172772
The beat frequency observed when two sound waves $y_{1}=0.5 \sin (410 t)$ and $y_{2}=0.5 \sin$ $(454 t)$ travel in the same direction is
1 5
2 3
3 7
4 2
5 4
Explanation:
C Given that, Equations of waves are- $\mathrm{y}_{1}=0.5 \sin (410 \mathrm{t})$ $\mathrm{y}_{2}=0.5 \sin (454 \mathrm{t})$ Comparing above equations with standard equation $\mathrm{y}=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx}) \text { we get }$ $\omega_{1}=410 \& \omega_{2}=454$ Now, frequency of $f_{1}=\frac{410}{2 \pi} \text { and } f_{2}=\frac{454}{2 \pi}$ $f_{1}=\frac{205 \times 7}{22} f_{2}=\frac{227 \times 7}{22}$ $f_{1}=65.227 \mathrm{~Hz} f_{2}=72.227 \mathrm{~Hz}$ Hence, the beat frequency $(f)=f_{1}-f_{2}$ $\mathrm{f}=72.227-65.227 \sqcup 7 \text { beats }$
Kerala CEE 2012
WAVES
172773
Two sources of sound placed closed to each other, are emitting progressive waves given by $Y_{1}=4 \sin 600 \pi t$ and $Y_{2}=5 \sin 608 \pi t$ An observer located near these two sources of sound will hear
1 4 beat/s with intensity ratio $25: 16$ between waxing and waning
2 8 beat/s with intensity ratio $25: 16$ between waxing and waning
3 8 beat/s with intensity ratio $81: 1$ between waxing and waning
4 4 beat/s with intensity ratio $81: 1$ between waxing and waning
Explanation:
D Given progressive wave equation $\mathrm{Y}_{1}=4 \sin 600 \pi \mathrm{t}$ $\mathrm{Y}_{2}=5 \sin 608 \pi \mathrm{t}$ $\text { standard equation }$ $\mathrm{Y}=\mathrm{A} \sin \omega \mathrm{t}$ $\text { from equation (i) }$ $\omega \mathrm{t}==600 \pi \mathrm{t}$ $2 \pi \mathrm{f}_{1} \mathrm{t}=600 \pi \mathrm{t}$ $\mathrm{f}_{1}=300 \mathrm{~Hz}$ $\text { from equation (ii) }$ $2 \pi \mathrm{f}_{2} \mathrm{t}=608 \pi \mathrm{t}$ $\mathrm{f}_{2}=304 \mathrm{~Hz}$ Number of beats $=f_{2}-f_{1}=304-300=4$ beats $/ \mathrm{sec}$ $\frac{I_{\text {max }}}{I_{\text {min }}}=\left(\frac{A_{1}+A_{2}}{A_{1}-A_{2}}\right)^{2}$ $\frac{I_{\text {max }}}{I_{\text {min }}}=\left(\frac{4+5}{4-5}\right)^{2}$ $I_{\text {max }}: I_{\text {min }}=81: 1$
AIPMT-2012
WAVES
172777
There are two waves having wavelengths 100 $\mathrm{cm}$ and $101 \mathrm{~cm}$ and same velocity $303 \mathrm{~m} / \mathrm{s}$. The beat frequency is
172842
The intensity of sound gets reduced by $10 \%$ on passing through a slab. The reduction in intensity on passage through three consecutive slabs is
1 $30 \%$
2 $27.1 \%$
3 $20 \%$
4 $36 \%$
Explanation:
B After passing through $1^{\text {st }}$ slab, Intensity $=90 \%$ of $\mathrm{I}=0.9 \mathrm{I}$ After second slab, intensity $=90 \%$ of $(0.9 \mathrm{I})$ $=81 \% \mathrm{I}=0.81 \mathrm{I}$ After third slab, intensity $=90 \%$ of $(0.81 \mathrm{I})$ $=72.9 \% \text { of } \mathrm{I}$ $\therefore \quad$ Reduction in intensity $=27.1 \%$
