NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI15:WAVES
354812
The ratio of intensities between two coherent sound sources is \(4: 1\). The difference of loudness in decibels \((\mathrm{dB})\) between maximum and minimum intensities when they interfere in space is:
1 \(10 \log 2\)
2 \(20 \log 3\)
3 \(10 \log 3\)
4 \(20 \log 2\)
Explanation:
Given that \(\dfrac{I_{1}}{I_{2}}=\dfrac{4}{1}\) \(\therefore \dfrac{I_{\max }}{I_{\max }}=\left[\dfrac{\sqrt{I_{1} / I_{2}}+1}{\sqrt{I_{1} / I_{2}}-1}\right]^{2}=\left[\dfrac{2+1}{2-1}\right]^{2}=9\) The difference of loudness in decibles is \(\begin{array}{r}\beta_{1}-\beta_{2}=\log \dfrac{I_{\max }}{I_{0}}-10 \log \dfrac{I_{\min }}{I_{0}} \\=10 \log \dfrac{I_{\max }}{I_{\min }}=10 \log (9)=20 \log (3)\end{array}\)
PHXI15:WAVES
354813
If \(T\) is the reverberation time of an auditorium of volume \(V\) then
1 \(T \propto \dfrac{1}{V^{2}}\)
2 \(T \propto \dfrac{1}{V}\)
3 \(T \propto V\)
4 \(T \propto V^{2}\)
Explanation:
Reverberation time \(T \propto V\).
PHXI15:WAVES
354814
The intensity of sound increases at night due to:
1 Increases in density of air
2 Decreases in density of air
3 High temperature
4 None of these
Explanation:
At night, amount of carbon dioxide in atmosphere increases which raises the density of atmosphere. Since, intensity is directly proportional to density, intensity of sound is more at night. Option (1) is correct
PHXI15:WAVES
354815
The intensity level of sound wave is said to be 4 decibel. If the intensity of wave is doubled, then the intensity level of sound as expressed in decibel would be
354812
The ratio of intensities between two coherent sound sources is \(4: 1\). The difference of loudness in decibels \((\mathrm{dB})\) between maximum and minimum intensities when they interfere in space is:
1 \(10 \log 2\)
2 \(20 \log 3\)
3 \(10 \log 3\)
4 \(20 \log 2\)
Explanation:
Given that \(\dfrac{I_{1}}{I_{2}}=\dfrac{4}{1}\) \(\therefore \dfrac{I_{\max }}{I_{\max }}=\left[\dfrac{\sqrt{I_{1} / I_{2}}+1}{\sqrt{I_{1} / I_{2}}-1}\right]^{2}=\left[\dfrac{2+1}{2-1}\right]^{2}=9\) The difference of loudness in decibles is \(\begin{array}{r}\beta_{1}-\beta_{2}=\log \dfrac{I_{\max }}{I_{0}}-10 \log \dfrac{I_{\min }}{I_{0}} \\=10 \log \dfrac{I_{\max }}{I_{\min }}=10 \log (9)=20 \log (3)\end{array}\)
PHXI15:WAVES
354813
If \(T\) is the reverberation time of an auditorium of volume \(V\) then
1 \(T \propto \dfrac{1}{V^{2}}\)
2 \(T \propto \dfrac{1}{V}\)
3 \(T \propto V\)
4 \(T \propto V^{2}\)
Explanation:
Reverberation time \(T \propto V\).
