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PHXI15:WAVES

354803 A two-fold increase in intensity of a wave implies an increases of (given: $\log_{10} 2 = 0.301$ )

1

10 d B

2

2 d B

3

0.5 d B

4

3.01 d B

Explanation:

$L_{1} = 10 \log_{10} \frac{I}{I_{0}}$ decibel
$L_{2} = 10 \log \frac{2 I}{I_{o}} \Rightarrow Δ L = 3.01 d B$

PHXI15:WAVES

354804 The sound intensity is $0.008 W / m^{2}$ at a distance of $10 m$ from an isotropic point source of sound. The power of the source is

1 0.8

w a t t

2 2.5

w a t t

3 10

w a t t

4 8

w a t t

Explanation:

$\frac{P}{4 π r^{2}} = I$
$\Rightarrow P = I \cdot 4 π r^{2}$
$= (0.008 W / m^{2}) (4 \cdot π \cdot 10^{2}) = 10 W$

PHXI15:WAVES

354805 The intensity of sound increases at night due to

1 increase in density of air

2 decrease in density of air

3 low temperature

4 None of the above

PHXI15:WAVES

354806 When we hear a sound, we can identify its source based on

1 Intensity of sound

2 Amplitude of sound

3 Overtones present in the sound

4 Wavelength of sound

PHXI15:WAVES

354807 An increase in intensity level of $1 d B$ implies an increase in intensity of:
(given antilog $0.1 = 1.2589$ )

1

3.01 %

2

1 %

3

0.1 %

4

26 %

Explanation:

Intensity level is decibel is given by
$\begin{matrix} L = 10 \log_{10} \frac{I_{1}}{I_{0}} \\ L + 1 = 10 \log_{10} \frac{I_{2}}{I_{0}} \\ \Rightarrow 1 = 10 \log_{10} \frac{I_{2}}{I_{0}} - 10 \log_{10} \frac{I_{1}}{I_{0}} \end{matrix}$
$\Rightarrow \frac{1}{10} = \log_{10} \frac{I_{2}}{I_{1}}$
$\Rightarrow \frac{I_{2}}{I_{1}} = 1.26 \Rightarrow I_{2} = 1.26 I_{1}$
$\frac{Δ I}{I} \times 100 = 26 %$

PHXI15:WAVES

354803 A two-fold increase in intensity of a wave implies an increases of (given: $\log_{10} 2 = 0.301$ )

1

10 d B

2

2 d B

3

0.5 d B

4

3.01 d B

Explanation:

$L_{1} = 10 \log_{10} \frac{I}{I_{0}}$ decibel
$L_{2} = 10 \log \frac{2 I}{I_{o}} \Rightarrow Δ L = 3.01 d B$

PHXI15:WAVES

354804 The sound intensity is $0.008 W / m^{2}$ at a distance of $10 m$ from an isotropic point source of sound. The power of the source is

1 0.8

w a t t

2 2.5

w a t t

3 10

w a t t

4 8

w a t t

Explanation:

$\frac{P}{4 π r^{2}} = I$
$\Rightarrow P = I \cdot 4 π r^{2}$
$= (0.008 W / m^{2}) (4 \cdot π \cdot 10^{2}) = 10 W$

PHXI15:WAVES

354805 The intensity of sound increases at night due to

1 increase in density of air

2 decrease in density of air

3 low temperature

4 None of the above

PHXI15:WAVES

354806 When we hear a sound, we can identify its source based on

1 Intensity of sound

2 Amplitude of sound

3 Overtones present in the sound

4 Wavelength of sound

PHXI15:WAVES

354807 An increase in intensity level of $1 d B$ implies an increase in intensity of:
(given antilog $0.1 = 1.2589$ )

1

3.01 %

2

1 %

3

0.1 %

4

26 %

Explanation:

Intensity level is decibel is given by
$\begin{matrix} L = 10 \log_{10} \frac{I_{1}}{I_{0}} \\ L + 1 = 10 \log_{10} \frac{I_{2}}{I_{0}} \\ \Rightarrow 1 = 10 \log_{10} \frac{I_{2}}{I_{0}} - 10 \log_{10} \frac{I_{1}}{I_{0}} \end{matrix}$
$\Rightarrow \frac{1}{10} = \log_{10} \frac{I_{2}}{I_{1}}$
$\Rightarrow \frac{I_{2}}{I_{1}} = 1.26 \Rightarrow I_{2} = 1.26 I_{1}$
$\frac{Δ I}{I} \times 100 = 26 %$

PHXI15:WAVES

354803 A two-fold increase in intensity of a wave implies an increases of (given: $\log_{10} 2 = 0.301$ )

1

10 d B

2

2 d B

3

0.5 d B

4

3.01 d B

Explanation:

$L_{1} = 10 \log_{10} \frac{I}{I_{0}}$ decibel
$L_{2} = 10 \log \frac{2 I}{I_{o}} \Rightarrow Δ L = 3.01 d B$

PHXI15:WAVES

354804 The sound intensity is $0.008 W / m^{2}$ at a distance of $10 m$ from an isotropic point source of sound. The power of the source is

1 0.8

w a t t

2 2.5

w a t t

3 10

w a t t

4 8

w a t t

Explanation:

$\frac{P}{4 π r^{2}} = I$
$\Rightarrow P = I \cdot 4 π r^{2}$
$= (0.008 W / m^{2}) (4 \cdot π \cdot 10^{2}) = 10 W$

