Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

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

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

  1. $C _{S} < C$
  2. $C _{s}=c$
  3. $C _{s}=C\left(\dfrac {\gamma}{3}\right)^{1/2}$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
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)

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

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

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

  1. increases with temperature

  2. decreases with temperature

  3. are independent of temperature

  4. none of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
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)

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

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:

  1. $18$ km
  2. $15$ km
  3. $12.38$ km
  4. $8$ km
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $=\dfrac{{V}^{2} _{rms}}{2g}$

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

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

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?

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.