J and K CET- 2004
WAVES
172779
A sound absorber attenuates the sound level by $20 \mathrm{~dB}$. The intensity decreases by a factor of
1 1000
2 10000
3 10
4 100
Explanation:
D The intensity level of the sound initially will be , $\beta_{1}=10 \log \left(\frac{\mathrm{I}}{\mathrm{I}_{0}}\right)$ Now, when the intensity of the sound is decreased, then the final intensity level of sound will be - $\beta_{2}=10 \log \left(\frac{I^{\prime}}{I_{0}}\right)$ As it is given that the sound is decreased by $20 \mathrm{~dB}$, which means that the difference between initial sound intensity level and final sound intensity level is $20 \mathrm{~dB}$, Therefore $\beta_{1}-\beta_{2}=10 \log \left(\frac{\mathrm{I}}{\mathrm{I}_{\circ}}\right)-10 \log \left(\frac{\mathrm{I}^{\prime}}{\mathrm{I}_{\circ}}\right)$ $20=10 \log \left(\frac{\mathrm{I}}{\mathrm{I}^{\prime}}\right)$ $2=\log \left(\frac{\mathrm{I}}{\mathrm{I}^{\prime}}\right)$ $-2=\log \left(\frac{\mathrm{I}^{\prime}}{\mathrm{I}}\right)$ $10^{-2}=\frac{\mathrm{I}^{\prime}}{\mathrm{I}}$ $\mathrm{I}^{\prime}=\frac{\mathrm{I}}{100}$ Therefore, the intensity of sound decreased by a factor of 100 .
172772
The beat frequency observed when two sound waves $y_{1}=0.5 \sin (410 t)$ and $y_{2}=0.5 \sin$ $(454 t)$ travel in the same direction is
1 5
2 3
3 7
4 2
5 4
Explanation:
C Given that, Equations of waves are- $\mathrm{y}_{1}=0.5 \sin (410 \mathrm{t})$ $\mathrm{y}_{2}=0.5 \sin (454 \mathrm{t})$ Comparing above equations with standard equation $\mathrm{y}=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx}) \text { we get }$ $\omega_{1}=410 \& \omega_{2}=454$ Now, frequency of $f_{1}=\frac{410}{2 \pi} \text { and } f_{2}=\frac{454}{2 \pi}$ $f_{1}=\frac{205 \times 7}{22} f_{2}=\frac{227 \times 7}{22}$ $f_{1}=65.227 \mathrm{~Hz} f_{2}=72.227 \mathrm{~Hz}$ Hence, the beat frequency $(f)=f_{1}-f_{2}$ $\mathrm{f}=72.227-65.227 \sqcup 7 \text { beats }$
Kerala CEE 2012
WAVES
172773
Two sources of sound placed closed to each other, are emitting progressive waves given by $Y_{1}=4 \sin 600 \pi t$ and $Y_{2}=5 \sin 608 \pi t$ An observer located near these two sources of sound will hear
1 4 beat/s with intensity ratio $25: 16$ between waxing and waning
2 8 beat/s with intensity ratio $25: 16$ between waxing and waning
3 8 beat/s with intensity ratio $81: 1$ between waxing and waning
4 4 beat/s with intensity ratio $81: 1$ between waxing and waning
Explanation:
