172791
The difference of sound levels between two points is $40 \mathrm{~dB}$. What is the ratio of pressure amplitudes between the two points?
1 10
2 200
3 100
4 400
Explanation:
C Given, $\mathrm{L}=10 \log \frac{\mathrm{I}}{\mathrm{I}_{\mathrm{o}}}$ $\mathrm{L}_{1}=10 \log \frac{\mathrm{I}_{1}}{\mathrm{I}_{\mathrm{o}}}$ $\mathrm{L}_{2}=10 \log \frac{\mathrm{I}_{2}}{\mathrm{I}_{\mathrm{o}}}$ $\mathrm{L}_{2}-\mathrm{L}_{1}=\Delta \mathrm{L}=40 \mathrm{~dB}$ Where, $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ be the two sound levels. Since, we know that, $\mathrm{L}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$ Now, $\Delta \mathrm{L}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$ $40=10 \log _{10}\left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$ $4=\log _{10}\left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$ $10^{4}=\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}$ We know, $\mathrm{P} \propto \sqrt{\mathrm{I}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\sqrt{\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\sqrt{10^{4}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=10^{2}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=100$
JCECE-2015
WAVES
172792
The time of reverberation of a room $A$ is one second. What will be the time (in seconds) of reverberation of a room, having all the dimensions double of those of room $A$ ?
1 2
2 4
3 $\frac{1}{2}$
4 1
Explanation:
A Reverberation time, $\mathrm{T}=\frac{0.16 \mathrm{~V}}{\Sigma \text { as }}$ Where, $\Sigma$ as $=a_{1} s_{1}+a_{2} s_{2}+\ldots . .=$ total absorption of the room. Where $\mathrm{V}$ is volume of the room $\mathrm{S}$ is the total surface area of room and $\bar{a}$ is average absorption coefficient Time of reverberation $\propto \frac{\mathrm{V}}{\mathrm{A}}$ (According to Sabine's formula) $\frac{\mathrm{T}^{\prime}}{\mathrm{T}}=\frac{\mathrm{V}^{\prime}}{\mathrm{S}^{\prime}} \times \frac{\mathrm{S}}{\mathrm{V}}$ $\frac{\mathrm{T}^{\prime}}{\mathrm{T}}=\frac{(2 \mathrm{a})^{3}}{(2 \mathrm{a})^{2}} \times \frac{\mathrm{a}^{2}}{\mathrm{a}^{3}}=2$ Hence, $\mathrm{T}^{\prime}=2 \mathrm{~T}=2 \times 1=2$
JCECE-2007
WAVES
172793
If equation of a sound wave is $\mathrm{y}=0.0015 \mathrm{sin}$ $(62.8 x+314 t)$ then its wavelength will be
1 0.1 unit
2 2 unit
3 0.3 unit
4 0.2 unit
Explanation:
A Given, $y=0.0015 \sin (62.8 x+314 t)$ Comparing above equation with the standard equation of sound, $y=a \sin 2 \pi\left(\frac{x}{\lambda}+\frac{t}{T}\right)$ we get, $2 \frac{\pi \mathrm{x}}{\lambda}=62.8 \mathrm{x}$ $\lambda=\frac{2 \pi}{62.8}=\frac{2 \times 3.14}{62.8}$ $\lambda=\frac{6.28}{62.8}$ $\lambda=0.1$ unit
JCECE-2007
WAVES
172795
The distance travelled by a sound wave when a tuning fork completes 25 vibrations is $16.5 \mathrm{~m}$. If the frequency of the tuning fork is $500 \mathrm{~Hz}$, find the velocity of sound.
1 $350 \mathrm{~ms}^{-1}$
2 $330 \mathrm{~ms}^{-1}$
3 $300 \mathrm{~ms}^{-1}$
4 $450 \mathrm{~ms}^{-1}$
Explanation:
B Given, Frequency of tuning fork $(n)=500 \mathrm{~Hz}$ Wavelength $(\lambda)=\frac{16.5}{25}$ Velocity of sound, $\mathrm{v}=\mathrm{n} \lambda$ $=500 \times \frac{16.5}{25}$ $\mathrm{v} =330 \mathrm{~m} / \mathrm{s} .$
172791
The difference of sound levels between two points is $40 \mathrm{~dB}$. What is the ratio of pressure amplitudes between the two points?
