MCQ 11 Mark
A train whistling at constant frequency is moving towards a station at a constant speed $V.$ The train goes past a stationary observer on the station. The frequency $n′$ of the sound as heard by the observer is plotted as a function of time $t$ Identify the expected curve:
AnswerWhen qbserver is at rest and source of sound id moving towards observer then observed frequency $n'.$
$\text{n}'=\Big(\frac{\text{v}}{\text{V}-\text{v}_\text{s}}\Big)\text{n}_0$
Where $n_0$ original frequency of source of sound
$v =$ speed of sound in medium
$\therefore\text{n}'>\text{n}_0\ \ \text{v}_\text{s}=$ speed of source
When source is moving away from observer
$\text{n}'=\frac{\text{v}}{(\text{v}+\text{v}_\text{s})}\text{n}_0\ \text{n}''<\text{n}_0$
Hence, the frequencies in both cases are same and $\text{n}'>\text{n}''.$
so graph $(c)$ verifies the answer.
View full question & answer→MCQ 21 Mark
During propagation of a plane progressive mechanical wave:
AnswerDuring propagation of mechanical wave each particles displaces from zero to maximum i.e., upto amplitude So amplitude of each particle is equal. Verifies the option $(b).$
For progressive wave medium particles oscillates about their mean position in which restoring force $\text{F}\propto(-\text{y}).$
So motion of medium particles is simple harmonic motion.
So verifies the option$(c)$. For progressive wave propagating in a medium of density $(\rho)$ and Bulk modulus $k$ the velocity $(\nu)$
As the depends on $k$ and $(\rho)$ and $k, (\rho)$ are different for different medium so of wave depends on nature of medium,
hence, verifies the option $(d).$ View full question & answer→MCQ 31 Mark
Two sine waves travel in the same direction in a medium. The amplitude of each wave is a and the phase difference between the two waves is $120^\circ.$ The resultant amplitude will be:
- ✓
$A$
- B
$2A$
- C
$4A$
- D
$\sqrt{2}\text{A}$
View full question & answer→MCQ 41 Mark
Which of the following statements are true for a stationary wave?
- A
Every particle has a fixed amplitude which is different from the amplitude of its nearest particle.
- B
All the particles cross their mean position at the same time.
- C
There is no net transfer of energy across any plane.
- ✓
AnswerIn stationary waves $[\text{y}(\text{x, t})=\text{a}\sin\text{kx}\cos\omega\text{t}]$ the particles between two nodes vibrates with different amplitude which increases fron nodes.
The amplitude of a particle will remain constant a $\cos kx,$ but varies with $\lambda$
$\because\text{k}=\frac{2\pi}{\lambda}.$
Hence verifies the option $(a)$
particles between two nodes are in same phase
i.e., motion of particles between two nodes will be either upward or downward and crosses the mean position at same time.
Hence the verifies option $(b)$
Hence the reject option $(c)$
As the particles at nodes are rest so energy does not transfer verifies option $(c)$
The amplitude of particles at nodes has amplitude zero verifies option $(c)$
View full question & answer→MCQ 51 Mark
Speed of sound waves in a fluid depends upon:
- A
Directty on density of the medium.
- B
Square of Bulk modulus of the medium.
- C
Inversly on the square root of density.
- ✓
Directly on the square root of bulk modulus of the medium.
AnswerCorrect option: D. Directly on the square root of bulk modulus of the medium.
Speed od sound wave in fluid of bulk modules $k$ and density $\rho$ is given by $\text{v}=\sqrt{\frac{\text{k}}{\rho}}$
so $\text{v}=\sqrt{\text{k}} ($if $\rho$ is constant$)$
And $\text{v}=\sqrt{\frac{1}{\rho}} ($if $k$ is constann$)$
so verifies the option $(d)$.
View full question & answer→MCQ 61 Mark
Which of the following statements is true?
- A
Both light and sound waves can travel in vacuum.
- B
Both light and sound waves in air are transverse.