D Given progressive wave equation $\mathrm{Y}_{1}=4 \sin 600 \pi \mathrm{t}$ $\mathrm{Y}_{2}=5 \sin 608 \pi \mathrm{t}$ $\text { standard equation }$ $\mathrm{Y}=\mathrm{A} \sin \omega \mathrm{t}$ $\text { from equation (i) }$ $\omega \mathrm{t}==600 \pi \mathrm{t}$ $2 \pi \mathrm{f}_{1} \mathrm{t}=600 \pi \mathrm{t}$ $\mathrm{f}_{1}=300 \mathrm{~Hz}$ $\text { from equation (ii) }$ $2 \pi \mathrm{f}_{2} \mathrm{t}=608 \pi \mathrm{t}$ $\mathrm{f}_{2}=304 \mathrm{~Hz}$ Number of beats $=f_{2}-f_{1}=304-300=4$ beats $/ \mathrm{sec}$ $\frac{I_{\text {max }}}{I_{\text {min }}}=\left(\frac{A_{1}+A_{2}}{A_{1}-A_{2}}\right)^{2}$ $\frac{I_{\text {max }}}{I_{\text {min }}}=\left(\frac{4+5}{4-5}\right)^{2}$ $I_{\text {max }}: I_{\text {min }}=81: 1$
AIPMT-2012
WAVES
172777
There are two waves having wavelengths 100 $\mathrm{cm}$ and $101 \mathrm{~cm}$ and same velocity $303 \mathrm{~m} / \mathrm{s}$. The beat frequency is
172842
The intensity of sound gets reduced by $10 \%$ on passing through a slab. The reduction in intensity on passage through three consecutive slabs is
1 $30 \%$
2 $27.1 \%$
3 $20 \%$
4 $36 \%$
Explanation:
B After passing through $1^{\text {st }}$ slab, Intensity $=90 \%$ of $\mathrm{I}=0.9 \mathrm{I}$ After second slab, intensity $=90 \%$ of $(0.9 \mathrm{I})$ $=81 \% \mathrm{I}=0.81 \mathrm{I}$ After third slab, intensity $=90 \%$ of $(0.81 \mathrm{I})$ $=72.9 \% \text { of } \mathrm{I}$ $\therefore \quad$ Reduction in intensity $=27.1 \%$
J and K CET- 2004
WAVES
172779
A sound absorber attenuates the sound level by $20 \mathrm{~dB}$. The intensity decreases by a factor of
1 1000
2 10000
3 10
4 100
Explanation:
D The intensity level of the sound initially will be , $\beta_{1}=10 \log \left(\frac{\mathrm{I}}{\mathrm{I}_{0}}\right)$ Now, when the intensity of the sound is decreased, then the final intensity level of sound will be - $\beta_{2}=10 \log \left(\frac{I^{\prime}}{I_{0}}\right)$ As it is given that the sound is decreased by $20 \mathrm{~dB}$, which means that the difference between initial sound intensity level and final sound intensity level is $20 \mathrm{~dB}$, Therefore $\beta_{1}-\beta_{2}=10 \log \left(\frac{\mathrm{I}}{\mathrm{I}_{\circ}}\right)-10 \log \left(\frac{\mathrm{I}^{\prime}}{\mathrm{I}_{\circ}}\right)$ $20=10 \log \left(\frac{\mathrm{I}}{\mathrm{I}^{\prime}}\right)$ $2=\log \left(\frac{\mathrm{I}}{\mathrm{I}^{\prime}}\right)$ $-2=\log \left(\frac{\mathrm{I}^{\prime}}{\mathrm{I}}\right)$ $10^{-2}=\frac{\mathrm{I}^{\prime}}{\mathrm{I}}$ $\mathrm{I}^{\prime}=\frac{\mathrm{I}}{100}$ Therefore, the intensity of sound decreased by a factor of 100 .