1 10
2 200
3 100
4 400
Explanation:
C Given, $\mathrm{L}=10 \log \frac{\mathrm{I}}{\mathrm{I}_{\mathrm{o}}}$ $\mathrm{L}_{1}=10 \log \frac{\mathrm{I}_{1}}{\mathrm{I}_{\mathrm{o}}}$ $\mathrm{L}_{2}=10 \log \frac{\mathrm{I}_{2}}{\mathrm{I}_{\mathrm{o}}}$ $\mathrm{L}_{2}-\mathrm{L}_{1}=\Delta \mathrm{L}=40 \mathrm{~dB}$ Where, $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ be the two sound levels. Since, we know that, $\mathrm{L}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$ Now, $\Delta \mathrm{L}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$ $40=10 \log _{10}\left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$ $4=\log _{10}\left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$ $10^{4}=\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}$ We know, $\mathrm{P} \propto \sqrt{\mathrm{I}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\sqrt{\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\sqrt{10^{4}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=10^{2}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=100$
JCECE-2015
WAVES
172792
The time of reverberation of a room $A$ is one second. What will be the time (in seconds) of reverberation of a room, having all the dimensions double of those of room $A$ ?
1 2
2 4
3 $\frac{1}{2}$
4 1
Explanation:
A Reverberation time, $\mathrm{T}=\frac{0.16 \mathrm{~V}}{\Sigma \text { as }}$ Where, $\Sigma$ as $=a_{1} s_{1}+a_{2} s_{2}+\ldots . .=$ total absorption of the room. Where $\mathrm{V}$ is volume of the room $\mathrm{S}$ is the total surface area of room and $\bar{a}$ is average absorption coefficient Time of reverberation $\propto \frac{\mathrm{V}}{\mathrm{A}}$ (According to Sabine's formula) $\frac{\mathrm{T}^{\prime}}{\mathrm{T}}=\frac{\mathrm{V}^{\prime}}{\mathrm{S}^{\prime}} \times \frac{\mathrm{S}}{\mathrm{V}}$ $\frac{\mathrm{T}^{\prime}}{\mathrm{T}}=\frac{(2 \mathrm{a})^{3}}{(2 \mathrm{a})^{2}} \times \frac{\mathrm{a}^{2}}{\mathrm{a}^{3}}=2$ Hence, $\mathrm{T}^{\prime}=2 \mathrm{~T}=2 \times 1=2$
JCECE-2007
WAVES
172793
If equation of a sound wave is $\mathrm{y}=0.0015 \mathrm{sin}$ $(62.8 x+314 t)$ then its wavelength will be
1 0.1 unit
2 2 unit
3 0.3 unit
4 0.2 unit
Explanation:
A Given, $y=0.0015 \sin (62.8 x+314 t)$ Comparing above equation with the standard equation of sound, $y=a \sin 2 \pi\left(\frac{x}{\lambda}+\frac{t}{T}\right)$ we get, $2 \frac{\pi \mathrm{x}}{\lambda}=62.8 \mathrm{x}$ $\lambda=\frac{2 \pi}{62.8}=\frac{2 \times 3.14}{62.8}$ $\lambda=\frac{6.28}{62.8}$ $\lambda=0.1$ unit
JCECE-2007
WAVES
172795
The distance travelled by a sound wave when a tuning fork completes 25 vibrations is $16.5 \mathrm{~m}$. If the frequency of the tuning fork is $500 \mathrm{~Hz}$, find the velocity of sound.
1 $350 \mathrm{~ms}^{-1}$
2 $330 \mathrm{~ms}^{-1}$
3 $300 \mathrm{~ms}^{-1}$
4 $450 \mathrm{~ms}^{-1}$
Explanation:
B Given, Frequency of tuning fork $(n)=500 \mathrm{~Hz}$ Wavelength $(\lambda)=\frac{16.5}{25}$ Velocity of sound, $\mathrm{v}=\mathrm{n} \lambda$ $=500 \times \frac{16.5}{25}$ $\mathrm{v} =330 \mathrm{~m} / \mathrm{s} .$
172791
The difference of sound levels between two points is $40 \mathrm{~dB}$. What is the ratio of pressure amplitudes between the two points?