- C
The sound waves in air are longitudinal, while the light waves are transverse.
- ✓
Both light and sound waves in air are longitudinal.
AnswerCorrect option: D. Both light and sound waves in air are longitudinal.
View full question & answer→MCQ 71 Mark
The picture of a progressive transverse wave at a particular instant of time gives:
- ✓
- B
Motion of the particle of the medium.
- C
- D
View full question & answer→MCQ 81 Mark
Water waves produced by a motor boat sailing in water are:
- A
Neither longitudinal nor transverse.
- ✓
Both longitudinal and transverse.
- C
- D
AnswerCorrect option: B. Both longitudinal and transverse.
As the waves are produced by motor boat on surface as well as inside water, the waves are both, transverse as well as longitudinal.
View full question & answer→MCQ 91 Mark
The wave generated from up and down jerk given to the string or by up and down motion of the piston at end of the pipe is:
- A
- B
- ✓
Both $(a)$ and $(b).$
- D
AnswerCorrect option: C. Both $(a)$ and $(b).$
View full question & answer→MCQ 101 Mark
When both the listener and source are moving towards each other, then which of the following is true regarding frequency and wavelength of wave observed by the observer?
- ✓
More frequency, less wavelength.
- B
More frequency, more wavelength.
- C
Less frequency, less wavelength.
- D
More frequency, constant wavelength.
AnswerCorrect option: A. More frequency, less wavelength.
View full question & answer→MCQ 111 Mark
In a longitudinal wave, the elastic property of the constituents of the medium that determines the stress under compressional strain is:
- A
Young's modulus $(Y).$
- ✓
Bulk modulus $(B).$
- C
Shear modulus $(S).$
- D
Either $(b)$ or $(C).$
AnswerCorrect option: B. Bulk modulus $(B).$
View full question & answer→MCQ 121 Mark
Two sound waves with wavelength $5.0m$ and $5.5m$ respectively, each propagate in a gas with velocity $330\ m/s.$ We expect the following number of beats/sec:
View full question & answer→MCQ 131 Mark
Two pulses having equal and opposite displacements moving in opposite directions overlap at $t =t_1s.$ The resultant displacement of the wave at $t = t_1s$ is:
- A
Twice the displacement of each pulse.
- B
Half the displacement of each pulse.
- ✓
- D
Either $(a)$ or $(c).$
AnswerThe displacement due to two pulses will exactly cancel out each other. Thus, there will be no displacement throughout.
View full question & answer→MCQ 141 Mark
Two sound waves of slightly different frequencies propagating in the same direction produce beats due to:
AnswerBeats are produced on account of interference of sound waves of slightly different frequencies.
View full question & answer→MCQ 151 Mark
When two waves of almost equal frequency $n_1$ and $n_2$ are produced simultaneously, then the time interval between successive maxima is:
- ✓
$\frac{1}{\text{n}_1-\text{n}_2}$
- B
$\frac{1}{\text{n}_1}-\frac{1}{\text{n}_2}$
- C
$\frac{1}{\text{n}_1}+\frac{1}{\text{n}_2}$
- D
$\frac{1}{\text{n}_1+\text{n}_2}$
AnswerCorrect option: A. $\frac{1}{\text{n}_1-\text{n}_2}$
Time interval between two successive maxima $=$ time interval between two successive beats
$=\frac{1}{\text{n}}=\frac{1}{\text{n}_1-\text{n}_2}$
View full question & answer→MCQ 161 Mark
In a sinusoidal wave, the time required for a particular point to move from maximum displacement to zero displacement is $0.17s.$ The frequency of the wave is:
- ✓
$1.47Hz$
- B
$0.36Hz$
- C
$0.73Hz$
- D
$2.94Hz$
AnswerCorrect option: A. $1.47Hz$
$T = 4\ \times$ Time from max. displacement to zero displacement.
