MCQ

MCQ 1011 Mark

A source of sound of frequency $n$ is moving towards a stationary observer with a speed $S$. If the speed of sound in air is $V$ and the frequency heard by the observer is $n_1$, the value of $n_1 / n$ is

A
$(V+S) / V$
B
$V /(V+S)$
C
$(V-S) / V$
✓
$V /(V-S)$

Answer

Correct option: D.

$V /(V-S)$

(d) By using $n^{\prime}=n \frac{v}{v-v_S} \Rightarrow \frac{n_1}{n}=\left(\frac{V}{V-S}\right)$

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MCQ 1021 Mark

Two tuning forks, $A$ and $B$, give 4 beats per second when sounded together. The frequency of $A$ is $320 \mathrm{~Hz}$. When some wax is added to $B$ and it is sounded with $A, 4$ beats per second are again heard. The frequency of $B$ is

A
$312 \mathrm{~Hz}$
B
$316 \mathrm{~Hz}$
✓
$324 \mathrm{~Hz}$
D
$328 \mathrm{~Hz}$

Answer

Correct option: C.

$324 \mathrm{~Hz}$

(c)$n-n \downarrow=x \text { (same) } $
$n \downarrow n=x \text{(same}\Rightarrow n=n+x)$
$=320+4=324 \mathrm{~Hz} .$

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MCQ 1031 Mark

In stationary waves, antinodes are the points where there is

A
Minimum displacement and minimum pressure change
B
Minimum displacement and maximum pressure change
C
Maximum displacement and maximum pressure change
✓
Maximum displacement and minimum pressure change

Answer

Correct option: D.

Maximum displacement and minimum pressure change

(d)

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MCQ 1041 Mark

The driver of a car travelling with speed 30 metres per second towards a hill sounds a horn of frequency $600 \mathrm{~Hz}$. If the velocity of sound in air is 330 metres per second, the frequency of the reflected sound as heard by the driver is

✓
$720 \mathrm{~Hz}$
B
$555.5 \mathrm{~Hz}$
C
$550 \mathrm{~Hz}$
D
$500 \mathrm{~Hz}$

Answer

Correct option: A.

$720 \mathrm{~Hz}$

(a) From the figure, it is clear that Frequency of reflected sound heard by the driver.$\begin{aligned}n^{\prime} & =n\left[\frac{v-\left(v_o\right{vv_s\right]=n\left[\frac{v+v_o}{v-v_s}\right]=n\left[\frac{v+v_{c a r}}{v-v_{c a r}}\right] \\& =600\left[\frac{330+30}{330-30}\right]=720 \mathrm{~Hz} .\end{aligned}$

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MCQ 1051 Mark

The equation $y=0.15 \sin 5 x \cos 300 t$, describes a stationary wave. The wavelength of the stationary wave is

A
Zero
✓
$1.256$ metres
C
$2.512$ metres
D
$0.628$ metre

Answer

Correct option: B.

$1.256$ metres

On comparing the given equation with standard equation $\frac{2 \pi}{\lambda}=5\Rightarrow \lambda=\frac{6.28}{5}=1.256 \mathrm{~m}$

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MCQ 1061 Mark

A boy is walking away from a wall towards an observer at a speed of $1$ metre/sec and blows a whistle whose frequency is $680 \mathrm{~Hz}$. The number of beats heard by the observer per second is (Velocity of sound in air $=340$ metres $/ \mathrm{sec}$

A
Zero
B
$2$
C
$8$
✓
$4$

Answer

Correct option: D.

$4$

Observer hears two frequencies(i) $n_1$ which is coming from the source directly(ii) $n_2$ which is coming from the reflection image of sourceso, $n_1=680\left(\frac{340}{340-1}\right)$ and $n_2=680\left(\frac{340}{340+1}\right)$$\Rightarrow n_1-n_2=4 \text {beats}$

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MCQ 1071 Mark

A source and listener are both moving towards each other with speed $v / 10$, where $v$ is the speed of sound. If the frequency of the note emitted by the source is $f$, the frequency heard by the listener would be nearly

A
$1.11 f$
✓
$1.22 f$
C
$f$
D
$1.27 f$

Answer

Correct option: B.