PHXI15:WAVES

354805 The intensity of sound increases at night due to

1 increase in density of air

2 decrease in density of air

3 low temperature

4 None of the above

PHXI15:WAVES

354806 When we hear a sound, we can identify its source based on

1 Intensity of sound

2 Amplitude of sound

3 Overtones present in the sound

4 Wavelength of sound

PHXI15:WAVES

354807 An increase in intensity level of $1 d B$ implies an increase in intensity of:
(given antilog $0.1 = 1.2589$ )

1

3.01 %

2

1 %

3

0.1 %

4

26 %

Explanation:

Intensity level is decibel is given by
$\begin{matrix} L = 10 \log_{10} \frac{I_{1}}{I_{0}} \\ L + 1 = 10 \log_{10} \frac{I_{2}}{I_{0}} \\ \Rightarrow 1 = 10 \log_{10} \frac{I_{2}}{I_{0}} - 10 \log_{10} \frac{I_{1}}{I_{0}} \end{matrix}$
$\Rightarrow \frac{1}{10} = \log_{10} \frac{I_{2}}{I_{1}}$
$\Rightarrow \frac{I_{2}}{I_{1}} = 1.26 \Rightarrow I_{2} = 1.26 I_{1}$
$\frac{Δ I}{I} \times 100 = 26 %$

PHXI15:WAVES

354803 A two-fold increase in intensity of a wave implies an increases of (given: $\log_{10} 2 = 0.301$ )

1

10 d B

2

2 d B

3

0.5 d B

4

3.01 d B

Explanation:

$L_{1} = 10 \log_{10} \frac{I}{I_{0}}$ decibel
$L_{2} = 10 \log \frac{2 I}{I_{o}} \Rightarrow Δ L = 3.01 d B$

PHXI15:WAVES

354804 The sound intensity is $0.008 W / m^{2}$ at a distance of $10 m$ from an isotropic point source of sound. The power of the source is

1 0.8

w a t t

2 2.5

w a t t

3 10

w a t t

4 8

w a t t

Explanation:

$\frac{P}{4 π r^{2}} = I$
$\Rightarrow P = I \cdot 4 π r^{2}$
$= (0.008 W / m^{2}) (4 \cdot π \cdot 10^{2}) = 10 W$

PHXI15:WAVES

354805 The intensity of sound increases at night due to

1 increase in density of air

2 decrease in density of air

3 low temperature

4 None of the above

PHXI15:WAVES

354806 When we hear a sound, we can identify its source based on

1 Intensity of sound

2 Amplitude of sound

3 Overtones present in the sound

4 Wavelength of sound

PHXI15:WAVES

354807 An increase in intensity level of $1 d B$ implies an increase in intensity of:
(given antilog $0.1 = 1.2589$ )

1

3.01 %

2

1 %

3

0.1 %

4

26 %

Explanation:

Intensity level is decibel is given by
$\begin{matrix} L = 10 \log_{10} \frac{I_{1}}{I_{0}} \\ L + 1 = 10 \log_{10} \frac{I_{2}}{I_{0}} \\ \Rightarrow 1 = 10 \log_{10} \frac{I_{2}}{I_{0}} - 10 \log_{10} \frac{I_{1}}{I_{0}} \end{matrix}$
$\Rightarrow \frac{1}{10} = \log_{10} \frac{I_{2}}{I_{1}}$
$\Rightarrow \frac{I_{2}}{I_{1}} = 1.26 \Rightarrow I_{2} = 1.26 I_{1}$
$\frac{Δ I}{I} \times 100 = 26 %$

PHXI15:WAVES

354803 A two-fold increase in intensity of a wave implies an increases of (given: $\log_{10} 2 = 0.301$ )

1

10 d B

2

2 d B

3

0.5 d B

4

3.01 d B

Explanation:

$L_{1} = 10 \log_{10} \frac{I}{I_{0}}$ decibel
$L_{2} = 10 \log \frac{2 I}{I_{o}} \Rightarrow Δ L = 3.01 d B$

PHXI15:WAVES

354804 The sound intensity is $0.008 W / m^{2}$ at a distance of $10 m$ from an isotropic point source of sound. The power of the source is

1 0.8

w a t t

2 2.5

w a t t

3 10

w a t t

4 8

w a t t

Explanation:

$\frac{P}{4 π r^{2}} = I$
$\Rightarrow P = I \cdot 4 π r^{2}$
$= (0.008 W / m^{2}) (4 \cdot π \cdot 10^{2}) = 10 W$

PHXI15:WAVES

354805 The intensity of sound increases at night due to

1 increase in density of air

2 decrease in density of air

3 low temperature

4 None of the above

PHXI15:WAVES

354806 When we hear a sound, we can identify its source based on

1 Intensity of sound

2 Amplitude of sound

3 Overtones present in the sound

4 Wavelength of sound

PHXI15:WAVES

354807 An increase in intensity level of $1 d B$ implies an increase in intensity of:
(given antilog $0.1 = 1.2589$ )

1

3.01 %

2

1 %

3

0.1 %

4

26 %

Explanation:

Intensity level is decibel is given by
$\begin{matrix} L = 10 \log_{10} \frac{I_{1}}{I_{0}} \\ L + 1 = 10 \log_{10} \frac{I_{2}}{I_{0}} \\ \Rightarrow 1 = 10 \log_{10} \frac{I_{2}}{I_{0}} - 10 \log_{10} \frac{I_{1}}{I_{0}} \end{matrix}$
$\Rightarrow \frac{1}{10} = \log_{10} \frac{I_{2}}{I_{1}}$
$\Rightarrow \frac{I_{2}}{I_{1}} = 1.26 \Rightarrow I_{2} = 1.26 I_{1}$
$\frac{Δ I}{I} \times 100 = 26 %$