$=4\times0.170=0.68\text{s}$
$\text{n}=\frac{1}{\text{T}}=\frac{1}{0.68}$
$=1.47\text{Hz}$
View full question & answer→MCQ 171 Mark
A siren placed at a railway platform is emitting sound of frequency $5kHz.$ A passenger sitting in a moving train $A$ records a frequency of $5.5kHz,$ while the train approaches the siren. During his return journey in a different train $B,$ he records a frequency of $6.0kHz.$ while approaching the same siren. The ratio of the velocity of train $B$ to that of train $A$ is:
- A
$\frac{242}{252}$
- ✓
$2$
- C
$\frac{5}{6}$
- D
$\frac{11}{6}$
AnswerWhen listener alone is moving towards the source.
$\text{n}'=\frac{(\nu+\nu_\text{L})\text{n}}{\nu}$
$\therefore 5.5=\Big(\frac{\nu+\nu_\text{A}}{\nu}\Big)5$
and $6.0=\Big(\frac{\nu+\nu_\text{B}}{\nu}\Big)5$
Solving $(i)$ and $(ii),$ we get
$\frac{\nu_\text{B}}{\nu_\text{A}}=2$
View full question & answer→MCQ 181 Mark
Two sound sources emitting sound each of wavelength $\lambda$ are fixed at a given distance apart. A listener moves with a velocity $u$ along the line joining the two sources. The number of beats heard by him per second is:
- ✓
$\frac{2\text{u}}{\lambda}$
- B
$\frac{\text{u}}{\lambda}$
- C
$\frac{\text{u}}{3\lambda}$
- D
$\frac{2\lambda}{\text{u}}$
AnswerCorrect option: A. $\frac{2\text{u}}{\lambda}$
Number of extra waves received/ sec. $=\pm\frac{\text{u}}{\lambda}$
$\therefore$ Number of beats/ sec. $=\frac{\text{u}}{\lambda}-\Big(\frac{-\text{u}}{\lambda}\Big)=\frac{2\text{u}}{\lambda}$
View full question & answer→MCQ 191 Mark
A string of mass $2.5\ kg$ is under a tension of $200N.$ The length of the stretched string is $20.0m.$ If the transverse jerk is struck at one end of the string, the disturbance will reach the other end in:
- A
- ✓
$0.5$ second
- C
$2$ seconds
- D
Data given is insufficient.
AnswerCorrect option: B. $0.5$ second
$M =$ mass string $2.5\ kg, l = 20m$
$M =$ mas per unit length $=\frac{\text{M}}{\text{l}}=\frac{2.5}{20}=0.125\text{kg/ m}$
$\text{v}=\sqrt{\frac{\text{T}}{\mu}}$
$=\sqrt{\frac{200}{0.125}}$
$=\sqrt{1600}$
$=40\text{m/ s}$
$\text{time} = \frac{\text{distance}}{\text{speed}}$
$=\frac{20\text{m}}{40\text{m/ s}}$
$=\frac{1}{2}\sec$
$=0.5\sec.$
View full question & answer→MCQ 201 Mark
Two pulses in a stretched string whose centres are initially $8 \ cm$ apart are moving towards each other as shown in figure. The speed of each pulse is $2 \ cms^{-1}$. After $2$ second, the total energy of the pulses will be:
- A
- ✓
- C
- D
Partly kinetic and partly potential.
