3 \sin \frac{2 \pi}{\lambda}(v t-x) $ and $ y_2=0.4 \sin \frac{2 \pi}{\lambda}(v t-x+\theta)$ interfere. The resultant amplitude at a place where phase difference is $\pi / 2$ will be
AnswerCorrect option: C. $0.5 \mathrm{~cm}$
$\text { Resultant amplitude }=\sqrt{a_1^2+a_2^2+2 a_1 a_2 \cos \phi}$
$ =\sqrt{0.3^2+0.4^2+2 \times 0.3 \times 0.4 \times \cos \frac{\pi}{2}}=0.5\ cm$
View full question & answer→MCQ 1551 Mark
A source of sound is travelling towards a stationary observer. The frequency of sound heard by the observer is of three times the original frequency. The velocity of sound is $v \mathrm{~m} / \mathrm{sec}$. The speed of source will be
- ✓
$\frac{2}{3} v$
- B
$v$
- C
$\frac{3}{2} v$
- D
$3 v$
AnswerCorrect option: A. $\frac{2}{3} v$
(a) $n^{\prime}=n\left(\frac{v}{v-v_S}\right) \Rightarrow \frac{n^{\prime}}{n}=\frac{v}{v-v_S} \Rightarrow \frac{v}{v-v_S}=3 \Rightarrow v_s=\frac{2 v}{3}$
View full question & answer→MCQ 1561 Mark
The equation of a transverse wave is given by
$
y=10 \sin \pi(0.01 x-2 t)
$
where $x$ and $y$ are in $\mathrm{cm}$ and $t$ is in second. Its frequency is
- A
$10 \mathrm{sec}^{-1}$
- B
$2 \mathrm{sec}^{-1}$
- ✓
$1 \mathrm{sec}^{-1}$
- D
$0.01 \mathrm{sec}^{-1}$
AnswerCorrect option: C. $1 \mathrm{sec}^{-1}$
Comparing with the standard equation,$y=A \sin \frac{2 \pi}{\lambda}(v t-x)$,
we have$v=200 \mathrm{~cm} / \mathrm{sec}, \lambda=200 \mathrm{~cm} ; \therefore n=\frac{v}{\lambda}=1 \mathrm{sec}^{-1}$
View full question & answer→MCQ 1571 Mark
A whistle sends out 256 waves in a second. If the whistle approaches the observer with velocity $1 / 3$ of the velocity of sound in air, the number of waves per second the observer will receive
Answer(a)
Wave number $=\frac{1}{\lambda}\text{ but } \frac{1}{\lambda^{\prime}}$
$=\frac{1}{\lambda}\left(\frac{v}{v-v_s}\right) \text { and } v_S$
$=\frac{v}{3} \therefore(\text { W.N. })^{\prime}=(\text { W.N. })\left(\frac{v}{v-v / 3}\right)=256 \times \frac{v}{2 v / 3} $
$=\frac{3}{2} \times 256=384 \quad \text { Activate Windo }$
View full question & answer→MCQ 1581 Mark
The equation of a wave travelling on a string is $y=4 \sin \frac{\pi}{2}\left(8 t-\frac{x}{8}\right)$. If $x$ and $y$ are in $\mathrm{cm}$, then velocity of wave is
- A
$64 \mathrm{~cm} / \mathrm{sec}$ in $-x$ direction
- B
$32 \mathrm{~cm} / \mathrm{sec}$ in $-x$ direction
- C
$32 \mathrm{~cm} / \mathrm{sec}$ in $+x$ direction
- ✓
$64 \mathrm{~cm} / \mathrm{sec}$ in $+x$ direction
AnswerCorrect option: D. $64 \mathrm{~cm} / \mathrm{sec}$ in $+x$ direction
(d) The given equation can be written as $y=4 \sin \left(4 \pi t-\frac{\pi x}{16}\right)$
$\Rightarrow(v)=\frac{\text { Co }- \text { efficient of } t(\omega)}{\text { Co }- \text { efficient of } x(K)} $
$\Rightarrow v=\frac{4 \pi}{\pi / 16}=64 \mathrm{~cm} / \mathrm{sec} \text { along }+x \text { direction. }$
View full question & answer→MCQ 1591 Mark
An observer moves towards a stationary source of sound of frequency $n$. The apparent frequency heard by him is $2 n$. If the velocity of sound in air is $332 \mathrm{~m} / \mathrm{sec}$, then the velocity of the observer is