$1.22 f$

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MCQ 1081 Mark

A transverse progressive wave on a stretched string has a velocity of $10 \mathrm{~ms}^{-1}$ and a frequency of $100 \mathrm{~Hz}$. The phase difference between two particles of the string which are $2.5 \mathrm{~cm}$ apart will be

A
$\frac{\pi}{8}$
B
$\frac{\pi}{4}$
C
$\frac{3 \pi}{8}$
✓
$\frac{\pi}{2}$

Answer

Correct option: D.

$\frac{\pi}{2}$

$(d) \ v=n \lambda $
$\Rightarrow \lambda=10 \mathrm{~cm}$
Phase difference $\frac{2\pi}{\lambda} \times$ Path difference $\frac{2 \pi}{10} \times 2.5=\frac{\pi}{2}$

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MCQ 1091 Mark

The path difference between the two waves $y_1=a_1 \sin \left(\omega t-\frac{2 \pi x}{\lambda}\right)$ and $y_2=a_2 \cos \left(\omega t-\frac{2 \pi x}{\lambda}+\phi\right)$ is

A
$\frac{\lambda}{2 \pi} \phi$
✓
$\frac{\lambda}{2 \pi}\left(\phi+\frac{\pi}{2}\right)$
C
$\frac{2 \pi}{\lambda}\left(\phi-\frac{\pi}{2}\right)$
D
$\frac{2 \pi}{\lambda} \phi$

Answer

Correct option: B.

$\frac{\lambda}{2 \pi}\left(\phi+\frac{\pi}{2}\right)$

$y_1=a_1 \sin \left(\omega t-\frac{2 \pi x}{\lambda}\right)$ and$y_2=a_2\cos\left(\omega t-\frac{2 \pi x}{\lambda}+\phi\right)=a_2 \sin \left(\omega t-\frac{2 \pi x}{\lambda}+\phi+\frac{\pi}{2}\right)$
So phase difference $=\phi+\frac{\pi}{2}$ and $\Delta=\frac{\lambda}{2 \pi}\left(\phi+\frac{\pi}{2}\right)$

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MCQ 1101 Mark

A plane wave is described by the equation $y=3 \cos \left(\frac{x}{4}-10 t-\frac{\pi}{2}\right)$. The maximum velocity of the particles of the medium due to this wave is

✓
$30$
B
$\frac{3 \pi}{2}$
C
$3 / 4$
D
$40$

Answer

Correct option: A.

$30$

$v_{\max }=a \omega=3 \times 10=30$

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MCQ 1111 Mark

In a stationary wave, all particles are

✓
At rest at the same time twice in every period of oscillation
B
At rest at the same time only once in every period of oscillation
C
Never at rest at the same time
D
Never at rest at all

Answer

Correct option: A.

At rest at the same time twice in every period of oscillation

(a)

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MCQ 1121 Mark

When a tuning fork $A$ of unknown frequency is sounded with another tuning fork $B$ of frequency $256 \mathrm{~Hz}$, then $3$ beats per second are observed. After that $A$ is loaded with wax and sounded, the again $3$ beats per second are observed. The frequency of the tuning fork $A$ is

A
$250 \mathrm{~Hz}$
B
$253 \mathrm{~Hz}$
✓
$259 \mathrm{~Hz}$
D
$262 \mathrm{~Hz}$

Answer

Correct option: C.

$259 \mathrm{~Hz}$

It is given that$n=$ Unknown frequency $=$ ?$n_{\text {}}=$ Known frequency $=256 \mathrm{~Hz}$
$x=3$ bps, which remains same after loadingUnknown tuning fork $A$ is loaded so $n\downarrow$
Hence $(n)\downarrow n_s=x(nn)\downarrow=x(\Rightarrow n_1=n_1+x=256+3=259 \mathrm{~Hz}$.