View full question & answer→MCQ 211 Mark
Two vibrating strings of the same material, but lengths $L$ and $2L$ have radii $2r$ and $r$ respectively. They are stretched under the same tension. But the strings vibrate in their fundamental modes, the one of length $L$ with frequency $v_1$ and the other with frequency $\nu_2.$ The ratio $\frac{\nu_1}{\nu_2}$ is given by
AnswerAs fundamental frequency of vibration is:
$\nu=\frac{1}{2\text{l}}\sqrt{\frac{\text{T}}{\text{m}}}$
$\therefore \nu\propto \frac{1}{\text{l}\sqrt{\text{m}}}$
$(\because T$ is constant$)$
$\text{m}=\pi\text{r}^2\rho$
$\therefore \text{m}\propto \text{r}^2$
$\nu=\frac{1}{\text{l}\sqrt{\text{m}}}\propto \frac{1}{\text{lr}}$
$\frac{\nu_1}{\nu_2}=\frac{\text{l}_2\text{r}_2}{\text{l}_1\text{r}_1}$
$=\Big(\frac{2\text{L}}{\text{L}}\Big)\Big(\frac{\text{r}}{2\text{r}}\Big)=1$
View full question & answer→MCQ 221 Mark
A vehicle with a horn of frequency $n$ is moving with a velocity of $30\ m/s$ in a direction perpendicular to the straight line joining the observer and the vehicle. The observer precedes the sound to have a frequency $(n + n_1)$. If velocity of sound in air is $300\ m/s, n,$ would be:
- A
$n_1 = 10n$
- ✓
$n_1 = 0$
- C
$n_1 = 0.1n$
- D
$n_1= -0.1n$
AnswerCorrect option: B. $n_1 = 0$
As source is moving at $90^\circ$ to the line joining the observer and the vehicle, therefore $n’ = n.$
Hence $n_1 = 0$
View full question & answer→MCQ 231 Mark
A transverse harmonic wave on a string is described by $\text{y}(\text{x},\text{t})=3.0\sin(36\ \text{t}+0.018\text{x}+\pi/ 4)$ where x and y are in $cm$ and $t$ is in $s$. The positive direction of $x$ is from left to right.
AnswerThe standard from of a wave propagated from left to right i. e., in $+ve$ direction
$\text{y}=\text{a}\sin(\omega\text{t}-\text{k}\text{x}+\phi)$ and
$\text{y}=3.0\sin\Big(36\text{t}+0.018\text{x}+\frac{\pi}{4}\Big) ($given$)$
$a.$ As in given equation $x$ is in positive sigh so given wave travalling from right to left verifies option $(a)$
$b. \text{k}=0.018=\frac{2\pi}{\lambda}=0.018\Rightarrow\lambda=\frac{2\pi}{0.018}$
$\therefore\text{v}=\text{v}\lambda=\frac{18}{\pi}\times\frac{2\pi}{0.018}=2000\text{cm}/ \ \text{s}=20\text{m/ s}$
Verifies the option $(b)$
$c. \omega=36$ or $2\pi\text{n}=36$ or $\text{v}=\frac{36}{2\pi}=\frac{18}{3.14}=5.7\text{Hz}$
Verifies option $(c) n = 5.7Hz$
View full question & answer→MCQ 241 Mark
With propagation of longitudinal waves through a medium, the quantity transmitted is:
- A
- ✓
- C
- D
Energy, matter and momentum.
AnswerDuring propagation of any wave in a medium only energy is transmitted from one point to another. Matter does not change its own position it, vibrates about its mean position only.
View full question & answer→MCQ 251 Mark
A steel wire has linear mass density $6.9 \times 10^{-3}\ kgm^{-1}$. If the wire is under a tension of $60N$, then the speed of the transverse waves on the wire is:
- A
$63\ ms^{-1}$
- B
$75\ ms^{-1}$
- C
$73\ ms^{-1}$
- ✓
$93\ ms^{-1}$
AnswerCorrect option: D. $93\ ms^{-1}$
Linear mass density $= 6.9 \times 10^{-3}kg m^{-1}$
Tension, $T = 60N$ Thus, speed of wave on the wire is given by
$\nu=\sqrt{\frac{\text{T}}{\mu}}=\sqrt{\frac{60\text{N}}{6.9\times10^{-3}\text{kg m}^{-1}}}$
$=93\text{ms}^{-1}$
View full question & answer→MCQ 261 Mark
Change in temperature of the medium changes:
- A
Frequency of sound waves.
- B
Amplitude of sound waves.
- ✓
Wavelength of sound waves.