- A
$166 \mathrm{~m} / \mathrm{sec}$
- B
$664 \mathrm{~m} / \mathrm{sec}$
- ✓
$332 \mathrm{~m} / \mathrm{sec}$
- D
$1328 \mathrm{~m} / \mathrm{sec}$
AnswerCorrect option: C. $332 \mathrm{~m} / \mathrm{sec}$
$in^{\prime}= n \frac{v+v_O}{v}$
$ \Rightarrow 2n=n \left(\frac{v+v_0}{v}\right) \Rightarrow \frac{v+v_0}{v}=2 $
$\Rightarrow v_O=v=332 \mathrm{~m}/ \mathrm{sec}$
View full question & answer→MCQ 1601 Mark
Tuning fork $F_1$ has a frequency of $256 \mathrm{~Hz}$ 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/second with $F_1$. The frequency of $F_2$ before loading was
- A
$253 \mathrm{~Hz}$
- ✓
$262 \mathrm{~Hz}$
- C
$250 \mathrm{~Hz}$
- D
$259 \mathrm{~Hz}$
AnswerCorrect option: B. $262 \mathrm{~Hz}$
(b) $n_1=$ Known frequency $=256 \mathrm{~Hz}, n_s=$ ?$x=6$ bps,
which remains the same after loading.
Unknown tuning fork $F$ is loaded
so $n \downarrow$
Hence $n \downarrow=x$
$n \downarrow-n_1=x \quad \text {... (ii) } \quad \longrightarrow \text { Correct } $
$\Rightarrow n_u=n_{+}+x=256+6=262 \mathrm{~Hz} .$
View full question & answer→MCQ 1611 Mark
Two tuning forks have frequencies $450 \mathrm{~Hz}$ and $454 \mathrm{~Hz}$ respectively. On sounding these forks together, the time interval between successive maximum intensities will be
- ✓
$1 / 4 \mathrm{sec}$
- B
$1 / 2 \mathrm{sec}$
- C
$1 \mathrm{sec}$
- D
$2 \mathrm{sec}$
AnswerCorrect option: A. $1 / 4 \mathrm{sec}$
(a) The time interval between successive maximum intensities will be $\frac{1}{n_1 \sim n_2}=\frac{1}{454-450}=\frac{1}{4} \mathrm{sec}$.
View full question & answer→MCQ 1621 Mark
When a tuning fork of frequency $341$ is sounded with another tuning fork, six beats per second are heard. When the second tuning fork is loaded with wax and sounded with the first tuning fork, the number of beats is two per second. The natural frequency of the second tuning fork is
Answer(d) $n_{-}=$Known frequency $=341 \mathrm{~Hz}, n_s=$ ?$x=6$bps,
which is decreasing (i.e. $x \downarrow$ ) after loading (from 6 to 1 bps)
Unknown tuning fork is loaded so $n \downarrow$
Hence $n-n \downarrow=x \downarrow$ Wrong $n \downarrow-n=x \downarrow \quad \ldots \text { (ii) } $
$\Rightarrow n=n+x=341+6=347 \mathrm{~Hz} $ $\longrightarrow \text { Correct }$
View full question & answer→MCQ 1631 Mark
- A
Strain is maximum at nodes
- B
Strain is maximum at antinodes
- ✓
Strain is minimum at nodes
- D
Amplitude is zero at all the points
AnswerCorrect option: C. Strain is minimum at nodes
(c) At nodes pressure change (strain) is maximum
View full question & answer→MCQ 1641 Mark
There is a destructive interference between the two waves of wavelength $\lambda$ coming from two different paths at a point. To get maximum sound or constructive interference at that point, the path of one wave is to be increased by
- A
$\frac{\lambda}{4}$
- ✓
$\frac{\lambda}{2}$
- C
$\frac{3 \lambda}{4}$
- D
$\lambda$
AnswerCorrect option: B. $\frac{\lambda}{2}$
(b) With path difference $\frac{\lambda}{2}$, waves are out of phase at the point of observation.