172772
The beat frequency observed when two sound waves $y_{1}=0.5 \sin (410 t)$ and $y_{2}=0.5 \sin$ $(454 t)$ travel in the same direction is
1 5
2 3
3 7
4 2
5 4
Explanation:
C Given that, Equations of waves are- $\mathrm{y}_{1}=0.5 \sin (410 \mathrm{t})$ $\mathrm{y}_{2}=0.5 \sin (454 \mathrm{t})$ Comparing above equations with standard equation $\mathrm{y}=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx}) \text { we get }$ $\omega_{1}=410 \& \omega_{2}=454$ Now, frequency of $f_{1}=\frac{410}{2 \pi} \text { and } f_{2}=\frac{454}{2 \pi}$ $f_{1}=\frac{205 \times 7}{22} f_{2}=\frac{227 \times 7}{22}$ $f_{1}=65.227 \mathrm{~Hz} f_{2}=72.227 \mathrm{~Hz}$ Hence, the beat frequency $(f)=f_{1}-f_{2}$ $\mathrm{f}=72.227-65.227 \sqcup 7 \text { beats }$
Kerala CEE 2012
WAVES
172773
Two sources of sound placed closed to each other, are emitting progressive waves given by $Y_{1}=4 \sin 600 \pi t$ and $Y_{2}=5 \sin 608 \pi t$ An observer located near these two sources of sound will hear
1 4 beat/s with intensity ratio $25: 16$ between waxing and waning
2 8 beat/s with intensity ratio $25: 16$ between waxing and waning
3 8 beat/s with intensity ratio $81: 1$ between waxing and waning
4 4 beat/s with intensity ratio $81: 1$ between waxing and waning
Explanation:
D Given progressive wave equation $\mathrm{Y}_{1}=4 \sin 600 \pi \mathrm{t}$ $\mathrm{Y}_{2}=5 \sin 608 \pi \mathrm{t}$ $\text { standard equation }$ $\mathrm{Y}=\mathrm{A} \sin \omega \mathrm{t}$ $\text { from equation (i) }$ $\omega \mathrm{t}==600 \pi \mathrm{t}$ $2 \pi \mathrm{f}_{1} \mathrm{t}=600 \pi \mathrm{t}$ $\mathrm{f}_{1}=300 \mathrm{~Hz}$ $\text { from equation (ii) }$ $2 \pi \mathrm{f}_{2} \mathrm{t}=608 \pi \mathrm{t}$ $\mathrm{f}_{2}=304 \mathrm{~Hz}$ Number of beats $=f_{2}-f_{1}=304-300=4$ beats $/ \mathrm{sec}$ $\frac{I_{\text {max }}}{I_{\text {min }}}=\left(\frac{A_{1}+A_{2}}{A_{1}-A_{2}}\right)^{2}$ $\frac{I_{\text {max }}}{I_{\text {min }}}=\left(\frac{4+5}{4-5}\right)^{2}$ $I_{\text {max }}: I_{\text {min }}=81: 1$
AIPMT-2012
WAVES
172777
There are two waves having wavelengths 100 $\mathrm{cm}$ and $101 \mathrm{~cm}$ and same velocity $303 \mathrm{~m} / \mathrm{s}$. The beat frequency is
172842
The intensity of sound gets reduced by $10 \%$ on passing through a slab. The reduction in intensity on passage through three consecutive slabs is
1 $30 \%$
2 $27.1 \%$
3 $20 \%$
4 $36 \%$
Explanation:
B After passing through $1^{\text {st }}$ slab, Intensity $=90 \%$ of $\mathrm{I}=0.9 \mathrm{I}$ After second slab, intensity $=90 \%$ of $(0.9 \mathrm{I})$ $=81 \% \mathrm{I}=0.81 \mathrm{I}$ After third slab, intensity $=90 \%$ of $(0.81 \mathrm{I})$ $=72.9 \% \text { of } \mathrm{I}$ $\therefore \quad$ Reduction in intensity $=27.1 \%$
J and K CET- 2004
WAVES
172779
A sound absorber attenuates the sound level by $20 \mathrm{~dB}$. The intensity decreases by a factor of
1 1000
2 10000
3 10
4 100
Explanation:
D The intensity level of the sound initially will be , $\beta_{1}=10 \log \left(\frac{\mathrm{I}}{\mathrm{I}_{0}}\right)$ Now, when the intensity of the sound is decreased, then the final intensity level of sound will be - $\beta_{2}=10 \log \left(\frac{I^{\prime}}{I_{0}}\right)$ As it is given that the sound is decreased by $20 \mathrm{~dB}$, which means that the difference between initial sound intensity level and final sound intensity level is $20 \mathrm{~dB}$, Therefore $\beta_{1}-\beta_{2}=10 \log \left(\frac{\mathrm{I}}{\mathrm{I}_{\circ}}\right)-10 \log \left(\frac{\mathrm{I}^{\prime}}{\mathrm{I}_{\circ}}\right)$ $20=10 \log \left(\frac{\mathrm{I}}{\mathrm{I}^{\prime}}\right)$ $2=\log \left(\frac{\mathrm{I}}{\mathrm{I}^{\prime}}\right)$ $-2=\log \left(\frac{\mathrm{I}^{\prime}}{\mathrm{I}}\right)$ $10^{-2}=\frac{\mathrm{I}^{\prime}}{\mathrm{I}}$ $\mathrm{I}^{\prime}=\frac{\mathrm{I}}{100}$ Therefore, the intensity of sound decreased by a factor of 100 .