- D
AnswerCorrect option: C. Wavelength of sound waves.
Speed of sound wave in amedium $\text{v}=\sqrt{\frac{\gamma\ \text{RT}}{\text{M}}}.$
here $\gamma,\text{R}$ and $M$ are constent.
Hence, $\text{v}\propto\sqrt{\text{T}} ($where $T$ is tempreature of the medium$)$
It mence when temprature changes, speed also changes.
As, $\text{v}=\text{f}\lambda,$ where $f$ is frequency and $\lambda$ is wavelength.
As frequency $(f)$ remains fixed, $\text{v}\propto\lambda$ or $\lambda$ hence wavelength $(\lambda)$changes.
Hence verifies the option $(c).$
View full question & answer→MCQ 271 Mark
A hospital uses an ultrasonic scanner to locate tumours in a tissue. The operating frequency of scanner is $4.2MHz.$ The speed of sound in the tissue is $1.7\ km s^{-1}$. The wavelength of sound in the tissue is close to:
- ✓
$4 \times 10^{-4}m$
- B
$8 \times 1^{-4}m$
- C
$4 \times 10^{-3}m$
- D
$8 \times 10^{-3}m$
AnswerCorrect option: A. $4 \times 10^{-4}m$
$\text{n}=4.2\text{MHz}=4.2\times10^6\text{Hz}$
$\nu=1.7\ \text{km s}^{-1}=1.7\times10^3\text{m s}^{-1}$
$\lambda =\frac{\nu}{\text{n}}=\frac{1.7\times10^3}{4.2\times10^6}$
$=4\times10^{-4}\text{m}$
View full question & answer→MCQ 281 Mark
Let a wave $\text{y(x, t)}=\text{a}\sin(\text{kx}-\omega\text{t})$ is reflected from an open boundary and then the incident and reflected waves overlaps. Then the amplitude of resultant wave:
- A
$2\text{a}\cos(\text{kx})$
- ✓
$2\text{a}\sin(\text{kx})$
- C
$2\text{a}\sin\Big(\frac{\text{kx}}{2}\Big)$
- D
$\text{a}\sin(\text{kx}) $
AnswerCorrect option: B. $2\text{a}\sin(\text{kx})$
We have incident wave $\text{y}_1=\text{a}\sin(\text{kx}-\omega\text{t})$
So the reflected wave is $\text{y}_2=\text{a}\sin(\text{kx}+\omega\text{t})$
From principle of superposition, the standing wave equation obtained after superimposing $y_1$ and $y_2$,
we get
$\text{y(x, t)}=2\text{a}\sin\text{kx}\cos\omega\text{t}$
Thus, the resultant amplitude is
$\text{A(x)}=2\text{a}\sin\text{kx}$
View full question & answer→MCQ 291 Mark
The displacement of the wave given by equation $\text{y (x, t)}=\text{a}\sin(\text{kx}-\omega\text{t}+\phi),$ where $\phi = 0$ at point $x$ and $t = 0$ is same as that at point:
AnswerCorrect option: B. $\text{x}+\frac{2\text{n}\pi}{\text{k}}$
$\text{y}(\text{x}, 0)=\text{a}\sin\text{kx}=\text{a}\sin (\text{kx}+2\text{n}\pi)$
$=\text{a}\sin\text{k}\Big(\text{x}+\frac{2\text{n}\pi}{\text{k}}\Big)$
$\Rightarrow $ The displacement at points $x$ and $\Big(\text{x}+\frac{2\text{n}\pi}{\text{k}}\Big)$
are the same where, $n = 1, 2, 3,......$
View full question & answer→MCQ 301 Mark
The time period of mass suspended from a spring is $T.$ If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be:
- A
$\frac{\text{T}}{4}$
- B
$\text{T}$
- C
$\frac{\text{T}}{2}$
- ✓
$2\text{T}$
AnswerCorrect option: D. $2\text{T}$
View full question & answer→MCQ 311 Mark
A source emits a sound of frequency of $400Hz,$ but the listener hears it to be $390Hz.$ 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.
