Tag: waves

Questions Related to waves

Ultrasonic, infrasonic and audio waves travel through a medium with speeds $V _{u}, V _{i}$ and $V _a$ respectively then,

  1. $V _{u}, V _{i}$ and $V _{a}$ are equal

  2. $V _{u} > V _{a}> V _{i}$

  3. $V _{u} < V _{a} < V _{i}$

  4. $ V _{a}< V _{u} $ and $V _{u} \approx V _{i} $


Correct Option: A
Explanation:

Velocity of sound wave in a medium is given by
$v= \sqrt{\frac{K}{\rho}}$ where K is the bulk modulus and $\rho$ is the density.
The classification of sound waves based on wavelength($\lambda$) is independent of speed of sound in the medium( speed depends on properties of a medium).
Hence, $V _u$, $V _i$ and $V _a$ are all equal.

The extension in a string obeying Hooke's law $v$ is $x$. The speed of sound in the stretched string is $v$. If the extension in the string is increased to $1.5\ x$, the speed of sound will be

  1. $1.22\ v$

  2. $0.61\ v$

  3. $1.50\ v$

  4. $0.75\ v$


Correct Option: A
Explanation:

Speed of sound in stretched string
$v = \dfrac {\overline {T}}{\mu} ..... (i)$
where $T$ is the tension in the string and $\mu$ is mass per unit length.
According to Hooke's law, $F\propto X$
$\therefore T\propto X$ .... (ii)
From Eqs. (i) and (ii)
$v\propto$
$\therefore v' = \overline {1.5V} = 1.22\ V$.

According to Newton's formula, the speed of sound in air at STP is:
(Take the mass of $1$ mole of are is $29 \times 10^{-3} \,\,kg)$

  1. $250 \,\, m \,\,s^{-1}$

  2. $260 \,\, m \,\,s^{-1}$

  3. $270 \,\, m \,\,s^{-1}$

  4. $280 \,\, m \,\,s^{-1}$


Correct Option: D
Explanation:

$1$ mole of any gas occupies $22.4$ liters at STP.
Therefore, the density of air at STP is
$\rho = \dfrac{\text{Mass of one mole of air}}{\text{Volume of one mole of air at STP}}$

$= \dfrac{29 \times 10^{-3} \,\,kg}{22.4 \times 10^{-3} \,\,m^3} = 1.29 \,\,kg \,\,m^{-3}$

At STP, $P = 1\,\,atm = 1.01 \times 10^5 \,\,N \,\,m^{-2}$

$V =\sqrt{\left( \dfrac { P }{ \rho  }\right)}=\sqrt { \dfrac {1.01 \times 10^5 \,\,N \,\,m^{-2}  }{ 1.29 \times kg \,\,m^{-3} }  } = 280 \,\,m \,\,s^{-1} $

The velocities of sound at the same temperature in two monoatomic gases of densities $p _1$ and $p _2$ are $v _1$ and $v _2$ respectively. If $p _1/p _2 = 4$, then the value of $v _1/v _2$ is

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

  2. $2$

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

  4. $4$


Correct Option: A,C
Explanation:

For velocity of sound in gas
$\displaystyle v=\sqrt {\frac {p\gamma }{p}}$

$[P$ is pressure and $p$ is density of gas, $\gamma$ is $C _p/C _v]$

Here, $\displaystyle v _1 = \sqrt {\frac {\gamma P}{p _1}}$ and $v _2 = \sqrt {\frac {\gamma P}{p _2}}$

$\displaystyle \frac {v _1}{v _2} = \sqrt {\frac {p _2}{p _1}} = \sqrt {\frac {1}{4}}=\frac {1}{2}$

The speed of sound through a gaseous medium bears a constant ratio with the rms speed of its molecules. What is this constant ratio ?

  1. $\sqrt{\dfrac{\gamma}{3}}$

  2. $\gamma -1$

  3. $\sqrt{\dfrac{2\gamma}{3}}$

  4. $\gamma$


Correct Option: A
Explanation:

$V _{sound}=\sqrt{\dfrac{\gamma RT}{M}}$ and $V _{rms}=\sqrt{\dfrac{3RT}{M}}$
$\Rightarrow\dfrac{V _{sound}}{V _{rms}}=\sqrt{\dfrac{\gamma}{3}}$
Hence (A) is correct.

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

  1. get doubled

  2. get halved

  3. remain same

  4. get quadrupled


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

  1. $C _{S} < C$

  2. $C _{s}=c$

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

  4. $None\ of\ these$


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

  1. increases with temperature

  2. decreases with temperature

  3. are independent of temperature

  4. none of the above


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:

  1. $18$ km

  2. $15$ km

  3. $12.38$ km

  4. $8$ km


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

  1. True

  2. False


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