MCQ 11 Mark
The walls of the hall built for music concerns should
- Aamplify sound
- BReflect sound
- Ctransmit sound
- ✓Absorb sound
Answer
View full question & answer→Correct option: D.
Absorb sound
Questions
M.C.Q (1 Marks)
Take a timed test
47 questions · 46 auto-graded MCQ + 1 self-marked written.
MCQ 21 Mark
Speed of sound is maximum in
- Aair
- Bwater
- Cvacuum
- ✓solid
Answer
View full question & answer→Correct option: D.
solid
MCQ 31 Mark
The Laplace’s correction in the expression for velocity of sound given by Newton is needed because sound waves
- Aare longitudinal
- Bpropagate isothermally
- ✓propagate adiabatically
- Dare of long wavelength
Answer
View full question & answer→Correct option: C.
propagate adiabatically
MCQ 41 Mark
When sound waves travel from air to water, which of these remains constant ?
- AVelocity
- ✓Frequency
- CWavelength
- DAll of above
Answer
View full question & answer→Correct option: B.
Frequency
MCQ 51 Mark
A sound carried by the air from a sitar to a listener is a wave of the following type.
- ALongitudinal stationary
- BTransverse progressive
- CTransverse stationery
- ✓Longitudinal progressive
Answer
View full question & answer→Correct option: D.
Longitudinal progressive
MCQ 61 Mark
When both source and listener are approaching each other the observed frequency of sound is given by ( $V_L$ and $V_S$ is the velocity of listener and source respectively, $n_0=$ radiated frequency)
- ✓$n=n_0\left[\frac{V+V_L}{V-V_S}\right]$
- B$n=n_0\left[\frac{V-V_L}{V+V_S}\right]$
- C$n=n_0\left[\frac{V-V_L}{V-V_S}\right]$
- D$n=n_0\left[\frac{V+V_L}{V+V_S}\right]$
Answer
View full question & answer→Correct option: A.
$n=n_0\left[\frac{V+V_L}{V-V_S}\right]$
(a) :As, both source and listener are approaching to each other.
So, by Doppler's effect
$
n=n_0\left(\frac{V+V_L}{V-V_S}\right)
$
So, by Doppler's effect
$
n=n_0\left(\frac{V+V_L}{V-V_S}\right)
$
MCQ 71 Mark
Sound waves of frequency $600 Hz$ fall normally on a perfectly reflecting wall. The shortest distance from the wall at which all particles will have maximum amplitude of vibration is (speed of sound $=300 m s ^{-1}$ )
- A$\frac{1}{4} m$
- ✓$\frac{1}{8} m$
- C$\frac{3}{8} m$
- D$\frac{7}{8} m$
Answer
View full question & answer→Correct option: B.
$\frac{1}{8} m$
(b) :The minimum distance of rarefaction is $\frac{\lambda}{4}$ As, $v=\lambda f$
$
\lambda=\frac{v}{f}=\frac{300}{600}=\frac{1}{2} m ; \quad d=\frac{\lambda}{4}=\frac{1}{8} m
$
$
\lambda=\frac{v}{f}=\frac{300}{600}=\frac{1}{2} m ; \quad d=\frac{\lambda}{4}=\frac{1}{8} m
$
MCQ 81 Mark
A source of sound is moving towards a stationary observer with $\left(\frac{1}{10}\right)^{\text {th }}$ of the speed of sound. The ratio of apparent to real frequency is
- ✓$10: 9$
- B$11: 10$
- C$(11)^2:(10)^2$
- D$(9)^2:(10)^2$
Answer
View full question & answer→Correct option: A.
$10: 9$
(a) : $V_{\text {sound }}=v, V_{\text {source }}=v / 10$
$
f^{\prime}=f\left(\frac{v-0}{v-\frac{v}{10}}\right)=\frac{10}{9} f ; \frac{f^{\prime}}{f}=\frac{10}{9}
$
$
f^{\prime}=f\left(\frac{v-0}{v-\frac{v}{10}}\right)=\frac{10}{9} f ; \frac{f^{\prime}}{f}=\frac{10}{9}
$
MCQ 91 Mark
A progressive wave is given by $y=12 \sin (5 t-4 x)$. On this wave, how far away are the two points having a phase difference of $90^{\circ}$ ?