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MCQ 1131 Mark

Two waves are given by $y_1=a \sin (\omega t-k x)$ and $y_2=a \cos (\omega t-k x)$ The phase difference between the two waves is

A
$\frac{\pi}{4}$
B
$\pi$
C
$\frac{\pi}{8}$
✓
$\frac{\pi}{2}$

Answer

Correct option: D.

$\frac{\pi}{2}$

(d) $y_1=a \sin (\omega t-k x)$$\text { and } y_2=a \cos (\omega t-k x)=a \sin\left(\omega t-k x+\frac{\pi}{2}\right)$Hence phase difference between these two is $\frac{\pi}{2}$.

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MCQ 1141 Mark

Doppler shift in frequency does not depend upon

A
The frequency of the wave produced
B
The velocity of the source
C
The velocity of the observer
✓
Distance from the source to the listener

Answer

Correct option: D.

Distance from the source to the listener

(d)

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MCQ 1151 Mark

The musical interval between two tones of frequencies $320 \mathrm{~Hz}$ and $240 \mathrm{~Hz}$ is

A
$80$
✓
$\left(\frac{4}{3}\right)$
C
$560$
D
$320 \times 240$

Answer

Correct option: B.

$\left(\frac{4}{3}\right)$

(b) Musical interval is the ratio of frequencies $=\frac{320}{240}=\frac{4}{3}$

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MCQ 1161 Mark

Learned Indian classical vocalists do not like the accompaniment of a harmonium because

A
Intensity of the notes of the harmonium is too large
B
Notes of the harmonium are too shrill
C
Diatonic scale is used in the harmonium
✓
Tempered scale is used in the harmonium

Answer

Correct option: D.

Tempered scale is used in the harmonium

(d) Indian classical vocalists don't like harmoniuim because it uses tempered scale.

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MCQ 1171 Mark

The relation between time and displacement for two particles is given by
$y_1=0.06 \sin 2 \pi\left(0.04 t+\phi_1\right), y_2=0.03 \sin 2 \pi\left(1.04 t+\phi_2\right)$
The ratio of the intensity of the waves produced by the vibrations of the two particles will be

A
$2: 1$
B
$1: 2$
✓
$4: 1$
D
$1: 4$

Answer

Correct option: C.

$4: 1$

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MCQ 1181 Mark

With what velocity an observer should move relative to a stationary source so that he hears a sound of double the frequency of source

✓
Velocity of sound towards the source
B
Velocity of sound away from the source
C
Half the velocity of sound towards the source
D
Double the velocity of sound towards the source

Answer

Correct option: A.

Velocity of sound towards the source

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MCQ 1191 Mark

Two passenger trains moving with a speed of $108 \mathrm{~km} / \mathrm{hour}$ cross each other. One of them blows a whistle whose frequency is $750 \mathrm{~Hz}$. If sound speed is $330 \mathrm{~m} / \mathrm{s}$, then passengers sitting in the other train, after trains cross each other will hear sound whose frequency will be

A
$900 \mathrm{~Hz}$
✓
$625 \mathrm{~Hz}$
C
$750 \mathrm{~Hz}$
D
$800 \mathrm{~Hz}$

Answer

Correct option: B.

$625 \mathrm{~Hz}$

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MCQ 1201 Mark

Two tuning forks $A$ and $B$ vibrating simultaneously produce 5 beats. Frequency of $B$ is 512 . It is seen that if one arm of $A$ is filed, then the number of beats increases. Frequency of $A$ will be

A
$502$
B
$507$
✓
$517$
D
$522$

Answer

Correct option: C.

