Question

An increase in the intensity level of one decibel implies an increase in intensity of
A.1%
B.3.01%
C.26%
D.0.1%

Answer 1

Hint: Use the formula for intensity of sound level in units of\[{\text{dB}}\].
Sound intensity level describes the level of sound relative to the reference sound. The formula for determining the intensity of sound level \[\beta \] is,
\[\beta = 10\,{\log _{10}}\left( {\dfrac{I}{{{I_0}}}} \right)\]
Here, I is the intensity of sound you heard and \[{I_0}\] is the threshold intensity of hearing. The threshold intensity of hearing is the faintest sound that a human can hear. It has value \[{10^{ - 12}}\,{\text{W}}\,{{\text{m}}^{ - 2}}\]. The corresponding sound intensity level for threshold intensity is 0 decibels.

Complete step by step answer:
Intensity of a sound wave is defined as the ratio of the power of waves transmitted through a given area. It has a unit \[{\text{W}}\,{{\text{m}}^{ - {\text{2}}}}\].
Suppose \[\beta \] is the intensity level of sound of intensity \[{I_1}\]. Therefore,
\[\beta = 10\,{\log _{10}}\left( {\dfrac{{{I_1}}}{{{I_0}}}} \right)\] …… (1)

An increase in the intensity level by one decibel implies,
\[\beta + 1 = 10\,{\log _{10}}\left( {\dfrac{{{I_2}}}{{{I_0}}}} \right)\] …… (2)



Subtract equation (1) from equation (2).
\[\beta + 1 - \beta = 10\,{\log _{10}}\left( {\dfrac{{{I_2}}}{{{I_0}}}} \right) - 10\,{\log _{10}}\left( {\dfrac{{{I_1}}}{{{I_0}}}} \right)\]
\[ \Rightarrow 1 = 10\,\left( {{{\log }_{10}}\left( {\dfrac{{{I_2}}}{{{I_0}}}} \right) - \,{{\log }_{10}}\left( {\dfrac{{{I_1}}}{{{I_0}}}} \right)} \right)\]
\[ \Rightarrow 1 = 10\,\left( {{{\log }_{10}}\left( {\dfrac{{{I_2}}}{{{I_0}}}} \right) - \,{{\log }_{10}}\left( {\dfrac{{{I_1}}}{{{I_0}}}} \right)} \right)\]
\[ \Rightarrow 1 = 10\,{\log _{10}}\left( {\dfrac{{{I_2}}}{{{I_1}}}} \right)\]
\[ \Rightarrow \,{\log _{10}}\left( {\dfrac{{{I_2}}}{{{I_1}}}} \right) = \dfrac{1}{{10}}\]
\[ \Rightarrow \dfrac{{{I_2}}}{{{I_1}}} = {10^{0.1}}\]
\[\therefore \dfrac{{{I_2}}}{{{I_1}}} = 1.26\]

Suppose \[{I_1} = 1\].
\[\therefore {I_2} = {I_1} + 0.26\]
\[ \Rightarrow {I_2} = {I_1} + 26\% \]

Therefore, one decibel increase in the intensity level implies 26% increase in the intensity of the sound.

So, the correct answer is Option C .

Note:
Assume the intensity of the first sound as 1 unit. Then if the intensity of the second sound is 1.26 times the intensity of the first sound, it is 26% greater than the intensity of the first sound.