- A$\frac{\pi}{4}$
- ✓$\frac{\pi}{8}$
- C$\frac{\pi}{16}$
- D$\frac{\pi}{32}$
Answer
View full question & answer→Correct option: B.
$\frac{\pi}{8}$
(b) :$y=12 \sin (5 t-4 x)$
Here, phase is $5 t-4 x$. Let two points $x_1$ and $x_2$ differ
$
\begin{aligned}
& \text { by } \frac{\pi}{2} \\
& \Rightarrow\left(5 t-4 x_1\right)-\left(5 t-4 x_2\right)=\frac{\pi}{2} \\
& 4\left(x_2-x_1\right)=\frac{\pi}{2} ; x_2-x_1=\frac{\pi}{8} \\
&
\end{aligned}
$
Here, phase is $5 t-4 x$. Let two points $x_1$ and $x_2$ differ
$
\begin{aligned}
& \text { by } \frac{\pi}{2} \\
& \Rightarrow\left(5 t-4 x_1\right)-\left(5 t-4 x_2\right)=\frac{\pi}{2} \\
& 4\left(x_2-x_1\right)=\frac{\pi}{2} ; x_2-x_1=\frac{\pi}{8} \\
&
\end{aligned}
$
MCQ 101 Mark
The equation of the wave is $y=10 \sin \left(\frac{2 \pi t}{30}+\alpha\right)$. If the displacement is $5 cm$ at $t=0$ then the total phase at $t=7.5 s$ will be $\left(\sin 30^{\circ}=0.5\right)$
- A$\frac{\pi}{3} rad$
- B$\frac{\pi}{2} rad$
- C$\frac{2 \pi}{5} rad$
- ✓$\frac{2 \pi}{3} rad$
Answer
View full question & answer→Correct option: D.
$\frac{2 \pi}{3} rad$
(d) : $y=10 \sin \left(\frac{2 \pi}{30}+\alpha\right)$
At $t=0, y=5 cm \quad \therefore 5=10 \sin \alpha$ $\sin \alpha=\frac{1}{2} \Rightarrow \alpha=\frac{\pi}{6}$ radian
$\begin{aligned} \text { Total phase } & =\frac{2 \pi t}{30}+\alpha=\frac{2 \pi}{30} \times 7.5+\frac{\pi}{6} \\ & =\frac{\pi}{2}+\frac{\pi}{6}=\frac{3 \pi+\pi}{6}=\frac{4 \pi}{6}=\frac{2 \pi}{3}\end{aligned}$
At $t=0, y=5 cm \quad \therefore 5=10 \sin \alpha$ $\sin \alpha=\frac{1}{2} \Rightarrow \alpha=\frac{\pi}{6}$ radian
$\begin{aligned} \text { Total phase } & =\frac{2 \pi t}{30}+\alpha=\frac{2 \pi}{30} \times 7.5+\frac{\pi}{6} \\ & =\frac{\pi}{2}+\frac{\pi}{6}=\frac{3 \pi+\pi}{6}=\frac{4 \pi}{6}=\frac{2 \pi}{3}\end{aligned}$
MCQ 111 Mark
The displacement of a wave travelling in the $x$ direction is $y=10^{-4} \sin \left[600 t-2 x+\frac{\pi}{3}\right] m$, where $x$ is in metre and $t$ in second. The speed of the wave is
- ✓$300 m / s$
- B$200 m / s$
- C$150 m / s$
- D$600 m / s$
Answer
View full question & answer→Correct option: A.