$517$

(c) After filling frequency increases, so $n_A$ decreases $(\downarrow)$.
Also it isgiven that beat frequency increases (i.e., $x \uparrow$ )
Hence $n \downarrow-n_{-}=x \uparrow$ $n-n \uparrow=x \uparrow$
correct $n_{-}-n_x=x \uparrow $
$\Rightarrow n_{+}=n_c+x=512+5=517 \mathrm{~Hz} .$

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MCQ 1211 Mark

A wave is reflected from a rigid support. The change in phase on reflection will be

A
$\pi / 4$
B
$\pi / 2$
✓
$\pi$
D
$2 \pi$

Answer

Correct option: C.

$\pi$

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MCQ 1221 Mark

Ten tuning forks are arranged in increasing order of frequency in such a way that any two nearest tuning forks produce 4 beats/sec. The highest frequency is twice of the lowest. Possible highest and the lowest frequencies are

A
$80$ and $40$
B
$100$ and $50$
C
$44$ and $22$
✓
$72$ and $36$

Answer

Correct option: D.

$72$ and $36$

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MCQ 1231 Mark

A source of sound emitting a note of frequency $200 \mathrm{~Hz}$ moves towards an observer with a velocity $v$ equal to the velocity of sound. If the observer also moves away from the source with the same velocity $v$, the apparent frequency heard by the observer is

A
$50 \mathrm{~Hz}$
B
$100 \mathrm{~Hz}$
C
$150 \mathrm{~Hz}$
✓
$200 \mathrm{~Hz}$

Answer

Correct option: D.

$200 \mathrm{~Hz}$

Since there is no relative motion between observer and source, therefore there is no apparent change in frequency.

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MCQ 1241 Mark

Consider ten identical sources of sound all giving the same frequency but having phase angles which are random. If the average intensity of each source is $I_0$, the average of resultant intensity $l$ due to all these ten sources will be

A
$I=100 I_0$
✓
$I=10 I_0$
C
$ I=I_0$
D
$I=\sqrt{10} I_0$

Answer

Correct option: B.

$I=10 I_0$

In case of interference of two waves resultant intensity $I=I_1+I_2+2 \sqrt{I_1 I_2}\cos\phi$
If $\phi$ varies randomly with time, so $(\cos \phi)_{a v}=0$ $\Rightarrow I=I_1+I_2$
For $n$ identical waves, $I=I_0+I_0+\ldots \ldots =n I_0$ here $I=10I_0$.

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MCQ 1251 Mark

A tuning fork and a sonometer wire were sounded together and produce $4$ beats per second. When the length of sonometer wire is $95 \mathrm{~cm}$ or $100 \mathrm{~cm}$, the frequency of the tuning fork is

✓
$156 \mathrm{~Hz}$
B
$152 \mathrm{~Hz}$
C
$148 \mathrm{~Hz}$
D
$160 \mathrm{~Hz}$

Answer

Correct option: A.

$156 \mathrm{~Hz}$

(a) Probable frequencies of tuning fork be $n+4$ or $n-4$Frequency of sonometer wire $n \propto \frac{1}{l}$
$\therefore \frac{n+4}{n-4}=\frac{100}{95} \text {or} 95(n+4)=100(n-4)$or $95 n+380=100 (n-400)$ or $5 n=780$ or $n=156$

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MCQ 1261 Mark

When two sound waves with a phase difference of $\pi / 2$, and each having amplitude $A$ and frequency $\omega$, are superimposed on each other, then the maximum amplitude and frequency of resultant wave is

A
$\frac{A}{\sqrt{2}}: \frac{\omega}{2}$
B
$\frac{A}{\sqrt{2}}: \omega$
C
$\sqrt{2} A: \frac{\omega}{2}$
✓
$\sqrt{2} A: \omega$

Answer

Correct option: D.

$\sqrt{2} A: \omega$

(d) $A_{\max }=\sqrt{A^2+A^2}=A \sqrt{2}$, frequency will remain same i.e. $\omega$

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MCQ 1271 Mark

The source producing sound and an observer both are moving along the direction of propagation of sound waves. If the respective velocities of sound, source and an observer are $v, v_s$ and $v_o$, then the apparent frequency heard by the observer will be ( $n=$ frequency of sound)

A
$\frac{n\left(v+v_o\right)}{v-v_o}$
✓
$\frac{n\left(v-v_o\right)}{v-v_s}$
C
$\frac{n\left(v-v_o\right)}{v+v_s}$
D
$\frac{n\left(v+v_o\right)}{v+v_s}$

Answer

Correct option: B.