View full question & answer→MCQ 1651 Mark
The distance between the nearest node and antinode in a stationary wave is
- A
$\lambda$
- B
$\frac{\lambda}{2}$
- ✓
$\frac{\lambda}{4}$
- D
$2 \lambda$
AnswerCorrect option: C. $\frac{\lambda}{4}$
View full question & answer→MCQ 1661 Mark
The relation between phase difference $(\Delta \phi)$ and path difference $(\Delta x)$ is
- ✓
$\Delta \phi=\frac{2 \pi}{\lambda} \Delta x$
- B
$\Delta \phi=2 \pi \lambda \Delta x$
- C
$\Delta \phi=\frac{2 \pi \lambda}{\Delta x}$
- D
$\Delta \phi=\frac{2 \Delta x}{\lambda}$
AnswerCorrect option: A. $\Delta \phi=\frac{2 \pi}{\lambda} \Delta x$
View full question & answer→MCQ 1671 Mark
The equation $\vec{\phi}(x, t)=\vec{j} \sin \left(\frac{2 \pi}{\lambda} v t\right) \cos \left(\frac{2 \pi}{\lambda} x\right)$ represents
- A
Transverse progressive wave
- B
Longitudinal progressive wave
- C
Longitudinal stationary wave
- ✓
Transverse stationary wave
AnswerCorrect option: D. Transverse stationary wave
View full question & answer→MCQ 1681 Mark
A sound source of frequency $170 \mathrm{~Hz}$ is placed near a wall. A man walking from a source towards the wall finds that there is a periodic rise and fall of sound intensity. If the speed of sound in air is 340 $\mathrm{m} / \mathrm{s}$ the distance (in metres) separating the two adjacent positions of minimum intensity is
Answer$v=n \lambda \Rightarrow \lambda=\frac{v}{n}=\frac{340}{170}$
$\Rightarrow \lambda=2$ Distance separating the position of minimum intensity = $\frac{\lambda}{2}=\frac{2}{2}=1 \mathrm{~m}$
View full question & answer→MCQ 1691 Mark
When a sound wave of frequency $300 \mathrm{~Hz}$ passes through a medium the maximum displacement of a particle of the medium is $0.1 \mathrm{~cm}$. The maximum velocity of the particle is equal to
- ✓
$60\ \pi\ \mathrm{cm} / \mathrm{sec}$
- B
$30\ \pi \ \mathrm{cm} / \mathrm{sec}$
- C
$30 \mathrm{~cm} / \mathrm{sec}$
- D
$60 \mathrm{~cm} / \mathrm{sec}$
AnswerCorrect option: A. $60\ \pi\ \mathrm{cm} / \mathrm{sec}$
(a) $v_{\max }=a \omega=a \times 2 \pi n=0.1 \times 2 \pi \times 300=60\ \pi\ \mathrm{cm} / \mathrm{sec}$
View full question & answer→MCQ 1701 Mark
The equation of a wave travelling in a string can be written as $y=3 \cos \pi(100 t-x)$. Its wavelength is
- A
$100 \mathrm{~cm}$
- ✓
$2 \mathrm{~cm}$
- C
$5 \mathrm{~cm}$
- D
AnswerCorrect option: B. $2 \mathrm{~cm}$
(b) Comparing the given equation with $y=a \cos (\omega t-k x)$We get $k=\frac{2 \pi}{\lambda}=\pi \Rightarrow \lambda=2 \mathrm{~cm}$
View full question & answer→MCQ 1711 Mark
A wave equation which gives the displacement along the $Y$ direction is given by the equation $y=10^4 \sin (60 t+2 x)$, where $x$ and $y$ are in metres and $t$ is time in seconds. This represents a wave
- ✓
Travelling with a velocity of $30 \mathrm{~m} / \mathrm{sec}$ in the negative $X$ direction
- B
Of wavelength $\pi$ metre
- C
Of frequency $30 / \pi \mathrm{Hz}$
- D
Of amplitude $10^4$ metre travelling along the negative $X$ direction
AnswerCorrect option: A. Travelling with a velocity of $30 \mathrm{~m} / \mathrm{sec}$ in the negative $X$ direction
On comparing the given equation with $y=a \sin (\omega t+k x)$, it is clear that wave is travelling in negative $x-$ direction.It's amplitude $a=10 m$ and $\omega=60, k=2$.
