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PHXI15:WAVES

354678 Newton assumed that sound propagation in a gas takes under

1 Isothermal condition

2 adiabatic condition

3 isobaric condition

4 Isentropic condition

PHXI15:WAVES

354679 The temperature of a mono-atomic gas in an uniform container of length ' $L$ ' varies linearly from $T_{0}$ to $T_{L}$ as shown in the figure. If the molecular weight of the gas is $M_{0}$ , then the time taken by a wave pulse in travelling from end $A$ to end $B$ is
supporting img

1

\frac{2 L}{(\sqrt{T_{L}} + \sqrt{T_{0}})} \sqrt{\frac{3 M_{0}}{5 R}}

2

\sqrt{\frac{3 (T_{L} - T_{0})}{5 R M_{0} L}}

3

\frac{2 L}{(\sqrt{T_{L}} - \sqrt{T_{0}})} \sqrt{\frac{3 M_{0}}{5 R}}

4

\sqrt{\frac{M_{0}}{2 R (T_{L} - T_{0})}}

Explanation:

Velocity of sound wave is
$v = \sqrt{\frac{γ R T}{M_{0}}} = \sqrt{\frac{5 R T}{3 M_{0}}}$
supporting img

The temperature as a function of position is
$T = T_{0} + \frac{(T_{L} - T_{0}) x}{L}$
$d x = v \cdot d t = \sqrt{\frac{5 R}{3 M_{0}} [T_{0} + (\frac{T_{L} - T_{0}}{L}) x]} d t$
$\int_{0}^{L} \frac{d x}{\sqrt{\frac{5 R}{3 M_{0}} [T_{0} + \frac{(T_{L} - T_{0}) x}{L}]}} = \int_{0}^{t} d t$
$\Rightarrow \sqrt{\frac{3 M_{0}}{5 R}} \int_{0}^{L} {[T_{0} + (\frac{T_{L} - T_{0}}{L}) x]}^{- 1 / 2} d x = t$
$\sqrt{\frac{3 M_{0}}{5 R}} (\frac{2 L}{T_{L} - T_{0}}) {[\sqrt{T_{0} + (\frac{(T_{L} - T_{0})}{L})} x]}_{0}^{L} = t$
$\Rightarrow \sqrt{\frac{3 M_{0}}{5 R}} (\frac{2 L}{T_{L} - T_{0}}) (\sqrt{T_{L}} - \sqrt{T_{0}}) = t$
$t = \frac{2 L}{(\sqrt{T_{L}} + \sqrt{T_{0}})} \sqrt{\frac{3 M_{0}}{5 R}}$

PHXI15:WAVES

354680 If at the same temperature and pressure, the densities for two diatomic gases are respectively $d_{1}$ and $d_{2}$ , then the ratio of velocities of sound in these gases will be

1

\sqrt{\frac{d_{1}}{d_{2}}}

2

\sqrt{\frac{d_{2}}{d_{1}}}

3

\sqrt{d_{1} d_{2}}

4

d_{1} d_{2}

Explanation:

Speed of sound $v = \sqrt{\frac{γ P}{d}} \Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{d_{2}}{d_{1}}}$
$[∵ P \approx constant]$

PHXI15:WAVES

354681 The velocity of sound is $v_{s}$ in air. If the density of air is increased to 4 times at constant pressure, then the new velocity of sound will be

1

\frac{v_{s}}{2}

2

\frac{3}{2} v_{s}^{2}

3

12 v_{s}

4

\frac{v_{s}}{12}

Explanation:

$v_{sound} α \frac{1}{\sqrt{ρ}} \Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{ρ_{2}}{ρ_{1}}} = \sqrt{\frac{4}{1}} = 2$
$\Rightarrow v_{2} = \frac{v_{1}}{2}$

PHXI15:WAVES

354678 Newton assumed that sound propagation in a gas takes under

1 Isothermal condition

2 adiabatic condition

3 isobaric condition

4 Isentropic condition

PHXI15:WAVES

354679 The temperature of a mono-atomic gas in an uniform container of length ' $L$ ' varies linearly from $T_{0}$ to $T_{L}$ as shown in the figure. If the molecular weight of the gas is $M_{0}$ , then the time taken by a wave pulse in travelling from end $A$ to end $B$ is
supporting img

1

\frac{2 L}{(\sqrt{T_{L}} + \sqrt{T_{0}})} \sqrt{\frac{3 M_{0}}{5 R}}

2

\sqrt{\frac{3 (T_{L} - T_{0})}{5 R M_{0} L}}

3

\frac{2 L}{(\sqrt{T_{L}} - \sqrt{T_{0}})} \sqrt{\frac{3 M_{0}}{5 R}}

4

\sqrt{\frac{M_{0}}{2 R (T_{L} - T_{0})}}

Explanation:

Velocity of sound wave is
$v = \sqrt{\frac{γ R T}{M_{0}}} = \sqrt{\frac{5 R T}{3 M_{0}}}$
supporting img