$\frac{n\left(v-v_o\right)}{v-v_s}$

(b)

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MCQ 1281 Mark

The sound carried by air from a sitar to a listener is a wave of the following type

A
Longitudinal stationary
B
Transverse progressive
C
Transverse stationary
✓
Longitudinal progressive

Answer

Correct option: D.

Longitudinal progressive

(d) Observer receives sound waves (music) which are longitudinal progressive waves.

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MCQ 1291 Mark

For the stationary wave $y=4 \sin \left(\frac{\pi x}{15}\right) \cos (96 \pi t)$, the distance between a node and the next antinode is

✓
$7.5$
B
$15$
C
$22.5$
D
$30$

Answer

Correct option: A.

$7.5$

Comparing given equation with standard equation $y=2 a \sin \frac{2 \pi x}{\lambda} \cos \frac{2 \pi v t}{\lambda}$ gives us $\frac{2\pi}{\lambda}=\frac{\pi}{15}\Rightarrow\lambda=30$
Distance between nearest node and antinodes = $\frac{\lambda}{4}=\frac{30}{4}=7.5$

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MCQ 1301 Mark

Which of the property makes difference between progressive and stationary waves

A
Amplitude
B
Frequency
✓
Propagation of energy
D
Phase of the wave

Answer

Correct option: C.

Propagation of energy

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MCQ 1311 Mark

The equation of a wave is $y=2 \sin \pi(0.5 x-200 t)$, where $x$ and $y$ are expressed in $\mathrm{cm}$ and $t$ in sec. The wave velocity is

A
$100 \mathrm{~cm} / \mathrm{sec}$
B
$200 \mathrm{~cm} / \mathrm{sec}$
C
$300 \mathrm{~cm} / \mathrm{sec}$
✓
$400 \mathrm{~cm} / \mathrm{sec}$

Answer

Correct option: D.

$400 \mathrm{~cm} / \mathrm{sec}$

(d) Comparing given equation with standard equation of progressive wave. The velocity of wave$v=\frac{\omega(\text { Co }- \text { efficient of } t)}{k(\text { Co }- \text{efficient of } x)}=\frac{200 \pi}{0.5 \pi}=400 \mathrm{~cm} / \mathrm{s}$

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MCQ 1321 Mark

If amplitude of waves at distance $r$ from a point source is $A$, the amplitude at a distance $2 r$ will be

A
$2 A$
B
$A$
✓
$A / 2$
D
$A / 4$

Answer

Correct option: C.

$A / 2$

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MCQ 1331 Mark

A tuning fork gives $5$ beats with another tuning fork of frequency $100 \mathrm{~Hz}$. When the first tuning fork is loaded with wax, then the number of beats remains unchanged, then what will be the frequency of the first tuning fork

A
$95 \mathrm{~Hz}$
B
$100 \mathrm{~Hz}$
✓
$105 \mathrm{~Hz}$
D
$110 \mathrm{~Hz}$

Answer

Correct option: C.

$105 \mathrm{~Hz}$

Suppose $n_{+}=$known frequency $=100 \mathrm{~Hz}, n_{+}=$? $x=5$ bps, which remains unchanged after loading Unknown tuning fork is loaded so $n \downarrow$
Hence $n-n \downarrow=x$$n \downarrow-n=x$
From equation (i), it is clear that as $n$. decreases, beat frequency. (i.e. $n-(n))$ can never be $x$ again.
From equation(ii),as $(n) \downarrow$, beat frequency (i.e. $(n)-n$ ) decreases as long as $(n)$ remains greater than $n$, if $(n)$ become lesser than $n$ the beat frequency will increase again and will be$(x)$.
Hence this is correct.So, $n=n+x=100+5=105 \mathrm{~Hz}$.

