Tag: intensity and loudness

Questions Related to intensity and loudness

The loudness of sound is measured in _______ .

intensity and loudness waves physics

  1. decibels

  2. metres

  3. kilograms

  4. seconds

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Correct Option: A
Explanation:

The loudness of sound is measured in units called decibels (dB). A decibel unit expresses the relative intensity of sounds on a scale from zero for the average least perceptible sound to about 100 dB, which is near the level most people find uncomfortably loud.

S. I. unit of intensity of sound is:

intensity and loudness waves physics

  1. joule s$^{-1}$ m$^{-2}$

  2. watt m$^{-2}$

  3. joule s/m$^2$

  4. both A and B

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Correct Option: D
Explanation:

In S. I. unit of intensity of sound is
joule s$^{-1}$ m$^{-2}$ or watt m$^{-2}$
Intensity = sound energy/time $\times$ area

Identify the scale which can be used to measure the intensity of sound?

intensity and loudness waves physics

  1. decibel

  2. acoustic

  3. ultrasound

  4. infrasound

  5. Hertz

Show answer
Correct Option: A
Explanation:

Intensity (or loudness) of sound         $L =10 \log _{10} \dfrac{I}{I _o}$

It is measured in decibel  (dB)

A normal human being can hear sound having an intensity level of maximum ...................

intensity and loudness waves physics

  1. 50 dB

  2. 80 dB

  3. 100 dB

  4. 150 dB

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Correct Option: B
Explanation:

Decibel levels help us to understand the loudness of a sound with respect to a reference level. A decibel level of zero dB represents the lowest level of sound and a level of 80 dB represents the level of pain for humans.

How many order of magnitude more powerful is 90 dB sound than 40 dB sound?

intensity and loudness waves physics

  1. 5

  2. 50

  3. 500

  4. $ {10}^ {5} $

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Correct Option: D
Explanation:

$90 = 10 log (\dfrac{I _{1}}{I})$

$40 = 10 log (\dfrac{I _{2}}{I})$

$90 - 40 = 10 log (\dfrac{I _{1}}{I _{2}})$

$\dfrac{I _{1}}{I _{2}} = 10^{5}$

If the intensity of sound is increased by a factor of 30 , by how many decibels is the sound level increased ?

intensity and loudness waves physics

  1. 12 dB

  2. 14.77 dB

  3. 10 dB

  4. 13 dB

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Correct Option: B
Explanation:
The intensity of sound in increased by factor $30$
Let $2$ be the intensity of previous sound
Intensity of recent sound $I=30I$
$\beta =10 \log\dfrac{I}{I _o}$
$\beta _1=10 \log\dfrac{I}{I _o}$ where intensity of sound is $ I$
$\beta _1=10 \log\dfrac{30I}{I _o}$ when intensity of sound is $I$
Increased in sound level $\Rightarrow \beta _2-\beta _1$
$=10 \log\dfrac{30I}{I _o}-10\log \dfrac{I}{I _o}$
$=10 \log 30$
$10 \log \dfrac{30 I}{I}$
$=14.77d\beta$.

When a person wears a hearing aid, the sound intensity level increases by 30 dB. The sound intensity increases by  

intensity and loudness waves physics

  1. e$^{3}$

  2. 10$^{3}$

  3. 30

  4. 10$^{2}$

Show answer
Correct Option: B
Explanation:

Let the intensity of the sound without hearing Aid is $I _0$, and the 


intensity after wearing hearing aid is $I$, then from the formula, 

$L _{db}=10 \log _{10}{\dfrac{I}{I _0}}$

$30=10 \log _{10}{\dfrac{I}{I _0}}$

$\dfrac{I}{I _0}=10^3$

$I=10^3 \times I _0$

Option "B" is correct.

Spherical sound waves are emitted uniformly in all directions from a point source. The variation in sound level SL as a function of distance 'r' from the source can be written as 

intensity and loudness waves physics

  1. SL = -b log r$^{a}$

  2. SL = a - b (log r)$^{2}$

  3. SL = a - b log r

  4. SL = a - b/r$^{2}$

Show answer
Correct Option: C
Explanation:

we know $SL = 10 log \dfrac{I}{I _o}$


$I = \dfrac{P}{4\pi r^2}$  where P is the power of the source

$SL = 10  log\dfrac{P}{4\pi r^2I _o} = 10  log \dfrac{P}{4\pi I _o} - 10  log  r^2 = 10  log \dfrac{P}{4\pi I _o} - 20  log  r = a - b  log  r$

The intensity level of two sounds are 100 dB and 50 dB. What is the ratio of their intensities?

intensity and loudness waves physics

  1. 10$^{1}$

  2. 10$^{3}$

  3. 10$^{5}$

  4. 10$^{10}$

Show answer
Correct Option: C
Explanation:

Loudness of two sounds is given as  $L _2 = 100 \ dB$ and $L _1 = 50 \ dB$

Loudness of sound  $L = 10 \log _{10}\dfrac{I}{I _o}$
$\implies$  $L _2 - L _1 = 10\log _{10}\dfrac{I _2}{I _1}$

Or  $100 - 50   = 10\log _{10}\dfrac{I _2}{I _1}$

Or  $\log _{10}\dfrac{I _2}{I _1} = 5$
Or  $\dfrac{I _2}{I _1} = 10^5 $

A source of sound emits 200 W power which is uniformly distributed over a sphere of radius 10 m. What is the loudness of sound on the surface of sphere? 

intensity and loudness waves physics

  1. 70dB

  2. 107dB

  3. 80dB

  4. 112dB

Show answer
Correct Option: D
Explanation:

Intensity is given by:
$I=\dfrac{W}{4\pi{r}^2}$
$I=\dfrac{200}{4\pi\times 100} =0.159W/m^2$

In terms of decibels, $I=10\log _{ 10 }{ I } +120 dB$
$I=112dB$