$300 m / s$
(a) :The given equation,
$
y=10^{-4} \sin \left[600 t-2 x+\frac{\pi}{3}\right] m
$
On comparing this equation with standard equation with standard equation, $y=A \sin (\omega t-k x)$, we get
$
\begin{aligned}
\omega & =600 ; k=2 \\
v & =\frac{\omega}{k}=\frac{600}{2}=300 m / s
\end{aligned}
$
$
y=10^{-4} \sin \left[600 t-2 x+\frac{\pi}{3}\right] m
$
On comparing this equation with standard equation with standard equation, $y=A \sin (\omega t-k x)$, we get
$
\begin{aligned}
\omega & =600 ; k=2 \\
v & =\frac{\omega}{k}=\frac{600}{2}=300 m / s
\end{aligned}
$
MCQ 121 Mark
Sound of frequency $1000 Hz$ from a stationary source is reflected from an object approaching the source at $30 m s ^{-1}$, back to a stationary observer located at the source. The speed of sound in air is $330 m s ^{-1}$. The frequency of the sound heard by the observer is
- ✓$1200 Hz$
- B$1000 Hz$
- C$1090 Hz$
- D$1100 Hz$
Answer
View full question & answer→Correct option: A.
$1200 Hz$
(a) :The frequency of sound from the stationary source reaches the object which is approaching towards the source is
$
n^{\prime}=n_0\left(\frac{v+v_o}{v}\right)
$
where, $n_0=$ frequency of source, $v=$ speed of sound in air and $v_o=$ speed of the object.
The frequency of reflected sound from the object heard by the stationary observer located at the source is
$
\begin{aligned}
& n^{\prime \prime}=n^{\prime}\left(\frac{v}{v-v_o}\right)=n_0\left(\frac{v+v_o}{v-v_o}\right) \\
& =1000\left(\frac{330+30}{330-30}\right)=1200 Hz
\end{aligned}
$
$
n^{\prime}=n_0\left(\frac{v+v_o}{v}\right)
$
where, $n_0=$ frequency of source, $v=$ speed of sound in air and $v_o=$ speed of the object.
The frequency of reflected sound from the object heard by the stationary observer located at the source is
$
\begin{aligned}
& n^{\prime \prime}=n^{\prime}\left(\frac{v}{v-v_o}\right)=n_0\left(\frac{v+v_o}{v-v_o}\right) \\
& =1000\left(\frac{330+30}{330-30}\right)=1200 Hz
\end{aligned}
$
MCQ 131 Mark
When source of sound moves towards a stationary observer, the wavelength of sound received by him
- ✓decreases while frequency increases.
- Bremains the same whereas frequency increases.
- Cincreases and frequency also increases.
- Ddecreases while frequency remains the same.
Answer
View full question & answer→Correct option: A.
decreases while frequency increases.
(a) :The apparent frequency of sound during the relative motion of source and observer is given by
$
n_a=n\left[\frac{v \pm v_o}{v \mp v_s}\right]
$
Since observer is at rest at $v_o=0$
$
n_a=n\left[\frac{v}{v-v_s}\right]
$
So, the frequency increase, as wavelength decreases.
$
n_a=n\left[\frac{v \pm v_o}{v \mp v_s}\right]
$
Since observer is at rest at $v_o=0$
$
n_a=n\left[\frac{v}{v-v_s}\right]
$
So, the frequency increase, as wavelength decreases.
MCQ 141 Mark
The observer is moving with velocity $v_0$ towards the stationary source of sound and then after crossing moves away from the source with velocity $v_0$. Assume that the medium through which the sound waves travel is at rest. If $v$ is the velocity of sound and $n$ is the frequency emitted by the source then the difference between apparent frequencies heard by the observer is
- ✓$\frac{2 n v_0}{v}$
- B$\frac{n v_0}{v}$
- C$\frac{v}{2 n v_0}$
- D$\frac{v}{n v_0}$
Answer
View full question & answer→Correct option: A.