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MCQ 1341 Mark

Two tuning forks of frequencies 256 and 258 vibrations/sec are sounded together, then time interval between consecutive maxima heard by the observer is

A
$2 \mathrm{sec}$
✓
$0.5 \mathrm{sec}$
C
$250 \mathrm{sec}$
D
$252 \mathrm{sec}$

Answer

Correct option: B.

$0.5 \mathrm{sec}$

(b) $T=\frac{1}{258-256}=0.5 \mathrm{sec}$

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MCQ 1351 Mark

The equation of the propagating wave is $y=25 \sin (20 t+5 x)$, where $y$ is displacement. Which of the following statement is not true

A
The amplitude of the wave is $25$ units
✓
The wave is propagating in positive $x$-direction
C
The velocity of the wave is $4$ units
D
The maximum velocity of the particles is $500$ units

Answer

Correct option: B.

The wave is propagating in positive $x$-direction

(b) Positive sign in the argument of sin indicating that wave is travelling in negative $(x)$-direction.

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MCQ 1361 Mark

A source and an observer are moving towards each other with a speed equal to $\frac{v}{2}$ where $v$ is the speed of sound. The source is emitting sound of frequency $n$. The frequency heard by the observer will be

A
Zero
B
$n$
C
$\frac{n}{3}$
✓
$3 n$

Answer

Correct option: D.

$3 n$

(d)$n^{\prime}=n\left(\frac{v+v_0}{v-v_s}\right)=n\left(\frac{v+v / 2}{v-v / 2}\right)=3 n$

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MCQ 1371 Mark

Two wires are in unison. If the tension in one of the wires is increased by $2 \%, 5$ beats are produced per second. The initial frequency of each wire is

A
$200 \mathrm{~Hz}$
B
$400 \mathrm{~Hz}$
✓
$500 \mathrm{~Hz}$
D
$1000 \mathrm{~Hz}$

Answer

Correct option: C.

$500 \mathrm{~Hz}$

$n \propto \frac{1}{2 \ell} \sqrt{\frac{T}{m}}$
$\therefore n \propto T $
$\frac{n_2}{n_1}=\sqrt{\frac{T_2}{T_1}}=\frac{\sqrt{1.02 T_1}}{T_1}=\sqrt{1.02}=1.01$
Also $n_2-n_1=5$
$\therefore 1.01 n_1-n_1=5 $
$\therefore 0.01 n _1=5$
$\therefore n_1=\frac{5}{0.01}=500\ Hz$

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MCQ 1381 Mark

The phase difference between the two particles situated on both the sides of a node is

A
$0^{\circ}$
B
$90^{\circ}$
✓
$180^{\circ}$
D
$360^{\circ}$

Answer

Correct option: C.

$180^{\circ}$

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MCQ 1391 Mark

A car sounding a horn of frequency $1000 \mathrm{~Hz}$ passes an observer. The ratio of frequencies of the horn noted by the observer before and after passing of the car is $11: 9$. If the speed of sound is $v$, the speed of the car is

✓
$\frac{1}{10} v$
B
$\frac{1}{2} v$
C
$\frac{1}{5} v$
D
$v$

Answer

Correct option: A.

$\frac{1}{10} v$

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MCQ 1401 Mark

A source of sound of frequency $500 \mathrm{~Hz}$ is moving towards an observer with velocity $30 \mathrm{~m} / \mathrm{s}$. The speed of sound is $330 \mathrm{~m} / \mathrm{s}$. the frequency heard by the observer will be

✓
$550 \mathrm{~Hz}$
B
$458.3 \mathrm{~Hz}$
C
$530 \mathrm{~Hz}$
D
$545.5 \mathrm{~Hz}$

Answer

Correct option: A.