Hence frequency $n=\frac{\omega}{2 \pi}=\frac{60}{2 \pi}=\frac{30}{\pi} \mathrm{Hz}$
$k=\frac{2 \pi}{\lambda}=2 $
$\Rightarrow \lambda=\pi \mathrm{m} $ and $ v=\frac{\omega}{k}=\frac{60}{2}=30 \mathrm{~m} / \mathrm{s}$
View full question & answer→MCQ 1721 Mark
There are three sources of sound of equal intensity with frequencies $400,401$ and $402 \mathrm{vib} / \mathrm{sec}$. The number of beats heard per second is
View full question & answer→MCQ 1731 Mark
A tuning fork whose frequency as given by manufacturer is $512 \mathrm{~Hz}$ is being tested with an accurate oscillator. lt is found that the fork produces a beat of $2 \mathrm{~Hz}$ when oscillator reads $514 \mathrm{~Hz}$ but produces a beat of $6 \mathrm{~Hz}$ when oscillator reads $510 \mathrm{~Hz}$. The actual frequency of fork is
- A
$508 \mathrm{~Hz}$
- B
$512 \mathrm{~Hz}$
- ✓
$516 \mathrm{~Hz}$
- D
$518 \mathrm{~Hz}$
AnswerCorrect option: C. $516 \mathrm{~Hz}$
(c) The tuning fork whose frequency is being tested produces $2$ beats with oscillator at $514 \mathrm{~Hz}$, therefore, frequency of tuning fork may either be $512$ or $516.$ Withoscillator frequency $510$ it gives $6$ beats $/ \mathrm{sec}$, therefore frequency of tuning fork may be either $516$ or $504.$ Therefore, the actual frequency is $516 \mathrm{~Hz}$ which gives $2$ beats/sec with $514 \mathrm{~Hz}$ and 6 beats/sec with $510 \mathrm{~Hz}$.
View full question & answer→MCQ 1741 Mark
The amplitude of two waves are in ratio $5: 2$. If all other conditions for the two waves are same, then what is the ratio of their energy densities
- A
$5: 2$
- B
$10: 4$
- C
$2.5: 1$
- ✓
$25: 4$
AnswerCorrect option: D. $25: 4$
$25: 4$
View full question & answer→MCQ 1751 Mark
In Melde's experiment, the string vibrates in $4$ loops when a $50$ gram weight is placed in the pan of weight $15$ gram. To make the string to vibrates in 6 loops the weight that has to be removed from the pan is
AnswerCorrect option: C. $0.036 \mathrm{~kg} \mathrm{wt}$
Frequency of vibration of string is given by
$ln=\frac{p}{2}l\sqrt{\frac{T}{m}} \Rightarrow p \sqrt{T}=\text { constant } \Rightarrow \frac{p_1}{p_2}=\sqrt{\frac{T_2}{T_1}} $
$\text { Hence } \frac{4}{6}=\sqrt{\frac{T_2}{(50+15)}} g m-\text { force}\Rightarrow T_2=28.8 \mathrm{gm}-f$
Hence weight removed from the pan$=T_1T_2=65-28.8=3.62\ \mathrm{gm} \ \text {-force }=0.036 \mathrm{~kg} \text {-f. }$
View full question & answer→MCQ 1761 Mark
The equation of a progressive wave is $y=8 \sin \left[\pi\left(\frac{t}{10}-\frac{x}{4}\right)+\frac{\pi}{3}\right]$. The wavelength of the wave is
- ✓
$8 \mathrm{~m}$
- B
$4 \mathrm{~m}$
- C
$2 \mathrm{~m}$
- D
$10 \mathrm{~m}$
AnswerCorrect option: A. $8 \mathrm{~m}$
(a) From the given equation $k=\frac{2 \pi}{\lambda}=$ Co-efficient of $x$$=\frac{\pi}{4} \Rightarrow \lambda=8\ m$
View full question & answer→MCQ 1771 Mark