$\frac{2 n v_0}{v}$
(a) :The frequency heard by the observer, when moving towards the stationary source
$
n^{\prime}=n\left(\frac{v+v_0}{v}\right)
$
The frequency heard by the observer, when moving away from the stationary source
$
n^{\prime \prime}=n\left(\frac{v-v_0}{v}\right)
$
The difference between apparent frequencies $=n^{\prime}-n^{\prime \prime}$
$
=\frac{n}{v}\left[v+v_0-v+v_0\right]=\frac{2 n v_0}{v}
$
$
n^{\prime}=n\left(\frac{v+v_0}{v}\right)
$
The frequency heard by the observer, when moving away from the stationary source
$
n^{\prime \prime}=n\left(\frac{v-v_0}{v}\right)
$
The difference between apparent frequencies $=n^{\prime}-n^{\prime \prime}$
$
=\frac{n}{v}\left[v+v_0-v+v_0\right]=\frac{2 n v_0}{v}
$
MCQ 151 Mark
When the observer moves towards the stationary source with velocity, $V_1$, the apparent frequency of emitted note is $F_1$. When the observer moves away from the source with velocity $V_1$, the apparent frequency is $F_2$. If $V$ is the velocity of sound in air and $\frac{F_1}{F_2}=2$ then $\frac{V}{V_1}=$ ?
- A2
- ✓3
- C4
- D5
Answer
View full question & answer→Correct option: B.
3
(b) :When the observer moves towards the stationary source, apparent frequency of emitted note,
$
F_1=\left(\frac{V+V_1}{V}\right) n ...(i)
$
When the observer moves away from the stationary source, apparent frequency of emitted note
$
F_2=\left(\frac{V-V_1}{V}\right) n
$ ...(ii)
Dividing eqn. (i) by eqn. (ii), we get
$
\begin{aligned}
& \frac{F_1}{F_2}=\frac{V+V_1}{V-V_1} \\
& \because \quad \frac{F_1}{F_2}=2 \text { (given) } \therefore \quad \frac{V+V_1}{V-V_1}=2 \\
& \Rightarrow \quad V+V_1=2 V-2 V_1 \Rightarrow V=3 V_1 \text { or } \frac{V}{V_1}=3
\end{aligned}
$
$
F_1=\left(\frac{V+V_1}{V}\right) n ...(i)
$
When the observer moves away from the stationary source, apparent frequency of emitted note
$
F_2=\left(\frac{V-V_1}{V}\right) n
$ ...(ii)
Dividing eqn. (i) by eqn. (ii), we get
$
\begin{aligned}
& \frac{F_1}{F_2}=\frac{V+V_1}{V-V_1} \\
& \because \quad \frac{F_1}{F_2}=2 \text { (given) } \therefore \quad \frac{V+V_1}{V-V_1}=2 \\
& \Rightarrow \quad V+V_1=2 V-2 V_1 \Rightarrow V=3 V_1 \text { or } \frac{V}{V_1}=3
\end{aligned}
$
MCQ 161 Mark
The pitch of the whistle of an engine appears to drop to $\left(\frac{5}{6}\right)^{\text {th }}$ of original value when it passes a stationary observer. If the speed of sound in air is $350 m / s$ then the speed of engine is
- A$35 m / s$
- ✓$70 m / s$
- C$105 m / s$
- D$140 m / s$
Answer
View full question & answer→Correct option: B.
$70 m / s$
(b) :Pitch of sound depends on its frequency.
Speed of sound, $v=350 m s ^{-1}$
Speed of engine, $u_s=$ ?
Apparent pitch (frequency) heard by observer $=\left(\frac{5}{6}\right)^{\text {th }}$ of original pitch (frequency) of engine.
Let original frequency of the engine be $v_0$.
Apparent frequency heard by stationary observer $=0$
Then $v=\left(\frac{5}{6}\right) v_0$
$\because \quad$ Engine is moving away from the source
$
\therefore \quad v=\left(\frac{v}{v+u_s}\right) v_0
$
So, $\frac{5}{6} v_0=\frac{350}{350+u_s} \times v_0$
$
350+u_s=420 \therefore u_s=70 m s ^{-1} .