Since, apparent frequency is lesser than the actual frequency, hence the relative separation between source and listener should be increasing.
View full question & answer→MCQ 321 Mark
A transverse wave propagating along $X-$axis is represented by
$\text{y (x, t)}=8.0\sin(0.5\pi\text{x}-4\pi\text{t}-\frac{\pi}{4})$
where $x$ is in metre and $t$ is in seconds. The speed of the wave is:
AnswerCorrect option: D. $\frac{\pi}{4}\ \text{m/s}$
View full question & answer→MCQ 331 Mark
A standing wave is generated on a string. Which of the following statement$(s)$ is/are correct for the standing waves?
AnswerThere is no phase difference between oscillations of different elements of the wave. However, the string as a whole vibrates in phase with different amplitudes at different points. Also, there is zero movement of the wave pattern. Hence, they are called standing or stationary waves.
View full question & answer→MCQ 341 Mark
Two waves of equal amplitude $A$ and equal frequency travel in the same direction in a medium. The amplitude of the resultant wave is:
- A
$0$
- B
$A$
- C
$2A$
- ✓
Between $0$ to $2A$
AnswerCorrect option: D. Between $0$ to $2A$
View full question & answer→MCQ 351 Mark
A resonating column has resonant frequencies as $100Hz, 300Hz, 500Hz.$ Then it may:
- A
- B
A pipe closed at both ends.
- ✓
- D
View full question & answer→MCQ 361 Mark
A string of length $0.4m$ and mass $10^{-2}\ kg$ is tightly clamped at the ends. The tension in the string is $1.6N.$ Identical wave pulses are produced at one end at equal intervals of time $\Delta\text{t}.$ The minimum value of $\Delta\text{t}$ which allows constructive interference between successive pulses is:
- ✓
$0.05s$
- B
$0.10s$
- C
$0.20s$
- D
$0.40s$
AnswerCorrect option: A. $0.05s$
$\text{n}=\frac{1}{27}\sqrt{\frac{\text{T}}{\text{m}}}=\frac{1}{2\times0.4}\sqrt{\frac{1.6}{\frac{10^{-2}}{0.4}}}=10\text{Hz}$
$\therefore$ Time period $t =\frac{1}{\text{n}}=\frac{1}{10}=0.1\text{s}$
As distance between two fixed ends $=\frac{\lambda}{2}$
$\therefore$ for constructive interference, min value of time
$\Delta \text{t}=\frac{\text{t}}{2}=\frac{0.1}{2}=0.05\text{s}$
View full question & answer→MCQ 371 Mark
To increase the frequency from $100Hz$ to $400Hz$ the tension in the string has to be changed by:
- A
$4$ times.
- ✓
$16$ times.
- C
$2$ times.
- D
AnswerCorrect option: B. $16$ times.
View full question & answer→MCQ 381 Mark
Two waves represented by $\text{y}_1=\text{a}\sin\omega\text{t}$ and $\text{y}_2=\text{a}\sin(\omega\text{t}+\phi)$ with $\phi=\frac{\pi}{2}$ are superposed at any point at a particular instant. The resultant amplitude is:
- A
$a$
- B
$4a$
- ✓
$\sqrt{2}\text{a}$
- D
AnswerCorrect option: C. $\sqrt{2}\text{a}$
View full question & answer→MCQ 391 Mark
Which of the following statements are true for wave motion:
- A
Mechanical transverse waves can propagate through all mediums.
- B
Longitudinal waves can propagate through solids only.
- ✓
Mechanical transverse waves can propagate through solids only.
- D
Longitudinal waves can propagate through vacuum.
AnswerCorrect option: C. Mechanical transverse waves can propagate through solids only.
In case of mechanical transverse wave propagates through a medium, the medium particles oscillate right angles to the direction of wave motion or energy propagation. It travels in the form of crests and troughs.