1 10
2 200
3 100
4 400
Explanation:
C Given, $\mathrm{L}=10 \log \frac{\mathrm{I}}{\mathrm{I}_{\mathrm{o}}}$ $\mathrm{L}_{1}=10 \log \frac{\mathrm{I}_{1}}{\mathrm{I}_{\mathrm{o}}}$ $\mathrm{L}_{2}=10 \log \frac{\mathrm{I}_{2}}{\mathrm{I}_{\mathrm{o}}}$ $\mathrm{L}_{2}-\mathrm{L}_{1}=\Delta \mathrm{L}=40 \mathrm{~dB}$ Where, $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ be the two sound levels. Since, we know that, $\mathrm{L}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$ Now, $\Delta \mathrm{L}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$ $40=10 \log _{10}\left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$ $4=\log _{10}\left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$ $10^{4}=\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}$ We know, $\mathrm{P} \propto \sqrt{\mathrm{I}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\sqrt{\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\sqrt{10^{4}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=10^{2}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=100$
JCECE-2015
WAVES
172792
The time of reverberation of a room $A$ is one second. What will be the time (in seconds) of reverberation of a room, having all the dimensions double of those of room $A$ ?
1 2
2 4
3 $\frac{1}{2}$
4 1
Explanation:
A Reverberation time, $\mathrm{T}=\frac{0.16 \mathrm{~V}}{\Sigma \text { as }}$ Where, $\Sigma$ as $=a_{1} s_{1}+a_{2} s_{2}+\ldots . .=$ total absorption of the room. Where $\mathrm{V}$ is volume of the room $\mathrm{S}$ is the total surface area of room and $\bar{a}$ is average absorption coefficient Time of reverberation $\propto \frac{\mathrm{V}}{\mathrm{A}}$ (According to Sabine's formula) $\frac{\mathrm{T}^{\prime}}{\mathrm{T}}=\frac{\mathrm{V}^{\prime}}{\mathrm{S}^{\prime}} \times \frac{\mathrm{S}}{\mathrm{V}}$ $\frac{\mathrm{T}^{\prime}}{\mathrm{T}}=\frac{(2 \mathrm{a})^{3}}{(2 \mathrm{a})^{2}} \times \frac{\mathrm{a}^{2}}{\mathrm{a}^{3}}=2$ Hence, $\mathrm{T}^{\prime}=2 \mathrm{~T}=2 \times 1=2$
JCECE-2007
WAVES
172793
If equation of a sound wave is $\mathrm{y}=0.0015 \mathrm{sin}$ $(62.8 x+314 t)$ then its wavelength will be
1 0.1 unit
2 2 unit
3 0.3 unit
4 0.2 unit
Explanation:
A Given, $y=0.0015 \sin (62.8 x+314 t)$ Comparing above equation with the standard equation of sound, $y=a \sin 2 \pi\left(\frac{x}{\lambda}+\frac{t}{T}\right)$ we get, $2 \frac{\pi \mathrm{x}}{\lambda}=62.8 \mathrm{x}$ $\lambda=\frac{2 \pi}{62.8}=\frac{2 \times 3.14}{62.8}$ $\lambda=\frac{6.28}{62.8}$ $\lambda=0.1$ unit
JCECE-2007
WAVES
172795
The distance travelled by a sound wave when a tuning fork completes 25 vibrations is $16.5 \mathrm{~m}$. If the frequency of the tuning fork is $500 \mathrm{~Hz}$, find the velocity of sound.
1 $350 \mathrm{~ms}^{-1}$
2 $330 \mathrm{~ms}^{-1}$
3 $300 \mathrm{~ms}^{-1}$
4 $450 \mathrm{~ms}^{-1}$
Explanation:
B Given, Frequency of tuning fork $(n)=500 \mathrm{~Hz}$ Wavelength $(\lambda)=\frac{16.5}{25}$ Velocity of sound, $\mathrm{v}=\mathrm{n} \lambda$ $=500 \times \frac{16.5}{25}$ $\mathrm{v} =330 \mathrm{~m} / \mathrm{s} .$
172791
The difference of sound levels between two points is $40 \mathrm{~dB}$. What is the ratio of pressure amplitudes between the two points?
