Tag: waves

Questions Related to waves

The velocity of sound in a gas is 300 m$s^{-1}$. The root means square velocity of the molecules is of the order of

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

  1. 4 m$s^{-1}$

  2. 40 m$s^{-1}$

  3. 400 m$s^{-1}$

  4. 4000 m$s^{-1}$

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Correct Option: C
Explanation:

$\displaystyle v _{mix} = v _{sound} \sqrt \frac{3}{\gamma} = 400 m/s$

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.

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

  1. $v _{rms} > v _{av} > v _{mp} $

  2. $v _{av} > v _{mp} > v _{rms}$

  3. $v _{mp} > v _{av} > v _{rms} $

  4. $v _{rms} > v _{av} = v _{mp}$

Show answer
Correct Option: A
Explanation:

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

so the answer is A.

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:-

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

  1. $ \dfrac {1}{4} $

  2. $ \dfrac {1}{2} $

  3. $2$

  4. $4$

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Correct Option: B

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 

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

  1. $3$

  2. $2$

  3. $1.5$

  4. $2.5$

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Correct Option: B

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

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

  1. $1$

  2. $3$

  3. $2$

  4. ${ 3 }/{ 2 }$

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Correct Option: C
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 } $
$\Rightarrow n=2$

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

intensity and loudness waves physics

  1. Producing sound of higher pitch

  2. Producing bond sound

  3. Producing sound of higher quality

  4. Decreasing purpose

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Correct Option: A

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

intensity and loudness waves physics

  1. 8

  2. 12

  3. 13

  4. 16

Show answer
Correct Option: D

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

intensity and loudness waves physics

  1. Amplitude

  2. Freqency

  3. Time period

  4. Pitch

Show answer
Correct Option: A
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.

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

intensity and loudness waves physics

  1. Loudness.

  2. Wavelength.

  3. Frequency.

  4. Waveform.

Show answer
Correct Option: A
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.

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

intensity and loudness waves physics

  1. waveform

  2. loudness

  3. wavelength

  4. intensity

Show answer
Correct Option: B
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.