Two closed pipe produce 10 beats per second when emitting their fundamental nodes. If their length are in ratio of $25: 26$. Then their fundamental frequency in $\mathrm{Hz}$, are
- A
$270,280$
- B
$260,270$
- ✓
$260,250$
- D
$260,280$
AnswerCorrect option: C. $260,250$
(c) $n_1-n_2=10$ Using $n_1,=\frac{v}{4 l_1}$ and $n_2=\frac{v}{4 l_2}$
$\Rightarrow \frac{n_1}{n_2}=\frac{l_2}{l_1}=\frac{26}{25}$
After solving these equation $n_1=260 \mathrm{~Hz}, n_2=250 \mathrm{~Hz}$
View full question & answer→MCQ 1781 Mark
If wavelength of a wave is $\lambda=6000 \mathring A$. Then wave number will be
- A
$166 \times 10^3 \mathrm{~m}$
- B
$16.6 \times 10^{-1} \mathrm{~m}$
- ✓
$1.66 \times 10^6 \mathrm{~m}$
- D
$1.66 \times 10^7 \mathrm{~m}$
AnswerCorrect option: C. $1.66 \times 10^6 \mathrm{~m}$
(c) $\bar{n}=\frac{1}{\lambda}=\frac{1}{6000 \times 10^{-10}}=1.66 \times 10^6 \mathrm{~m}^{-1}$
View full question & answer→MCQ 1791 Mark
If the equation of transverse wave is $y=5 \sin 2 \pi\left[\frac{t}{0.04}-\frac{x}{40}\right]$, where distance is in $\mathrm{cm}$ and time in second, then the wavelength of the wave is
- A
$60 \mathrm{~cm}$
- ✓
$40 \mathrm{~cm}$
- C
$35 \mathrm{~cm}$
- D
$25 \mathrm{~cm}$
AnswerCorrect option: B. $40 \mathrm{~cm}$
(b) Comparing with $y=a \sin 2 \pi\left[\frac{t}{T}-\frac{x}{\lambda}\right] \Rightarrow \lambda=40 \mathrm{~cm}$
View full question & answer→MCQ 1801 Mark
If the ratio of amplitude of two waves is $4: 3$. Then the ratio of maximum and minimum intensity will be
- A
$16: 18$
- B
$18: 16$
- ✓
$49: 1$
- D
$1: 49$
AnswerCorrect option: C. $49: 1$
(c) $\frac{I_{\max }}{I_{\min }}=\left(\frac{\frac{a_1}{a_2}+1}{\frac{a_1}{a_2}-1}\right)^2=\left(\frac{\frac{4}{3}+1}{\frac{4}{3}-1}\right)^2=\frac{49}{1}$
View full question & answer→MCQ 1811 Mark
If the ratio of amplitude of wave is $2: 1$, then the ratio of maximum and minimum intensity is
- ✓
$9: 1$
- B
$1: 9$
- C
$4: 1$
- D
$1: 4$
AnswerCorrect option: A. $9: 1$
(a) $\frac{I_{\max }}{I_{\min }}=\left(\frac{\frac{a_1}{a_2}+1}{\frac{a_1}{a_2}-1}\right)^2=\left(\frac{2+1}{2-1}\right)^2=9 / 1$
View full question & answer→MCQ 1821 Mark
- A
- B
Does not transport energy
- C
- ✓
View full question & answer→MCQ 1831 Mark
An observer standing near the sea shore observes 54 waves per minute. If the wavelength of the water wave is $10 \mathrm{~m}$ then the velocity of water wave is
- A
$540 \mathrm{~ms}$
- B
$5.4 \mathrm{~ms}$
- C
$0.184 \mathrm{~ms}$
- ✓
$9 \mathrm{~ms}$
AnswerCorrect option: D. $9 \mathrm{~ms}$
Number of waves per minute $=54$
$\therefore$ Number of waves per second $=54 / 60$
Now $v=n \lambda \Rightarrow n=\frac{54}{60} \times 10=9 \mathrm{~m} \mathrm{s}.$
View full question & answer→MCQ 1841 Mark
When a stationary wave is formed then its frequency is
- ✓
Same as that of the individual waves
- B
Twice that of the individual waves
- C
Half that of the individual waves
- D
AnswerCorrect option: A. Same as that of the individual waves