The temperature as a function of position is
$T = T_{0} + \frac{(T_{L} - T_{0}) x}{L}$
$d x = v \cdot d t = \sqrt{\frac{5 R}{3 M_{0}} [T_{0} + (\frac{T_{L} - T_{0}}{L}) x]} d t$
$\int_{0}^{L} \frac{d x}{\sqrt{\frac{5 R}{3 M_{0}} [T_{0} + \frac{(T_{L} - T_{0}) x}{L}]}} = \int_{0}^{t} d t$
$\Rightarrow \sqrt{\frac{3 M_{0}}{5 R}} \int_{0}^{L} {[T_{0} + (\frac{T_{L} - T_{0}}{L}) x]}^{- 1 / 2} d x = t$
$\sqrt{\frac{3 M_{0}}{5 R}} (\frac{2 L}{T_{L} - T_{0}}) {[\sqrt{T_{0} + (\frac{(T_{L} - T_{0})}{L})} x]}_{0}^{L} = t$
$\Rightarrow \sqrt{\frac{3 M_{0}}{5 R}} (\frac{2 L}{T_{L} - T_{0}}) (\sqrt{T_{L}} - \sqrt{T_{0}}) = t$
$t = \frac{2 L}{(\sqrt{T_{L}} + \sqrt{T_{0}})} \sqrt{\frac{3 M_{0}}{5 R}}$

PHXI15:WAVES

354680 If at the same temperature and pressure, the densities for two diatomic gases are respectively $d_{1}$ and $d_{2}$ , then the ratio of velocities of sound in these gases will be

1

\sqrt{\frac{d_{1}}{d_{2}}}

2

\sqrt{\frac{d_{2}}{d_{1}}}

3

\sqrt{d_{1} d_{2}}

4

d_{1} d_{2}

Explanation:

Speed of sound $v = \sqrt{\frac{γ P}{d}} \Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{d_{2}}{d_{1}}}$
$[∵ P \approx constant]$

PHXI15:WAVES

354681 The velocity of sound is $v_{s}$ in air. If the density of air is increased to 4 times at constant pressure, then the new velocity of sound will be

1

\frac{v_{s}}{2}

2

\frac{3}{2} v_{s}^{2}

3

12 v_{s}

4

\frac{v_{s}}{12}

Explanation:

$v_{sound} α \frac{1}{\sqrt{ρ}} \Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{ρ_{2}}{ρ_{1}}} = \sqrt{\frac{4}{1}} = 2$
$\Rightarrow v_{2} = \frac{v_{1}}{2}$

PHXI15:WAVES

354678 Newton assumed that sound propagation in a gas takes under

1 Isothermal condition

2 adiabatic condition

3 isobaric condition

4 Isentropic condition

PHXI15:WAVES

354679 The temperature of a mono-atomic gas in an uniform container of length ' $L$ ' varies linearly from $T_{0}$ to $T_{L}$ as shown in the figure. If the molecular weight of the gas is $M_{0}$ , then the time taken by a wave pulse in travelling from end $A$ to end $B$ is
supporting img

1

\frac{2 L}{(\sqrt{T_{L}} + \sqrt{T_{0}})} \sqrt{\frac{3 M_{0}}{5 R}}

2

\sqrt{\frac{3 (T_{L} - T_{0})}{5 R M_{0} L}}

3

\frac{2 L}{(\sqrt{T_{L}} - \sqrt{T_{0}})} \sqrt{\frac{3 M_{0}}{5 R}}

4

\sqrt{\frac{M_{0}}{2 R (T_{L} - T_{0})}}

Explanation:

Velocity of sound wave is
$v = \sqrt{\frac{γ R T}{M_{0}}} = \sqrt{\frac{5 R T}{3 M_{0}}}$
supporting img

The temperature as a function of position is
$T = T_{0} + \frac{(T_{L} - T_{0}) x}{L}$
$d x = v \cdot d t = \sqrt{\frac{5 R}{3 M_{0}} [T_{0} + (\frac{T_{L} - T_{0}}{L}) x]} d t$
$\int_{0}^{L} \frac{d x}{\sqrt{\frac{5 R}{3 M_{0}} [T_{0} + \frac{(T_{L} - T_{0}) x}{L}]}} = \int_{0}^{t} d t$
$\Rightarrow \sqrt{\frac{3 M_{0}}{5 R}} \int_{0}^{L} {[T_{0} + (\frac{T_{L} - T_{0}}{L}) x]}^{- 1 / 2} d x = t$
$\sqrt{\frac{3 M_{0}}{5 R}} (\frac{2 L}{T_{L} - T_{0}}) {[\sqrt{T_{0} + (\frac{(T_{L} - T_{0})}{L})} x]}_{0}^{L} = t$
$\Rightarrow \sqrt{\frac{3 M_{0}}{5 R}} (\frac{2 L}{T_{L} - T_{0}}) (\sqrt{T_{L}} - \sqrt{T_{0}}) = t$
$t = \frac{2 L}{(\sqrt{T_{L}} + \sqrt{T_{0}})} \sqrt{\frac{3 M_{0}}{5 R}}$