$550 \mathrm{~Hz}$

(a)$n^{\prime}=n\left(\frac{v}{v-v_S}\right) \Rightarrow n^{\prime}=500\left(\frac{330}{330-30}\right)=550 \mathrm{~Hz}$

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MCQ 1411 Mark

Two sound waves of slightly different frequencies propagating in the same direction produce beats due to

✓
Interference
B
Diffraction
C
Polarization
D
Refraction

Answer

Correct option: A.

Interference

(a)

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MCQ 1421 Mark

A man is watching two trains, one leaving and the other coming in with equal speeds of $4 \mathrm{~m} / \mathrm{sec}$. If they sound their whistles, each of frequency $240 \mathrm{~Hz}$, the number of beats heard by the man (velocity of sound in air $=320 \mathrm{~m} / \mathrm{sec}$ ) will be equal to

✓
$6$
B
$3$
C
$9$
D
$12$

Answer

Correct option: A.

$6$

Frequency of sound heard by the man from approaching train
$n_a=n\left(\frac{v}{v-v_s}\right)=240\left(\frac{320}{320-4}\right)=243 \ Hz$
Frequency of sound heard by the man from receding train
$n_r=n\left(\frac{v}{v+v_s}\right)=240\left(\frac{320}{320+4}\right)=237 \ Hz$
Hence, number of beats heard by man per sec
$=n_a-n_r=243-237=6$
Short trick : Number of beats heard per sec
$=\frac{2 n v v_S}{v^2-v_S^2}=\frac{2 n v v_S}{\left(v-v_S\right)\left(v+v_S\right)}=\frac{2 \times 240 \times 320 \times 4}{(320-4)(320+4)}=6$

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MCQ 1431 Mark

The equation of a transverse wave travelling on a rope is given by $y=10 \sin \pi(0.01 x-2.00 t)$ where $y$ and $x$ are in $\mathrm{cm}$ and $t$ in seconds. The maximum transverse speed of a particle in the rope is about

✓
$63 \mathrm{~cm} / \mathrm{s}$
B
$75 \mathrm{~cm} / \mathrm{s}$
C
$100 \mathrm{~cm} / \mathrm{s}$
D
$121 \mathrm{~cm} / \mathrm{s}$

Answer

Correct option: A.

$63 \mathrm{~cm} / \mathrm{s}$

The given equation is $y=10 \sin (0.01 \pi x-2 \pi t)$
Hence $\omega=$ coefficient of $t=2 \pi$
$\Rightarrow$ Maximum speed of the particle $v_{\max }=a \omega=10 \times 2 \pi$$=10 \times 2 \times 3.14=62.8 \approx 63 \mathrm{~cm} / \mathrm{s}$

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MCQ 1441 Mark

$41$ forks are so arranged that each produces $5$ beats per sec when sounded with its near fork. If the frequency of last fork is double the frequency of first fork, then the frequencies of the first and last fork are respectively

✓
$200,400$
B
$205,410$
C
$195,390$
D
$100,200$

Answer

Correct option: A.

$200,400$

(a) Similar to previous question
$n_{\infty}=n_{-}+(N-1) x $
$2 n=n \quad+(41-1) \times 5 $
$\Rightarrow n_{\infty}=200 \mathrm{~Hz} \text { and } n_{-}=400 \mathrm{~Hz}$

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MCQ 1451 Mark

A motor car blowing a horn of frequency $124 \mathrm{vib} / \mathrm{sec}$ moves with a velocity $72 \mathrm{~km} / \mathrm{hr}$ towards a tall wall. The frequency of the reflected sound heard by the driver will be (velocity of sound in air is 330 $\mathrm{m} / \mathrm{s})$

A
$109 \mathrm{vib} / \mathrm{sec}$
B
$132 \mathrm{vib} / \mathrm{sec}$
✓
$140 \mathrm{vib} / \mathrm{sec}$
D
$248 \mathrm{vib} / \mathrm{sec}$

Answer

Correct option: C.