PHXI15:WAVES
354814
The intensity of sound increases at night due to:
1 Increases in density of air
2 Decreases in density of air
3 High temperature
4 None of these
Explanation:
At night, amount of carbon dioxide in atmosphere increases which raises the density of atmosphere. Since, intensity is directly proportional to density, intensity of sound is more at night. Option (1) is correct
PHXI15:WAVES
354815
The intensity level of sound wave is said to be 4 decibel. If the intensity of wave is doubled, then the intensity level of sound as expressed in decibel would be
354812
The ratio of intensities between two coherent sound sources is \(4: 1\). The difference of loudness in decibels \((\mathrm{dB})\) between maximum and minimum intensities when they interfere in space is:
1 \(10 \log 2\)
2 \(20 \log 3\)
3 \(10 \log 3\)
4 \(20 \log 2\)
Explanation:
Given that \(\dfrac{I_{1}}{I_{2}}=\dfrac{4}{1}\) \(\therefore \dfrac{I_{\max }}{I_{\max }}=\left[\dfrac{\sqrt{I_{1} / I_{2}}+1}{\sqrt{I_{1} / I_{2}}-1}\right]^{2}=\left[\dfrac{2+1}{2-1}\right]^{2}=9\) The difference of loudness in decibles is \(\begin{array}{r}\beta_{1}-\beta_{2}=\log \dfrac{I_{\max }}{I_{0}}-10 \log \dfrac{I_{\min }}{I_{0}} \\=10 \log \dfrac{I_{\max }}{I_{\min }}=10 \log (9)=20 \log (3)\end{array}\)
PHXI15:WAVES
354813
If \(T\) is the reverberation time of an auditorium of volume \(V\) then
1 \(T \propto \dfrac{1}{V^{2}}\)
2 \(T \propto \dfrac{1}{V}\)
3 \(T \propto V\)
4 \(T \propto V^{2}\)
Explanation:
Reverberation time \(T \propto V\).
PHXI15:WAVES
354814
The intensity of sound increases at night due to:
1 Increases in density of air
2 Decreases in density of air
3 High temperature
4 None of these
Explanation:
At night, amount of carbon dioxide in atmosphere increases which raises the density of atmosphere. Since, intensity is directly proportional to density, intensity of sound is more at night. Option (1) is correct
PHXI15:WAVES
354815
The intensity level of sound wave is said to be 4 decibel. If the intensity of wave is doubled, then the intensity level of sound as expressed in decibel would be
354812
The ratio of intensities between two coherent sound sources is \(4: 1\). The difference of loudness in decibels \((\mathrm{dB})\) between maximum and minimum intensities when they interfere in space is:
1 \(10 \log 2\)
2 \(20 \log 3\)
3 \(10 \log 3\)
4 \(20 \log 2\)
Explanation:
Given that \(\dfrac{I_{1}}{I_{2}}=\dfrac{4}{1}\) \(\therefore \dfrac{I_{\max }}{I_{\max }}=\left[\dfrac{\sqrt{I_{1} / I_{2}}+1}{\sqrt{I_{1} / I_{2}}-1}\right]^{2}=\left[\dfrac{2+1}{2-1}\right]^{2}=9\) The difference of loudness in decibles is \(\begin{array}{r}\beta_{1}-\beta_{2}=\log \dfrac{I_{\max }}{I_{0}}-10 \log \dfrac{I_{\min }}{I_{0}} \\=10 \log \dfrac{I_{\max }}{I_{\min }}=10 \log (9)=20 \log (3)\end{array}\)
PHXI15:WAVES
354813
If \(T\) is the reverberation time of an auditorium of volume \(V\) then
1 \(T \propto \dfrac{1}{V^{2}}\)
2 \(T \propto \dfrac{1}{V}\)
3 \(T \propto V\)
4 \(T \propto V^{2}\)
Explanation:
Reverberation time \(T \propto V\).
PHXI15:WAVES
354814
The intensity of sound increases at night due to:
1 Increases in density of air
2 Decreases in density of air
3 High temperature
4 None of these
Explanation:
At night, amount of carbon dioxide in atmosphere increases which raises the density of atmosphere. Since, intensity is directly proportional to density, intensity of sound is more at night. Option (1) is correct
PHXI15:WAVES
354815
The intensity level of sound wave is said to be 4 decibel. If the intensity of wave is doubled, then the intensity level of sound as expressed in decibel would be