172772
The beat frequency observed when two sound waves $y_{1}=0.5 \sin (410 t)$ and $y_{2}=0.5 \sin$ $(454 t)$ travel in the same direction is
1 5
2 3
3 7
4 2
5 4
Explanation:
C Given that, Equations of waves are- $\mathrm{y}_{1}=0.5 \sin (410 \mathrm{t})$ $\mathrm{y}_{2}=0.5 \sin (454 \mathrm{t})$ Comparing above equations with standard equation $\mathrm{y}=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx}) \text { we get }$ $\omega_{1}=410 \& \omega_{2}=454$ Now, frequency of $f_{1}=\frac{410}{2 \pi} \text { and } f_{2}=\frac{454}{2 \pi}$ $f_{1}=\frac{205 \times 7}{22} f_{2}=\frac{227 \times 7}{22}$ $f_{1}=65.227 \mathrm{~Hz} f_{2}=72.227 \mathrm{~Hz}$ Hence, the beat frequency $(f)=f_{1}-f_{2}$ $\mathrm{f}=72.227-65.227 \sqcup 7 \text { beats }$
Kerala CEE 2012
WAVES
172773
Two sources of sound placed closed to each other, are emitting progressive waves given by $Y_{1}=4 \sin 600 \pi t$ and $Y_{2}=5 \sin 608 \pi t$ An observer located near these two sources of sound will hear
1 4 beat/s with intensity ratio $25: 16$ between waxing and waning
2 8 beat/s with intensity ratio $25: 16$ between waxing and waning
3 8 beat/s with intensity ratio $81: 1$ between waxing and waning
4 4 beat/s with intensity ratio $81: 1$ between waxing and waning
Explanation:
D Given progressive wave equation $\mathrm{Y}_{1}=4 \sin 600 \pi \mathrm{t}$ $\mathrm{Y}_{2}=5 \sin 608 \pi \mathrm{t}$ $\text { standard equation }$ $\mathrm{Y}=\mathrm{A} \sin \omega \mathrm{t}$ $\text { from equation (i) }$ $\omega \mathrm{t}==600 \pi \mathrm{t}$ $2 \pi \mathrm{f}_{1} \mathrm{t}=600 \pi \mathrm{t}$ $\mathrm{f}_{1}=300 \mathrm{~Hz}$ $\text { from equation (ii) }$ $2 \pi \mathrm{f}_{2} \mathrm{t}=608 \pi \mathrm{t}$ $\mathrm{f}_{2}=304 \mathrm{~Hz}$ Number of beats $=f_{2}-f_{1}=304-300=4$ beats $/ \mathrm{sec}$ $\frac{I_{\text {max }}}{I_{\text {min }}}=\left(\frac{A_{1}+A_{2}}{A_{1}-A_{2}}\right)^{2}$ $\frac{I_{\text {max }}}{I_{\text {min }}}=\left(\frac{4+5}{4-5}\right)^{2}$ $I_{\text {max }}: I_{\text {min }}=81: 1$
AIPMT-2012
WAVES
172777
There are two waves having wavelengths 100 $\mathrm{cm}$ and $101 \mathrm{~cm}$ and same velocity $303 \mathrm{~m} / \mathrm{s}$. The beat frequency is
172842
The intensity of sound gets reduced by $10 \%$ on passing through a slab. The reduction in intensity on passage through three consecutive slabs is
1 $30 \%$
2 $27.1 \%$
3 $20 \%$
4 $36 \%$
Explanation:
B After passing through $1^{\text {st }}$ slab, Intensity $=90 \%$ of $\mathrm{I}=0.9 \mathrm{I}$ After second slab, intensity $=90 \%$ of $(0.9 \mathrm{I})$ $=81 \% \mathrm{I}=0.81 \mathrm{I}$ After third slab, intensity $=90 \%$ of $(0.81 \mathrm{I})$ $=72.9 \% \text { of } \mathrm{I}$ $\therefore \quad$ Reduction in intensity $=27.1 \%$