1 10
2 200
3 100
4 400
Explanation:
C Given, $\mathrm{L}=10 \log \frac{\mathrm{I}}{\mathrm{I}_{\mathrm{o}}}$ $\mathrm{L}_{1}=10 \log \frac{\mathrm{I}_{1}}{\mathrm{I}_{\mathrm{o}}}$ $\mathrm{L}_{2}=10 \log \frac{\mathrm{I}_{2}}{\mathrm{I}_{\mathrm{o}}}$ $\mathrm{L}_{2}-\mathrm{L}_{1}=\Delta \mathrm{L}=40 \mathrm{~dB}$ Where, $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ be the two sound levels. Since, we know that, $\mathrm{L}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$ Now, $\Delta \mathrm{L}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$ $40=10 \log _{10}\left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$ $4=\log _{10}\left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$ $10^{4}=\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}$ We know, $\mathrm{P} \propto \sqrt{\mathrm{I}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\sqrt{\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\sqrt{10^{4}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=10^{2}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=100$
JCECE-2015
WAVES
172792
The time of reverberation of a room $A$ is one second. What will be the time (in seconds) of reverberation of a room, having all the dimensions double of those of room $A$ ?
1 2
2 4
3 $\frac{1}{2}$
4 1
Explanation:
A Reverberation time, $\mathrm{T}=\frac{0.16 \mathrm{~V}}{\Sigma \text { as }}$ Where, $\Sigma$ as $=a_{1} s_{1}+a_{2} s_{2}+\ldots . .=$ total absorption of the room. Where $\mathrm{V}$ is volume of the room $\mathrm{S}$ is the total surface area of room and $\bar{a}$ is average absorption coefficient Time of reverberation $\propto \frac{\mathrm{V}}{\mathrm{A}}$ (According to Sabine's formula) $\frac{\mathrm{T}^{\prime}}{\mathrm{T}}=\frac{\mathrm{V}^{\prime}}{\mathrm{S}^{\prime}} \times \frac{\mathrm{S}}{\mathrm{V}}$ $\frac{\mathrm{T}^{\prime}}{\mathrm{T}}=\frac{(2 \mathrm{a})^{3}}{(2 \mathrm{a})^{2}} \times \frac{\mathrm{a}^{2}}{\mathrm{a}^{3}}=2$ Hence, $\mathrm{T}^{\prime}=2 \mathrm{~T}=2 \times 1=2$
JCECE-2007
WAVES
172793
If equation of a sound wave is $\mathrm{y}=0.0015 \mathrm{sin}$ $(62.8 x+314 t)$ then its wavelength will be
1 0.1 unit
2 2 unit
3 0.3 unit
4 0.2 unit
Explanation:
A Given, $y=0.0015 \sin (62.8 x+314 t)$ Comparing above equation with the standard equation of sound, $y=a \sin 2 \pi\left(\frac{x}{\lambda}+\frac{t}{T}\right)$ we get, $2 \frac{\pi \mathrm{x}}{\lambda}=62.8 \mathrm{x}$ $\lambda=\frac{2 \pi}{62.8}=\frac{2 \times 3.14}{62.8}$ $\lambda=\frac{6.28}{62.8}$ $\lambda=0.1$ unit
JCECE-2007
WAVES
172795
The distance travelled by a sound wave when a tuning fork completes 25 vibrations is $16.5 \mathrm{~m}$. If the frequency of the tuning fork is $500 \mathrm{~Hz}$, find the velocity of sound.
1 $350 \mathrm{~ms}^{-1}$
2 $330 \mathrm{~ms}^{-1}$
3 $300 \mathrm{~ms}^{-1}$
4 $450 \mathrm{~ms}^{-1}$
Explanation:
B Given, Frequency of tuning fork $(n)=500 \mathrm{~Hz}$ Wavelength $(\lambda)=\frac{16.5}{25}$ Velocity of sound, $\mathrm{v}=\mathrm{n} \lambda$ $=500 \times \frac{16.5}{25}$ $\mathrm{v} =330 \mathrm{~m} / \mathrm{s} .$
