354678
Newton assumed that sound propagation in a gas takes under
1 Isothermal condition
2 adiabatic condition
3 isobaric condition
4 Isentropic condition
Explanation:
Conceptual Question
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
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 \(\sqrt{\dfrac{d_{1}}{d_{2}}}\)
2 \(\sqrt{\dfrac{d_{2}}{d_{1}}}\)
3 \(\sqrt{d_{1} d_{2}}\)
4 \(d_{1} d_{2}\)
Explanation:
Speed of sound \(v=\sqrt{\dfrac{\gamma P}{d}} \Rightarrow \dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{d_{2}}{d_{1}}}\) \([\because P \approx \text { constant }]\)
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
354678
Newton assumed that sound propagation in a gas takes under
1 Isothermal condition
2 adiabatic condition
3 isobaric condition
4 Isentropic condition
Explanation:
Conceptual Question
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
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 \(\sqrt{\dfrac{d_{1}}{d_{2}}}\)
2 \(\sqrt{\dfrac{d_{2}}{d_{1}}}\)
3 \(\sqrt{d_{1} d_{2}}\)
4 \(d_{1} d_{2}\)
Explanation:
Speed of sound \(v=\sqrt{\dfrac{\gamma P}{d}} \Rightarrow \dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{d_{2}}{d_{1}}}\) \([\because P \approx \text { constant }]\)
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
354678
Newton assumed that sound propagation in a gas takes under
1 Isothermal condition
2 adiabatic condition
3 isobaric condition
4 Isentropic condition
Explanation:
Conceptual Question
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
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 \(\sqrt{\dfrac{d_{1}}{d_{2}}}\)
2 \(\sqrt{\dfrac{d_{2}}{d_{1}}}\)
3 \(\sqrt{d_{1} d_{2}}\)
4 \(d_{1} d_{2}\)
Explanation:
Speed of sound \(v=\sqrt{\dfrac{\gamma P}{d}} \Rightarrow \dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{d_{2}}{d_{1}}}\) \([\because P \approx \text { constant }]\)
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
354678
Newton assumed that sound propagation in a gas takes under
1 Isothermal condition
2 adiabatic condition
3 isobaric condition
4 Isentropic condition
Explanation:
Conceptual Question
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
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 \(\sqrt{\dfrac{d_{1}}{d_{2}}}\)
2 \(\sqrt{\dfrac{d_{2}}{d_{1}}}\)
3 \(\sqrt{d_{1} d_{2}}\)
4 \(d_{1} d_{2}\)
Explanation:
Speed of sound \(v=\sqrt{\dfrac{\gamma P}{d}} \Rightarrow \dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{d_{2}}{d_{1}}}\) \([\because P \approx \text { constant }]\)
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