Tag: physics

Questions Related to physics

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

Show answer
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$s^{-1}$

Show answer
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. ${ 3 }/{ 2 }$

Show answer
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 } $
$\Rightarrow n=2$

Consider two conducting plates $A$ and $B$, between which the potential difference is $5 V$, plate $A$ being at a higher potential. A proton and an electron are released at plates $A$ and $B$ respectively. The two particles then move towards the opposite plates - the proton to plate $B$ and the electron to plate $A$. Which one will have a larger velocity when they reach their respective destination plates?

physics magnetic fields and electromagnetism contact and non-contact forces comparing force in magnetic, electric and gravitational fields identifying forces

  1. Both will have the same velocity

  2. The electron will have the larger velocity

  3. The proton will have the larger velocity

  4. None will be able to reach the destination point

Show answer
Correct Option: B
Explanation:

Between two conducting plates, there is the uniform electric field and therefore the same charge is applicable on both proton and neutron. Since Proton mass is more than that of the electron, it will have less acceleration and hence less speed achieved by it. Therefore electron will reach plate with larger velocity.