$
Speed of sound, $v=350 m s ^{-1}$
Speed of engine, $u_s=$ ?
Apparent pitch (frequency) heard by observer $=\left(\frac{5}{6}\right)^{\text {th }}$ of original pitch (frequency) of engine.
Let original frequency of the engine be $v_0$.
Apparent frequency heard by stationary observer $=0$
Then $v=\left(\frac{5}{6}\right) v_0$
$\because \quad$ Engine is moving away from the source
$
\therefore \quad v=\left(\frac{v}{v+u_s}\right) v_0
$
So, $\frac{5}{6} v_0=\frac{350}{350+u_s} \times v_0$
$
350+u_s=420 \therefore u_s=70 m s ^{-1} .
$
MCQ 171 Mark
A plane sound wave travelling with velocity $v$ in a medium $A$ reaches a point on the interface of medium $A$ and medium $B$. If velocity of sound in medium $B$ is $2 v$, the angle of incidence for total internal reflection of the wave will be greater than $\left(\sin 30^{\circ}=0.5\right.$ and $\left.\sin 90^{\circ}=1\right)$
- A$15^{\circ}$
- ✓$30^{\circ}$
- C$45^{\circ}$
- D$90^{\circ}$
Answer
View full question & answer→Correct option: B.
$30^{\circ}$
(b) :Let $i$ be angle of incidence and $i_c$ be the critical angle.
For total internal reflection, $i>i_c$
Here, $\sin i_c=\frac{v}{2 v}=\frac{1}{2}$
or $i_c=\sin ^{-1}\left(\frac{1}{2}\right)=\sin ^{-1}(0.5)=30^{\circ} \quad \therefore \quad i>30^{\circ}$
For total internal reflection, $i>i_c$
Here, $\sin i_c=\frac{v}{2 v}=\frac{1}{2}$
or $i_c=\sin ^{-1}\left(\frac{1}{2}\right)=\sin ^{-1}(0.5)=30^{\circ} \quad \therefore \quad i>30^{\circ}$
MCQ 181 Mark
The formula for speed of a transverse wave on a stretched spring is ……………… (m = linear mass density, T = tension in Spring)
- A$V =\sqrt{\frac{m}{T}}$
- ✓$V =\sqrt{\frac{T}{m}}$
- C$V =\left(\frac{m}{T}\right)^{\frac{3}{2}}$
- D$V =\left(\frac{T}{m}\right)^{\frac{3}{2}}$
Answer
View full question & answer→Correct option: B.
$V =\sqrt{\frac{T}{m}}$
$V =\sqrt{\frac{T}{m}}$
MCQ 191 Mark
The working of SONAR is based on …………………
- Aresonance
- Bspeed of a star
- ✓Doppler effect
- Dspeed of rotation of sun
Answer
View full question & answer→Correct option: C.
Doppler effect
Doppler effect
MCQ 201 Mark
If the source is moving away from the observer, then the apparent frequency …………..
- Awill increase
- Bwill remain the same
- Cwill be zero
- ✓will decrease
Answer
View full question & answer→Correct option: D.
will decrease
will decrease
MCQ 211 Mark
If the velocity of sound in hydrogen is $1248 \ m/s,$ the velocity of sound in oxygen is $[$Given $: M_O = 32$ and $M_H = 2]$
- A$1248 \ m/s$
- B$624 \ m/s$
- ✓$312 \ m/s$
- D$300 \ m/s$
Answer
View full question & answer→Correct option: C.
$312 \ m/s$
$312 \ m/s$
MCQ 221 Mark
The speed of sound in air at $\text{NTP}$ is $330 \ m/s.$ The period of sound wave of wavelength $66 \ cm$ is $………$
- A$0.2 s$
- B$0.1$
- C$0.1 \times 10^{-2} s$
- ✓$0.2 \times 10^{-2} s$
Answer
View full question & answer→Correct option: D.