PHXI15:WAVES

354680 If at the same temperature and pressure, the densities for two diatomic gases are respectively $d_{1}$ and $d_{2}$ , then the ratio of velocities of sound in these gases will be

1

\sqrt{\frac{d_{1}}{d_{2}}}

2

\sqrt{\frac{d_{2}}{d_{1}}}

3

\sqrt{d_{1} d_{2}}

4

d_{1} d_{2}

Explanation:

Speed of sound $v = \sqrt{\frac{γ P}{d}} \Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{d_{2}}{d_{1}}}$
$[∵ P \approx constant]$

PHXI15:WAVES

354681 The velocity of sound is $v_{s}$ in air. If the density of air is increased to 4 times at constant pressure, then the new velocity of sound will be

1

\frac{v_{s}}{2}

2

\frac{3}{2} v_{s}^{2}

3

12 v_{s}

4

\frac{v_{s}}{12}

Explanation:

$v_{sound} α \frac{1}{\sqrt{ρ}} \Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{ρ_{2}}{ρ_{1}}} = \sqrt{\frac{4}{1}} = 2$
$\Rightarrow v_{2} = \frac{v_{1}}{2}$

PHXI15:WAVES

354678 Newton assumed that sound propagation in a gas takes under

1 Isothermal condition

2 adiabatic condition

3 isobaric condition

4 Isentropic condition

PHXI15:WAVES

354679 The temperature of a mono-atomic gas in an uniform container of length ' $L$ ' varies linearly from $T_{0}$ to $T_{L}$ as shown in the figure. If the molecular weight of the gas is $M_{0}$ , then the time taken by a wave pulse in travelling from end $A$ to end $B$ is
supporting img

1

\frac{2 L}{(\sqrt{T_{L}} + \sqrt{T_{0}})} \sqrt{\frac{3 M_{0}}{5 R}}

2

\sqrt{\frac{3 (T_{L} - T_{0})}{5 R M_{0} L}}

3

\frac{2 L}{(\sqrt{T_{L}} - \sqrt{T_{0}})} \sqrt{\frac{3 M_{0}}{5 R}}

4

\sqrt{\frac{M_{0}}{2 R (T_{L} - T_{0})}}

Explanation:

Velocity of sound wave is
$v = \sqrt{\frac{γ R T}{M_{0}}} = \sqrt{\frac{5 R T}{3 M_{0}}}$
supporting img

The temperature as a function of position is
$T = T_{0} + \frac{(T_{L} - T_{0}) x}{L}$
$d x = v \cdot d t = \sqrt{\frac{5 R}{3 M_{0}} [T_{0} + (\frac{T_{L} - T_{0}}{L}) x]} d t$
$\int_{0}^{L} \frac{d x}{\sqrt{\frac{5 R}{3 M_{0}} [T_{0} + \frac{(T_{L} - T_{0}) x}{L}]}} = \int_{0}^{t} d t$
$\Rightarrow \sqrt{\frac{3 M_{0}}{5 R}} \int_{0}^{L} {[T_{0} + (\frac{T_{L} - T_{0}}{L}) x]}^{- 1 / 2} d x = t$
$\sqrt{\frac{3 M_{0}}{5 R}} (\frac{2 L}{T_{L} - T_{0}}) {[\sqrt{T_{0} + (\frac{(T_{L} - T_{0})}{L})} x]}_{0}^{L} = t$
$\Rightarrow \sqrt{\frac{3 M_{0}}{5 R}} (\frac{2 L}{T_{L} - T_{0}}) (\sqrt{T_{L}} - \sqrt{T_{0}}) = t$
$t = \frac{2 L}{(\sqrt{T_{L}} + \sqrt{T_{0}})} \sqrt{\frac{3 M_{0}}{5 R}}$

PHXI15:WAVES

354680 If at the same temperature and pressure, the densities for two diatomic gases are respectively $d_{1}$ and $d_{2}$ , then the ratio of velocities of sound in these gases will be

1

\sqrt{\frac{d_{1}}{d_{2}}}

2

\sqrt{\frac{d_{2}}{d_{1}}}

3

\sqrt{d_{1} d_{2}}

4

d_{1} d_{2}

Explanation:

Speed of sound $v = \sqrt{\frac{γ P}{d}} \Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{d_{2}}{d_{1}}}$
$[∵ P \approx constant]$

PHXI15:WAVES

354681 The velocity of sound is $v_{s}$ in air. If the density of air is increased to 4 times at constant pressure, then the new velocity of sound will be

1

\frac{v_{s}}{2}

2

\frac{3}{2} v_{s}^{2}

3

12 v_{s}

4

\frac{v_{s}}{12}

Explanation:

$v_{sound} α \frac{1}{\sqrt{ρ}} \Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{ρ_{2}}{ρ_{1}}} = \sqrt{\frac{4}{1}} = 2$
$\Rightarrow v_{2} = \frac{v_{1}}{2}$