Tag: waves

Quizzes (32) Questions (233)

Questions Related to waves

Multiple choice speed of sound in gas speed of a travelling wave oscillation and waves waves physics

If $v _{rms}$ = root mean square speed of molecules
$v _{av}$ = average speed of molecules
$v _{mp}$ = most probable speed of molecules
Then, identify the correct relation between these speeds.

  1. $v _{rms} > v _{av} > v _{mp} $
  2. $v _{av} > v _{mp} > v _{rms}$
  3. $v _{mp} > v _{av} > v _{rms} $
  4. $v _{rms} > v _{av} = v _{mp}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

root mean square speed of molecules > average speed of molecules > most probable speed of molecules 

so the answer is A.

Multiple choice speed of sound in gas speed of a travelling wave oscillation and waves waves physics

The velocity of sound at the same pressure in two monoatomic gases of densities $ \rho _1$  and $\rho _2$ are $v _1$ and $v _2 $ respectively. If $ \dfrac {\rho _1}{\rho _2} = 4 $ then the value of $ \dfrac {v _1}{v _2} $ is:-

  1. $ \dfrac {1}{4} $
  2. $ \dfrac {1}{2} $
  3. $2$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Velocity v = sqrt(gamma * P / rho). Since pressure P and gamma are constant, v is inversely proportional to sqrt(rho). Thus, v1 / v2 = sqrt(rho2 / rho1). Given rho1 / rho2 = 4, then rho2 / rho1 = 1/4. So v1 / v2 = sqrt(1/4) = 1/2.

Multiple choice speed of sound in gas speed of a travelling wave oscillation and waves waves physics

Two moles of hydrogen are mixed with n moles of helium. The root mean square speed of gas molecules in the mixture is $\sqrt2$ times the speed of sound in the mixture. Then n is 

  1. $3$
  2. $2$
  3. $1.5$
  4. $2.5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

v_rms = sqrt(3RT/M_mix). v_sound = sqrt(gamma_mix * RT / M_mix). Given v_rms = sqrt(2) * v_sound, then 3RT/M_mix = 2 * gamma_mix * RT / M_mix, so gamma_mix = 1.5. For a mixture, gamma = (n1Cp1 + n2Cp2) / (n1Cv1 + n2Cv2). With 2 moles H2 (gamma=1.4, Cv=2.5R) and n moles He (gamma=1.67, Cv=1.5R), solving for n yields 2.

Multiple choice speed of sound in gas speed of a travelling wave oscillation and waves waves physics

Two moles of helium are mixed with $n$ moles of hydrogen. The root mean square $\left( rms \right) $ speed of gas molecules in the mixture is $\sqrt { 2 } $ times the speed of sound in the mixture. Then, the value of $n$ is

  1. $1$
  2. $3$
  3. $2$
  4. ${ 3 }/{ 2 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\because { v } _{ rms }=\sqrt { \dfrac { 3RT }{ M }  } $ and ${ v } _{ sound }=\sqrt { \dfrac { \gamma RT }{ M }  } $,
${ v } _{ rms }=2{ v } _{ sound }$
i.e. $\gamma =\dfrac { 3 }{ 2 } =$ ratio of $\dfrac { { C } _{ p } }{ { C } _{ V } } $ for the mixture
${ C } _{ V }=\dfrac { { n } _{ 1 }{ C } _{ { V } _{ 1 } }+{ n } _{ 2 }{ C } _{ { V } _{ 2 } } }{ { n } _{ 1 }+{ n } _{ 2 } } $
and ${ C } _{ p }=\dfrac { { n } _{ 1 }{ c } _{ { p } _{ 1 } }+{ n } _{ 2 }{ C } _{ { p } _{ 2 } } }{ { n } _{ 1 }+{ n } _{ 2 } } $
$\therefore \gamma =\dfrac { { C } _{ p } }{ { C } _{ V } } =\dfrac { { n } _{ 1 }{ C } _{ { p } _{ 1 } }+{ n } _{ 2 }{ C } _{ { p } _{ 2 } } }{ { n } _{ 1 }{ C } _{ { V } _{ 1 } }+{ n } _{ 2 }{ C } _{ { V } _{ 2 } } } $
$\therefore \dfrac { 3 }{ 2 } =\dfrac { 2\left( \dfrac { 5 }{ 2 } R \right) +n\left( \dfrac { 7 }{ 2 } R \right)  }{ 2\left( \dfrac { 3 }{ 2 } R \right) +n\left( \dfrac { 5 }{ 2 } R \right)  } $
$\Rightarrow \dfrac { 3 }{ 2 } =\dfrac { 10+7n }{ 6+5n } $
$\Rightarrow n=2$

Multiple choice intensity and loudness waves physics

The balls of a walleye or sample are made of large size. It is of for 

  1. Producing sound of higher pitch

  2. Producing bond sound

  3. Producing sound of higher quality

  4. Decreasing purpose
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice intensity and loudness waves physics

The power of sound from speaker of radio is 10 W, the power of sound from the speaker of radio is 400 W by increasing the volume of radio. The power increased in dB as compared to original power is nearly

  1. 8

  2. 12

  3. 13

  4. 16
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The increase in decibels is calculated as 10 * log10(P2/P1). Here, 10 * log10(400/10) = 10 * log10(40) = 10 * 1.602 = 16.02 dB.

Multiple choice intensity and loudness waves physics

State the factors that determine the loudness of the sound heard.

  1. Amplitude

  2. Freqency

  3. Time period

  4. Pitch
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Amplitude is the size of the vibration, and this determines how loud the sound is Larger vibrations make a louder sound.The intensity or loudness of a sound depends upon the extent to which the sounding body vibrates, i.e., the amplitude of vibration. A sound is louder as the amplitude of vibration is greater, and the intensity decreases as the distance from the source increases. Hence, loudness is dependent on amplitude.

Multiple choice intensity and loudness waves physics

Name the characteristic of the sound affected due to a change in its amplitude.

  1. Loudness.

  2. Wavelength.

  3. Frequency.

  4. Waveform.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Amplitude is the size of the vibration, and this determines how loud the sound is.  Larger vibrations make a louder sound.
The intensity or loudness of a sound depends upon the extent to which the sounding body vibrates, i.e., the amplitude of vibration. A sound is louder as the amplitude of vibration is greater, and the intensity decreases as the distance from the source increases.
Hence, loudness of the sound is affected due to the change in its amplitude.

Multiple choice intensity and loudness waves physics

As we move away from the source along with the amplitude which of the following quantity decreases as well ?

  1. waveform

  2. loudness

  3. wavelength

  4. intensity
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The intensity or loudness of a sound depends upon the extent to which the sounding body vibrates, i.e., the amplitude of vibration. A sound is louder as the amplitude of vibration is greater, and the intensity decreases as the distance from the source increases. Loudness is measured in units called decibels. 
Loudness depends on the square of the amplitude of the wave. That is, $loudness{ \propto \left( amplitude \right)  }^{ 2 }$.
Hence, As it moves away from the source its amplitude as well as its loudness decreases.