When mechanical transverse wave propagates through a medium element of the medium is subjected to shearing stress. Solids and strings have shear modulus, that is why, sustain shearing stress. Fluids have no shape of their own, they yield to shearing stress. Transverse waves can be transmitted through solids, they can be setup on the surface of liquids. But they cannot be transmitted into liquids and gases.
View full question & answer→MCQ 401 Mark
A transverse wave propagating along $X-$axis is represented by
$\text{y}(\text{x,t})=8.0\sin\Big(0.5\pi\text{x}-4\pi\text{t}-\frac{\pi}{4}\Big)$
where $x$ is in metre and $t$ is in seconds. The speed of the wave is:
AnswerCorrect option: A. $8\ \text{m/s}$
Comparing with the standard wave equation
$\text{y}=\text{r}\sin \Big[\frac{2\pi\text{x}}{\lambda}-\frac{2\pi\text{t}}{\text{T}}-\phi\Big]$, we get
$\frac{2\pi}{\lambda}=0.5\pi,\lambda =4\text{m},\frac{2\pi}{\text{T}}=4\pi,\text{T}=0.5\text{s}$
Speed of wave $= 8m/s$
View full question & answer→MCQ 411 Mark
Sound waves of wavelength $\pi$ travelling in a medium with a speed of $v\ m/ s$ enter into another medium where its speed is $2v\ m/ s$. Wavelength of sound waves in the second medium is:
- A
$\pi$
- B
$\frac{\pi}{2}$
- ✓
$2\pi$
- D
$4\pi$
AnswerCorrect option: C. $2\pi$
When wave passes from one medium to another its frequency $(v)$ does not chang but its velocity and wavelength changs.
$\text{v}=\text{v}\lambda$ or $\text{v}=\frac{\text{v}}{\lambda}$
$\frac{\text{v}}{\lambda}=\frac{2\text{v}}{\lambda_2}\Rightarrow\lambda_2=2\lambda.$
Hence verifies the correct option $(c)$
View full question & answer→MCQ 421 Mark
A car is moving towards a high cliff. The driver sounds a horn of frequency $f.$ The reflected sound heard by the driver has a frequency $2f.$ If $y$ be the velocity of sound, then the velocity of the car, in the same velocity units would be:
- A
$\frac{\nu}{4}$
- B
$\frac{\nu}{2}$
- C
$\frac{\nu}{\sqrt{2}}$
- ✓
$\frac{\nu}{3}$
AnswerCorrect option: D. $\frac{\nu}{3}$
The frequency of sound incident on the cliff.
$\text{f}_1=\frac{\nu}{\nu-\nu'}\text{f}$
where $v'$ is velocity of car $($source$)$
For the sound reflected from the cliff, driver is the listener, moving towards the source
$\therefore$ Frequency of sound heard, $\text{f}_2=\frac{\nu+\nu'}{\nu}\text{f}_1$
$\text{f}_2=\frac{\nu+\nu'}{\nu}\times\frac{\nu}{\nu-\nu'}\text{f}=\frac{\nu+\nu'}{\nu-\nu'}\text{f}$
As $\text{f}_2=2\text{f}$
$\therefore \frac{\nu+\nu'}{\nu-\nu'}=2=\nu'=\frac{\nu}3{}$
View full question & answer→MCQ 431 Mark
A train, standing in a station yard, blows a whistle of frequency $400Hz$ in still air. The wind starts blowing in the direction from the yard to the station with a speed of $10\ m/ s$. Given that the speed of sound in still air is $340\ m/ s,$
AnswerCorrect option: C. Both $A$ and $B$
As the disatance between listener and source does not change so frequency of sound does not change the as heard listener. i.e.,he heard $\text{v}_0=400\text{Hz}.$ verifies option $(a)$
$\text{v}_0=400\text{Hz}$ frequency of source of sound.
Velocity of wind $\text{v}_\text{w}=10\text{m/ s}$ from source of listenr.