(a) If $y_{\text {incident }}=a \sin (\omega t-k x)$ and $y_{\text {stationary }}=a \sin (\omega t) \cos k x$ then it is clear that frequency of both is same $(\omega)$
View full question & answer→MCQ 1851 Mark
A person carrying a whistle emitting continuously a note of $272 \mathrm{~Hz}$ is running towards a reflecting surface with a speed of $18 \mathrm{~km} / \mathrm{hour}$. The speed of sound in air is $345 \mathrm{~ms}^{-1}$. The number of beats heard by him is
AnswerAccording the concept of sound image$n^{\prime}=\frac{v+v_{\text {person }}}{v-v_{\text {person }}} .272$
$=\frac{345+5}{345-5} \times 272=280 \mathrm{~Hz}$
$280 - 272= 8$
View full question & answer→MCQ 1861 Mark
A micro-wave and an ultrasonic sound wave have the same wavelength. Their frequencies are in the ratio (approximately)
- ✓
$10^6: 1$
- B
$10^8: 1 \quad$
- C
$10^3: 1$
- D
$10^9: 1$
AnswerCorrect option: A. $10^6: 1$
(a)$n=\frac{v}{\lambda} \propto v \Rightarrow \frac{n_{M W}}{n_{U S}} \approx \frac{3 \times 10^8}{3 \times 10^2} \approx 10^6: 1$
View full question & answer→MCQ 1871 Mark
25 tunning forks are arranged in series in the order of decreasing frequency. Any two successive forks produce 3 beats/sec. If the frequency of the first turning fork is the octave of the last fork, then the frequency of the 21 fork is
- A
$72 \mathrm{~Hz}$
- B
$288 \mathrm{~Hz}$
- ✓
$84 \mathrm{~Hz}$
- D
$87 \mathrm{~Hz}$
AnswerCorrect option: C. $84 \mathrm{~Hz}$
View full question & answer→MCQ 1881 Mark
Velocity of sound in air
$I$. Increases with temperature
$II$. Decreases with temperature
$III$. Increase with pressure
$IV$. Is independent of pressure
$V$. Is independent of temperature
Choose the correct answer.
- A
Only $I$ and $II$ are true
- B
Only $I$ and $III$ are true
- C
Only $II$ and $III$ are true
- ✓
Only $I$ and $IV$ are true
AnswerCorrect option: D. Only $I$ and $IV$ are true
Only $I$ and $IV$ are true
View full question & answer→MCQ 1891 Mark
When we hear a sound, we can identify its source from
- A
- B
- C
- ✓
Overtones present in the sound
AnswerCorrect option: D. Overtones present in the sound
(d) The sounds of different source are said to differ in quality. The number of overtones and their relative intensities determines the quality of any musical sound.
View full question & answer→MCQ 1901 Mark
A particle on the trough of a wave at any instant will come to the mean position after a time ( $T=$ time period)
- A
$T / 2$
- ✓
$T / 4$
- C
$T$
- D
$2 T$
AnswerCorrect option: B. $T / 4$
(b) The particle will come after a time $\frac{T}{4}$ to its mean position.
View full question & answer→MCQ 1911 Mark
The disc of a siren containing $60$ holes rotates at a constant speed of $360\ \mathrm{rpm}$. The emitted sound is in unison with a tuning fork of frequency
- A
$10 \mathrm{~Hz}$
- ✓
$360 \mathrm{~Hz}$
- C
$216 \mathrm{~Hz}$
- D
$6 \mathrm{~Hz}$
AnswerCorrect option: B. $360 \mathrm{~Hz}$
(b) Frequency $=\frac{360}{60} \times 60=360 \mathrm{~Hz}$.