J and K CET- 2004
WAVES
172779
A sound absorber attenuates the sound level by $20 \mathrm{~dB}$. The intensity decreases by a factor of
1 1000
2 10000
3 10
4 100
Explanation:
D The intensity level of the sound initially will be , $\beta_{1}=10 \log \left(\frac{\mathrm{I}}{\mathrm{I}_{0}}\right)$ Now, when the intensity of the sound is decreased, then the final intensity level of sound will be - $\beta_{2}=10 \log \left(\frac{I^{\prime}}{I_{0}}\right)$ As it is given that the sound is decreased by $20 \mathrm{~dB}$, which means that the difference between initial sound intensity level and final sound intensity level is $20 \mathrm{~dB}$, Therefore $\beta_{1}-\beta_{2}=10 \log \left(\frac{\mathrm{I}}{\mathrm{I}_{\circ}}\right)-10 \log \left(\frac{\mathrm{I}^{\prime}}{\mathrm{I}_{\circ}}\right)$ $20=10 \log \left(\frac{\mathrm{I}}{\mathrm{I}^{\prime}}\right)$ $2=\log \left(\frac{\mathrm{I}}{\mathrm{I}^{\prime}}\right)$ $-2=\log \left(\frac{\mathrm{I}^{\prime}}{\mathrm{I}}\right)$ $10^{-2}=\frac{\mathrm{I}^{\prime}}{\mathrm{I}}$ $\mathrm{I}^{\prime}=\frac{\mathrm{I}}{100}$ Therefore, the intensity of sound decreased by a factor of 100 .
172772
The beat frequency observed when two sound waves $y_{1}=0.5 \sin (410 t)$ and $y_{2}=0.5 \sin$ $(454 t)$ travel in the same direction is
1 5
2 3
3 7
4 2
5 4
Explanation:
C Given that, Equations of waves are- $\mathrm{y}_{1}=0.5 \sin (410 \mathrm{t})$ $\mathrm{y}_{2}=0.5 \sin (454 \mathrm{t})$ Comparing above equations with standard equation $\mathrm{y}=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx}) \text { we get }$ $\omega_{1}=410 \& \omega_{2}=454$ Now, frequency of $f_{1}=\frac{410}{2 \pi} \text { and } f_{2}=\frac{454}{2 \pi}$ $f_{1}=\frac{205 \times 7}{22} f_{2}=\frac{227 \times 7}{22}$ $f_{1}=65.227 \mathrm{~Hz} f_{2}=72.227 \mathrm{~Hz}$ Hence, the beat frequency $(f)=f_{1}-f_{2}$ $\mathrm{f}=72.227-65.227 \sqcup 7 \text { beats }$
Kerala CEE 2012
WAVES
172773
Two sources of sound placed closed to each other, are emitting progressive waves given by $Y_{1}=4 \sin 600 \pi t$ and $Y_{2}=5 \sin 608 \pi t$ An observer located near these two sources of sound will hear
1 4 beat/s with intensity ratio $25: 16$ between waxing and waning
2 8 beat/s with intensity ratio $25: 16$ between waxing and waning
3 8 beat/s with intensity ratio $81: 1$ between waxing and waning
4 4 beat/s with intensity ratio $81: 1$ between waxing and waning
Explanation:
D Given progressive wave equation $\mathrm{Y}_{1}=4 \sin 600 \pi \mathrm{t}$ $\mathrm{Y}_{2}=5 \sin 608 \pi \mathrm{t}$ $\text { standard equation }$ $\mathrm{Y}=\mathrm{A} \sin \omega \mathrm{t}$ $\text { from equation (i) }$ $\omega \mathrm{t}==600 \pi \mathrm{t}$ $2 \pi \mathrm{f}_{1} \mathrm{t}=600 \pi \mathrm{t}$ $\mathrm{f}_{1}=300 \mathrm{~Hz}$ $\text { from equation (ii) }$ $2 \pi \mathrm{f}_{2} \mathrm{t}=608 \pi \mathrm{t}$ $\mathrm{f}_{2}=304 \mathrm{~Hz}$ Number of beats $=f_{2}-f_{1}=304-300=4$ beats $/ \mathrm{sec}$ $\frac{I_{\text {max }}}{I_{\text {min }}}=\left(\frac{A_{1}+A_{2}}{A_{1}-A_{2}}\right)^{2}$ $\frac{I_{\text {max }}}{I_{\text {min }}}=\left(\frac{4+5}{4-5}\right)^{2}$ $I_{\text {max }}: I_{\text {min }}=81: 1$
AIPMT-2012
WAVES
172777
There are two waves having wavelengths 100 $\mathrm{cm}$ and $101 \mathrm{~cm}$ and same velocity $303 \mathrm{~m} / \mathrm{s}$. The beat frequency is
172842
The intensity of sound gets reduced by $10 \%$ on passing through a slab. The reduction in intensity on passage through three consecutive slabs is
1 $30 \%$
2 $27.1 \%$
3 $20 \%$
4 $36 \%$
Explanation:
B After passing through $1^{\text {st }}$ slab, Intensity $=90 \%$ of $\mathrm{I}=0.9 \mathrm{I}$ After second slab, intensity $=90 \%$ of $(0.9 \mathrm{I})$ $=81 \% \mathrm{I}=0.81 \mathrm{I}$ After third slab, intensity $=90 \%$ of $(0.81 \mathrm{I})$ $=72.9 \% \text { of } \mathrm{I}$ $\therefore \quad$ Reduction in intensity $=27.1 \%$
J and K CET- 2004
WAVES
172779
A sound absorber attenuates the sound level by $20 \mathrm{~dB}$. The intensity decreases by a factor of
1 1000
2 10000
3 10
4 100
Explanation:
D The intensity level of the sound initially will be , $\beta_{1}=10 \log \left(\frac{\mathrm{I}}{\mathrm{I}_{0}}\right)$ Now, when the intensity of the sound is decreased, then the final intensity level of sound will be - $\beta_{2}=10 \log \left(\frac{I^{\prime}}{I_{0}}\right)$ As it is given that the sound is decreased by $20 \mathrm{~dB}$, which means that the difference between initial sound intensity level and final sound intensity level is $20 \mathrm{~dB}$, Therefore $\beta_{1}-\beta_{2}=10 \log \left(\frac{\mathrm{I}}{\mathrm{I}_{\circ}}\right)-10 \log \left(\frac{\mathrm{I}^{\prime}}{\mathrm{I}_{\circ}}\right)$ $20=10 \log \left(\frac{\mathrm{I}}{\mathrm{I}^{\prime}}\right)$ $2=\log \left(\frac{\mathrm{I}}{\mathrm{I}^{\prime}}\right)$ $-2=\log \left(\frac{\mathrm{I}^{\prime}}{\mathrm{I}}\right)$ $10^{-2}=\frac{\mathrm{I}^{\prime}}{\mathrm{I}}$ $\mathrm{I}^{\prime}=\frac{\mathrm{I}}{100}$ Therefore, the intensity of sound decreased by a factor of 100 .