$0.2 \times 10^{-2} s$
$0.2 \times 10^{-2} s$
MCQ 231 Mark
A thunder clap was heard 6 seconds after a lightening flash was seen. If the speed of sound in air is 340 m/s at the time of observation, the distance of the listener from the thunder clap is ………………
- A56.6 m
- B346 m
- C1020 m
- ✓2040 m
Answer
View full question & answer→Correct option: D.
2040 m
2040 m
MCQ 241 Mark
A sound note emitted from a certain source has a velocity of 300 m/s in air and 1050 m/s in water. If the wavelength of sound note in air is 2 m, the wavelength in water is …………
- A2 m
- B6 m
- ✓7 m
- D12 m
Answer
View full question & answer→Correct option: C.
7 m
7 m
MCQ 251 Mark
Choose the correct statement.
- ✓For 1 °C rise in temperature, velocity of sound increases by 0.61 m/s.
- BFor 1 °C rise in temperature, velocity of sound decreases by 0.61 m/s.
- CFor $1^{\circ} C$ rise in temperature, velocity of sound decreases by $\frac{1}{273} m / s$.
- DFor $1^{\circ} C$ rise in temperature, velocity of sound increases by $\frac{1}{273} m / s$.
Answer
View full question & answer→Correct option: A.
For 1 °C rise in temperature, velocity of sound increases by 0.61 m/s.
For 1 °C rise in temperature, velocity of sound increases by 0.61 m/s.
MCQ 261 Mark
The velocity of sound in a gas is 340 m/s at the pressure P, what will be the velocity of the gas when only pressure is doubled and temperature same?
- A170 m/s
- B243 m/s
- ✓340 m/s
- D680 m/s
Answer
View full question & answer→Correct option: C.
340 m/s
340 m/s
MCQ 271 Mark
The wavelength of sound in air is 1.5 m and that in liquid is 2 m. If the velocity of sound in air is 330 m/s, the velocity of sound in liquid is
- A330 m/s
- ✓440 m/’s
- C495 m/s
- D660 m/s
Answer
View full question & answer→Correct option: B.
440 m/’s
440 m/’s
MCQ 281 Mark
Wavelength of the transverse wave is 30 cm. If the particle at some instant has displacement 2 cm, find the displacement of the particle 15 cm away at the same instant.
- A2 cm
- B17 cm
- ✓-2 cm
- D-17 cm
Answer
View full question & answer→Correct option: C.
-2 cm
-2 cm
MCQ 291 Mark
Longitudinal waves CANNOT be …………………
- Areflected
- Brefracted
- Cscattered
- ✓polarised
Answer
View full question & answer→Correct option: D.
polarised
polarised
MCQ 301 Mark
The speed of the mechanical wave depends upon ………………
- Aelastic properties of the medium only.
- Bdensity of the medium only.
- ✓elastic properties and density of the medium
- Dinitial speed.
Answer
View full question & answer→Correct option: C.
elastic properties and density of the medium
elastic properties and density of the medium
MCQ 311 Mark
Wave motion has ……………………
- Asingle periodicity.
- ✓double periodicity.
- Conly periodicity in space.
- Donly periodicity in time.
Answer
View full question & answer→Correct option: B.
double periodicity.
double periodicity.
MCQ 321 Mark
If pressure of air gets doubled at constant temperature then velocity of sound in air ……………….
- Agets doubled
- ✓remains unchanged
- C√2 times initial velocity
- Dbecomes half
Answer
View full question & answer→Correct option: B.
remains unchanged
remains unchanged
MCQ 331 Mark
In a transverse wave, are regions of negative displacement.
- Ararefactions
- Bcompressions
- Ccrests
- ✓troughs
Answer
View full question & answer→Correct option: D.
troughs
troughs
MCQ 341 Mark
At normal temperature, for an echo to be heard the reflecting surface should be at a minimum distance of ………………. m.
