354833
A sound absorber attenuates the sound level by \(20\;dB\). The intensity decreases by a factor of
1 10000
2 1000
3 10
4 100
Explanation:
Let \({I_1}\) & \({I_2}\) are initial and final intensities \(\beta_{1}=10 \log \left(\dfrac{I_{1}}{I_{0}}\right)\) Later, \(\beta_{2}=10 \log \left(\dfrac{I_{2}}{I_{0}}\right)\) Given, \(\beta_{1}-\beta_{2}=20\) \(\therefore 20 = 10\log \left( {\frac{{{I_1}}}{{{I_2}}}} \right)\) \({I_1} = 100\,{I_2}\)
PHXI15:WAVES
354834
A point source emits sound equally in all directions in a non-absorbing medium. Two points \(P\) and \(Q\) are at distance of \(2\;m\) and \(3\;m\) respectively from the source. The ratio of the intensities of the waves at \(P\) and \(Q\) is
1 \(9: 4\)
2 \(2: 3\)
3 \(3: 2\)
4 \(4: 9\)
Explanation:
The average power per unit area that is incident perpendicular to the direction of propagation is called the intensity, i.e. \(I=\dfrac{P}{4 \pi r^{2}} \Rightarrow I \propto \dfrac{1}{r^{2}}\) \(\Rightarrow \dfrac{I_{1}}{I_{2}}=\left(\dfrac{r_{2}}{r_{1}}\right)^{2}\) Here, \(r_{1}=2 m, r_{2}=3 m\) \(\therefore \dfrac{I_{1}}{I_{2}}=\left(\dfrac{3}{2}\right)^{2}=\dfrac{9}{4}=9: 4\)
AIIMS - 2006
PHXI15:WAVES
354835
A source of sound emits \(200\,\pi W\) power which is uniformly distributed over a sphere of radius \(10\;m\). What is the loudness of sound on the surface of the sphere?
1 \(74\;dB\)
2 \(70\;dB\)
3 \(84\;dB\)
4 \(80\;dB\)
Explanation:
Intensity, \(I=\dfrac{P}{A}=\dfrac{200 \pi}{4 \pi \times(10)^{2}}=0.5\) No. of decibels is given by \(\therefore 10{\log _{10}}\frac{I}{{{I_0}}} = 10{\log _{10}}\frac{{0.5}}{{{{10}^{ - 12}}}}\) \( = 10{\log _{10}}\left( {5 \times {{10}^{11}}} \right) = 84\;dB\)
354833
A sound absorber attenuates the sound level by \(20\;dB\). The intensity decreases by a factor of
1 10000
2 1000
3 10
4 100
Explanation:
Let \({I_1}\) & \({I_2}\) are initial and final intensities \(\beta_{1}=10 \log \left(\dfrac{I_{1}}{I_{0}}\right)\) Later, \(\beta_{2}=10 \log \left(\dfrac{I_{2}}{I_{0}}\right)\) Given, \(\beta_{1}-\beta_{2}=20\) \(\therefore 20 = 10\log \left( {\frac{{{I_1}}}{{{I_2}}}} \right)\) \({I_1} = 100\,{I_2}\)
PHXI15:WAVES
354834
A point source emits sound equally in all directions in a non-absorbing medium. Two points \(P\) and \(Q\) are at distance of \(2\;m\) and \(3\;m\) respectively from the source. The ratio of the intensities of the waves at \(P\) and \(Q\) is
1 \(9: 4\)
2 \(2: 3\)
3 \(3: 2\)
4 \(4: 9\)
Explanation:
The average power per unit area that is incident perpendicular to the direction of propagation is called the intensity, i.e. \(I=\dfrac{P}{4 \pi r^{2}} \Rightarrow I \propto \dfrac{1}{r^{2}}\) \(\Rightarrow \dfrac{I_{1}}{I_{2}}=\left(\dfrac{r_{2}}{r_{1}}\right)^{2}\) Here, \(r_{1}=2 m, r_{2}=3 m\) \(\therefore \dfrac{I_{1}}{I_{2}}=\left(\dfrac{3}{2}\right)^{2}=\dfrac{9}{4}=9: 4\)
AIIMS - 2006
PHXI15:WAVES
354835
A source of sound emits \(200\,\pi W\) power which is uniformly distributed over a sphere of radius \(10\;m\). What is the loudness of sound on the surface of the sphere?
1 \(74\;dB\)
2 \(70\;dB\)
3 \(84\;dB\)
4 \(80\;dB\)
Explanation:
Intensity, \(I=\dfrac{P}{A}=\dfrac{200 \pi}{4 \pi \times(10)^{2}}=0.5\) No. of decibels is given by \(\therefore 10{\log _{10}}\frac{I}{{{I_0}}} = 10{\log _{10}}\frac{{0.5}}{{{{10}^{ - 12}}}}\) \( = 10{\log _{10}}\left( {5 \times {{10}^{11}}} \right) = 84\;dB\)
354833
A sound absorber attenuates the sound level by \(20\;dB\). The intensity decreases by a factor of
1 10000
2 1000
3 10
4 100
Explanation:
Let \({I_1}\) & \({I_2}\) are initial and final intensities \(\beta_{1}=10 \log \left(\dfrac{I_{1}}{I_{0}}\right)\) Later, \(\beta_{2}=10 \log \left(\dfrac{I_{2}}{I_{0}}\right)\) Given, \(\beta_{1}-\beta_{2}=20\) \(\therefore 20 = 10\log \left( {\frac{{{I_1}}}{{{I_2}}}} \right)\) \({I_1} = 100\,{I_2}\)
PHXI15:WAVES
354834
A point source emits sound equally in all directions in a non-absorbing medium. Two points \(P\) and \(Q\) are at distance of \(2\;m\) and \(3\;m\) respectively from the source. The ratio of the intensities of the waves at \(P\) and \(Q\) is
1 \(9: 4\)
2 \(2: 3\)
3 \(3: 2\)
4 \(4: 9\)
Explanation:
The average power per unit area that is incident perpendicular to the direction of propagation is called the intensity, i.e. \(I=\dfrac{P}{4 \pi r^{2}} \Rightarrow I \propto \dfrac{1}{r^{2}}\) \(\Rightarrow \dfrac{I_{1}}{I_{2}}=\left(\dfrac{r_{2}}{r_{1}}\right)^{2}\) Here, \(r_{1}=2 m, r_{2}=3 m\) \(\therefore \dfrac{I_{1}}{I_{2}}=\left(\dfrac{3}{2}\right)^{2}=\dfrac{9}{4}=9: 4\)
AIIMS - 2006
PHXI15:WAVES
354835
A source of sound emits \(200\,\pi W\) power which is uniformly distributed over a sphere of radius \(10\;m\). What is the loudness of sound on the surface of the sphere?
1 \(74\;dB\)
2 \(70\;dB\)
3 \(84\;dB\)
4 \(80\;dB\)
Explanation:
Intensity, \(I=\dfrac{P}{A}=\dfrac{200 \pi}{4 \pi \times(10)^{2}}=0.5\) No. of decibels is given by \(\therefore 10{\log _{10}}\frac{I}{{{I_0}}} = 10{\log _{10}}\frac{{0.5}}{{{{10}^{ - 12}}}}\) \( = 10{\log _{10}}\left( {5 \times {{10}^{11}}} \right) = 84\;dB\)