$140 \mathrm{vib} / \mathrm{sec}$

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MCQ 1461 Mark

A wave is represented by the equation $y=7 \sin \left(7 \pi t-0.04 x \pi+\frac{\pi}{3}\right)$ $x$ is in metres and $t$ is in seconds. The speed of the wave is

✓
$175 \mathrm{~m} / \mathrm{sec}$
B
$49 \pi \mathrm{m} / \mathrm{sec}$
C
$49 \pi \mathrm{m} / \mathrm{sec}$
D
$0.28 \mathrm{\pi m} / \mathrm{sec}$

Answer

Correct option: A.

$175 \mathrm{~m} / \mathrm{sec}$

(a) $\quad v=\frac{\text { Co - efficient of } t}{\text { Co - efficient of } x}=\frac{7 \pi}{0.04}=175 \mathrm{~m} / \mathrm{s}$.

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MCQ 1471 Mark

When two sound waves are superimposed, beats are produced when they have

A
Different amplitudes and phases
B
Different velocities
C
Different phases
✓
Different frequencies

Answer

Correct option: D.

Different frequencies

(d) For producing beats, their must be small difference in frequency.

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MCQ 1481 Mark

The length of two open organ pipes are $l$ and $(l+\Delta l)$ respectively. Neglecting end correction, the frequency of beats between them will be approximately (Here $v$ is the speed of sound)

A
$\frac{v}{2 l}$
B
$\frac{v}{4 l}$
✓
$\frac{v \Delta l}{2 l^2}$
D
$\frac{v \Delta l}{l}$

Answer

Correct option: C.

$\frac{v \Delta l}{2 l^2}$

(c)$\lambda_1=2 l, \lambda_2=2 l+2 \Delta l$
$ \Rightarrow n_1=\frac{v}{2l}\text { and } n_2=\frac{v}{2 l+2 \Delta l} $
$\Rightarrow \text { No. of beats }=n_1-n_2=\frac{v}{2}\left(\frac{1}{l}-\frac{1}{l+\Delta l}\right)=\frac{v \Delta l}{2 l^2}$

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MCQ 1491 Mark

The equation of a stationary wave is $y=0.8 \cos \left(\frac{\pi x}{20}\right) \sin 200 \pi t$, where $x$ is in $\mathrm{cm}$ and $t$ is in sec. The separation between consecutive nodes will be

✓
$20 \mathrm{~cm}$
B
$10 \mathrm{~cm}$
C
$40 \mathrm{~cm}$
D
$30 \mathrm{~cm}$

Answer

Correct option: A.

$20 \mathrm{~cm}$

(a) On comparing the given equation with standard equation. $\left[y=2 a \sin \frac{2 \pi x}{\lambda} \cos \frac{2 \pi v t}{\lambda}\right]$
We get $\frac{2 \pi}{\lambda}=\frac{\pi}{20} \Rightarrow \lambda=40$
Separation between two consecutive nodes = $\frac{\lambda}{2}=\frac{40}{2}=20 \mathrm{~cm}$

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MCQ 1501 Mark

A source of sound is moving with constant velocity of $20 \mathrm{~m} / \mathrm{s}$ emitting a note of frequency $1000 \mathrm{~Hz}$. The ratio of frequencies observed by a stationary observer while the source is approaching him and after it crosses him will be(Speed of sound $v=340 \mathrm{~m} / \mathrm{s}$ )

✓
$9: 8$
B
$8: 9$
C
$1: 1$
D
$9: 10$

Answer

Correct option: A.

$9: 8$

(a) When source is approaching the observer, the frequency heard
$n_a=\left(\frac{v}{v-v_S}\right) \times n=\left(\frac{340}{340-20}\right) \times 1000=1063 \mathrm{~Hz}$
When source is receding, the frequency heard
$n_r=\left(\frac{v}{v+v_S}\right) \times n=\frac{340}{340+20} \times 1000=944 $
$\Rightarrow n_a: n_r=9: 8$
Short tricks : $\frac{n_a}{n_r}=\frac{v+v_S}{v-v_S}=\frac{340+20}{340-20}=\frac{9}{8}$.

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JEE physics MCQ