- A34.4
- ✓17.2
- C10
- D20
Answer
View full question & answer→Correct option: B.
17.2
17.2
MCQ 351 Mark
For a progressive wave, in the usual notation
- A$V=\lambda T$
- ✓$n =\frac{V}{\lambda}$
- C$T =\lambda V$
- D$\lambda=\frac{1}{n}$
Answer
View full question & answer→Correct option: B.
$n =\frac{V}{\lambda}$
$n =\frac{V}{\lambda}$
MCQ 361 Mark
If speed of sound in air at 0°C is 331 m/s. What will be its value at 35° C?
- A331 m/s
- B366 m/s
- ✓351.6 m/s
- D332 m/s.
Answer
View full question & answer→Correct option: C.
351.6 m/s
351.6 m/s
MCQ 371 Mark
If the bulk modulus of water is 2100 MPa, what is the speed of sound in water?
- ✓1450 m/s
- B2100 m/s
- C0.21 m/s
- D21 m/s
Answer
View full question & answer→Correct option: A.
1450 m/s
1450 m/s
MCQ 381 Mark
A radio station broadcasts at 760 kHz. What is the wavelength of the station?
- ✓395 m
- B790 m
- C760 m
- D197.5 m
Answer
View full question & answer→Correct option: A.
395 m
395 m
MCQ 391 Mark
A series of ocean waves, each 5.0 m from crest to crest, moving past the observer at a rate of 2 waves per second have wave velocity
- A2.5 m/s
- B5.0 m/s
- C8.0 m/s
- ✓10.0 m/s
Answer
View full question & answer→Correct option: D.
10.0 m/s
10.0 m/s
MCQ 401 Mark
The velocity of sound in air at NTP is 330 m/s. What will be its value when temperature is doubled and pressure is halved?
- A330 m/s
- B165 m/s
- C330 √2 m/s
- ✓$\frac{330}{\sqrt{ } 2} m / s$
Answer
View full question & answer→Correct option: D.
$\frac{330}{\sqrt{ } 2} m / s$
$\frac{330}{\sqrt{ } 2} m / s$
MCQ 411 Mark
The temperature at which speed of sound in air becomes double its value at 0 °C is ……………….
- A546 °C
- ✓819 °C
- C273 °C
- D1092 °C
Answer
View full question & answer→Correct option: B.
819 °C
819 °C
MCQ 421 Mark
With decrease in water vapour content in air, velocity of sound …………………..
- Aincreases
- ✓decreases
- Cremains constant
- Dcannot say
Answer
View full question & answer→Correct option: B.
decreases
decreases
MCQ 431 Mark
At a given temperature, velocity of sound in oxygen and in hydrogen has the ratio …………………
- A4:1
- B1:4
- C1:1
- D2:1
Answer
View full question & answer→1 : 4
MCQ 441 Mark
In a gas, velocity of sound varies directly as ………………
- Asquare root of isothermal elasticity.
- Bsquare of isothermal elasticity.
- ✓square root of adiabatic elasticity.
- Dadiabatic elasticity.
Answer
View full question & answer→Correct option: C.
square root of adiabatic elasticity.
square root of adiabatic elasticity.
MCQ 451 Mark
At room temperature, velocity of sound in air at 10 atmospheric pressure and at 1 atmospheric pressure will be in the ratio ……………..
- A10:1
- B1 : 10
- ✓1 : 1
- Dcannot say
Answer
View full question & answer→Correct option: C.
1 : 1
1 : 1
MCQ 461 Mark
Sound travels fastest in ……………..
- Awater
- Bair
- ✓steel
- Dkerosene oil
Answer
View full question & answer→Correct option: C.
steel
steel
MCQ 471 Mark
Water waves are …………….
- Alongitudinal
- Btransverse
- ✓both longitudinal and transverse
- Dneither longitudinal nor transverse
Answer
View full question & answer→Correct option: C.
both longitudinal and transverse
both longitudinal and transverse