View full question & answer→MCQ 1921 Mark
An open tube is in resonance with string (frequency of vibration of tube is $n$ ). If tube is dipped in water so that $75 \%$ of length of tube is inside water, then the ratio of the frequency of tube to string now will be
- A
$1$
- ✓
$2$
- C
$\frac{2}{3}$
- D
$\frac{3}{2}$
AnswerFor open tube, $n_0=\frac{v}{2 l}$For closed tube length available for resonance is
$(l^{\prime}=l \times \frac{25}{100}=\frac{l}{4})$
$ \therefore$ Fundamental frequency of water filled tube $n=\frac{v}{4 l^{\prime}}=\frac{v}{4 \times(l / 4)}=\frac{v}{l}=2n_0\Rightarrow\frac{n}{n_0}=2$
View full question & answer→MCQ 1931 Mark
In a resonance pipe the first and second resonances are obtained at depths $22.7 \mathrm{~cm}$ and $70.2 \mathrm{~cm}$ respectively. What will be the end correction
- ✓
$1.05 \mathrm{~cm}$
- B
$115.5 \mathrm{~cm}$
- C
$92.5 \mathrm{~cm}$
- D
$113.5 \mathrm{~cm}$
AnswerCorrect option: A. $1.05 \mathrm{~cm}$
(a)$\text { For end correction } x, \frac{l_2+x}{l_1+x}=\frac{3 \lambda / 4}{\lambda / 4}=3$
$x=\frac{l_2-3 l_1}{2}=\frac{70.2-3 \times22.7}{2}=1.05\mathrm{~cm}$
View full question & answer→MCQ 1941 Mark
In one metre long open pipe what is the harmonic of resonance obtained with a tuning fork of frequency $480 \mathrm{~Hz}$
View full question & answer→MCQ 1951 Mark
A wave has velocity $u$ in medium $P$ and velocity $2 u$ in medium $Q$. If the wave is incident in medium $P$ at an angle of $30^{\circ}$ then the angle of refraction will be
- A
$30^{\circ}$
- B
$45^{\circ}$
- C
$60^{\circ}$
- ✓
$90^{\circ}$
AnswerCorrect option: D. $90^{\circ}$
(d)$v=\frac{\sin i}{\sin r}=\frac{v_1}{v_2} $
$\Rightarrow \sin r=\sin 30^{\circ} \times \frac{2 u}{u} $
$\Rightarrow \sin r=\frac{1}{2} \times 2 \times 1$
$ \Rightarrow r=90^{\circ} $
View full question & answer→MCQ 1961 Mark
The loudness and pitch of a sound depends on
- A
- B
- ✓
- D
Frequency and number of harmonics
Answer(c) Loudness depends upon intensity while pitch depends upon frequency.
View full question & answer→MCQ 1971 Mark
In Melde's experiment in the transverse mode, the frequency of the tuning fork and the frequency of the waves in the strings are in the ratio
- ✓
$1: 1$
- B
$1: 2$
- C
$2: 1$
- D
$4: 1$
AnswerCorrect option: A. $1: 1$
(a) Because both tuning fork and string are in resonance condition.
View full question & answer→MCQ 1981 Mark
Oxygen is $16$ times heavier than hydrogen. Equal volumes of hydrogen and oxygen are mixed. The ratio of speed of sound in the mixture to that in hydrogen is
- ✓
$\sqrt{\frac{1}{8}}$
- B
$ \sqrt{\frac{32}{17}}$
- C
$\sqrt{8}$
- D
$\sqrt{\frac{2}{17}}$
AnswerCorrect option: A. $\sqrt{\frac{1}{8}}$
$\sqrt{\frac{1}{8}}$
View full question & answer→MCQ 1991 Mark
A man, standing between two cliffs, claps his hands and starts hearing a series of echoes at intervals of one second. If the speed of sound in air is $340 \mathrm{~ms}$, the distance between the cliffs is
- ✓
$340 \mathrm{~m}$
- B
$1620 \mathrm{~m}$
- C
$680 \mathrm{~m}$
- D
$1700 \mathrm{~m}$
AnswerCorrect option: A. $340 \mathrm{~m}$
View full question & answer→MCQ 2001 Mark
Two wires are fixed in a sonometer. Their tensions are in the ratio 8 $: 1$. The lengths are in the ratio $36: 35$. The diameters are in the ratio $4: 1$. Densities of the materials are in the ratio $1: 2$. If the lower frequency in the setting is $360 \mathrm{~Hz}$. the beat frequency when the two wires are sounded together is
AnswerFrequency in a stretched string is given by
$n=\frac{1}{2}l\sqrt{\frac{T}{\pi r^2 \rho}}=\frac{1}{l} \sqrt{\frac{T}{\pi d^2 \rho}} \quad(d=\text { Diameter of string })$
$\Rightarrow \frac{n_1}{n_2}=\frac{l_2}{l_1} \sqrt{\frac{T_1}{T_2}\times\left(\frac{d_2}{d_1}\right)^2 \times\left(\frac{\rho_2}{\rho_1}\right)} $
$=\frac{35}{36} \sqrt{\frac{8}{1} \times\left(\frac{1}{4}\right)^2 \times \frac{2}{1}}=\frac{35}{36} \Rightarrow n_2=\frac{36}{35} \times 360=370$
Hence beat frequency $=n_2-n_1=10$
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