Speed of sound in still air $= \text{v}_\text{s}=340\ \text{m/ s}$
As the listenner is standing on platfrom.
Speed od sound with respect to listener $\text{v}_\text{s}+\text{v}_\text{w}=340+10=350\ \text{m/ s}$
Verifies the option $(b)$
Reject the $(d)$ as it constant $\text{v}_0=400\text{Hz}$
View full question & answer→MCQ 441 Mark
Speed of sound wave in air:
- A
Is independent of temperature.
- B
- ✓
Increases with increase in humidity.
- D
Decreases with increase in humidity.
AnswerCorrect option: C. Increases with increase in humidity.
Speed of sound $($longitudinal$)$ wave in air is $\text{v}=\sqrt{\frac{\lambda\text{p}}{\text{p}}}.$ the density of water vapours is less $($rises up$)$ than the air so on increasing humidity, the density of medium decrease in turn increases the speed of sound in air by $\text{v}=\propto\frac{1}{\sqrt{\text{p}}} ($relation$).$
Hence verifies the option $(c).$
View full question & answer→MCQ 451 Mark
Following two wave trains are approaching each other.
$\text{y}_1=\text{a}\sin2000\pi\text{t}$, $\text{y}_2=\text{a}\sin2008\pi\text{t}$ The number of beats heard per second is:
AnswerBeat frequency $=\text{f}_2-\text{f}_1$
$=\frac{\omega_2-\omega_1}{2\pi}$
$=\frac{2008\pi-2000\pi}{2\pi}$
$=4\text{Hz}$
View full question & answer→MCQ 461 Mark
A whistle producing sound waves of frequencies $9500Hz$ and above is approaching a stationary person with speed $v\ ms^{-1}$. The velocity of sound in air is $300\ ms^{-1}$. If the person can hear frequencies upto a maximum of $10000Hz$ the maximum value of $v$ upto which he can hear the whistle is:
AnswerCorrect option: B. $15\ \text{ms}^{-1}$
Here, $f = 9500Hz, \nu_\text{s}=?, $
$v = 300ms^{-1}, f' = 1000Hz$
As the source is approaching a stationary listener,
$\therefore \text{f}'=\frac{\nu\times\text{f}}{\nu-\nu_\text{s}}$
$1000=\frac{300\times9500}{300-\nu_\text{s}}$
$285=300=\nu_\text{s}$
$\nu_\text{s}=300-285=15\text{ms}^{-1}$
View full question & answer→MCQ 471 Mark
A wave equation is given by $\text{y}=4\sin\Bigg[\pi\Big(\frac{\text{t}}{5}-\frac{\text{x}}{9}+\frac{1}{6}\Big)\Bigg]$ where, $x$ is in $cm$ and $t$ is in second. The wavelength of the wave is:
- ✓
$18\ cm$
- B
$9\ cm$
- C
$36\ cm$
- D
$6\ cm$
AnswerCorrect option: A. $18\ cm$
View full question & answer→MCQ 481 Mark
A student plotted the following four graphs representing the variation of velocity of sound in a gas with the pressure $p$ at constant temperature. Which one is correct?
View full question & answer→MCQ 491 Mark
A source $X$ of unknown frequency produces $8$ beats per second with a source of $250Hz$ and $12$ beats per second with a source of $270Hz.$ The frequency of the source $X$ is:
- A
$242Hz$
- ✓
$258Hz$
- C
$282Hz$
- D
$262Hz$
AnswerCorrect option: B. $258Hz$
View full question & answer→MCQ 501 Mark
If a propagating wave meets a boundary which is not completely rigid or is an interface between two different elastic media, then which of the statements is/ are correct?
- ✓
A part of the incident wave is reflected and a part is transmitted into the second medium.
- B
The incident wave is completely reflected from the boundary.
- C
Only part of the wave is reflected and the remaining part disappears art disappears.
- D
AnswerCorrect option: A. A part of the incident wave is reflected and a part is transmitted into the second medium.
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