Tag: speed of a travelling wave

Questions Related to speed of a travelling wave

If the pressure of a fixed quantity of a gas is increased 4 times keeping the temperature constant, the r.m.s velocity will :

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

  1. get doubled

  2. get halved

  3. remain same

  4. get quadrupled

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Correct Option: C
Explanation:

RMS velocity is independent of pressure. Hence velocity does not change with variations in pressure

The correct option is (c)

If $C _{s}$ be the velocity of sound in air and $C$ be the rms velocity, then

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

  1. $C _{S} < C$

  2. $C _{s}=c$

  3. $C _{s}=C\left(\dfrac {\gamma}{3}\right)^{1/2}$

  4. $None\ of\ these$

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Correct Option: C
Explanation:

Speed of sound in air, ${{C} _{s}}=\sqrt{\dfrac{\gamma P}{\rho }}\,\ldots \ldots \,(1)$

 $ Where, $

$ \gamma =specific\,heat\,ratio $

$ P=\,pressure $

$ \rho =\,density $

RMS velocity of air molecule, $C=\sqrt{\dfrac{3\overline{R}T}{{{M} _{o}}}}=\sqrt{\dfrac{3P}{\rho }}\,\ldots \ldots \,(2)$

$ where,\, $

$ \overline{R}=\text{universal}\,\text{gas}\,\text{constant} $

$ {{M} _{o}}=Molecular\,mass $

$ T=temperature $

From (1) and (2)

${{C} _{s}}=C{{\left( \dfrac{\gamma }{3} \right)}^{1/2}}$ 

With increase in temperature, the rms speed and wave speed in a gas

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

  1. increases with temperature

  2. decreases with temperature

  3. are independent of temperature

  4. none of the above

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Correct Option: A
Explanation:

Both RMS speed and speed of sound in gas are directly proportional to temperature. Thus, both the speeds increases with temperature

The correct option is (a)

If nitrogen gas molecule goes straight up with its rms speed at $0^o$C from the surface of the earth and there are no collisions with other molecules, then it will rise to an approximate height of:

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

  1. $18$ km

  2. $15$ km

  3. $12.38$ km

  4. $8$ km

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Correct Option: C
Explanation:

Molecular mass of Nitrogen molecule=$14$ g/mol

As nitrogen exists as ${N} _{2}$=28 g/mol=$0.028$ kg/mol
Also we know ${ v } _{ rms }=\sqrt { \dfrac { 3RT }{ M }  } $  where R= gas constant=8.31 bar/(K mol)=8.31$\times{10}^{5}$ Pa/(K mol)
T= temperature=${0}^{0}$ C=273 K
Also height $=\dfrac{{V}^{2} _{rms}}{2g}$

$=\dfrac { 3\times 8.31\times { 10 }^{ 5 }\times 273 }{ 2\times 9.81\times 0.028 } \ =12388\quad m=12.38\quad km$

RMS speed of sound varies with change in composition of the medium (diatomic, monoatomic, etc), in which the wave travels

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

  1. True

  2. False

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Correct Option: B
Explanation:

The RMS speed of sound is directly proportional to square root of temperature. Hence, it is independent of the properties of the medium

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 } $
$\Rightarrow n=2$