
A point source emits sound equally in all directions in a non-absorbing medium. Two points P and Q are at distances of 2m and 3m respectively from the source. The ratio of the intensities of the waves at P and Q is
A. 9 : 4
B. 2 : 3
C. 3 : 2
D. 4 : 9
Answer
161.1k+ views
Hint:In this question, we need to determine the ratio of the intensities of the waves at P and Q. For this, we need to use the concept that the intensity of the wave is inversely proportional to the distance. By the process of simplification, we get the desired ratio.
Formula used:
The intensity of a sound is given below.
\[I = \dfrac{P}{{4\pi {r^2}}}\]
Here, \[I\] is the intensity of sound, \[P\] is the power of a source and \[r\] is the distance of a point from the source.
That means the intensity of the wave is inversely proportional to the distance of a point from the source. Mathematically, it is denoted by \[Intensity{\text{ }}\alpha {\text{ }}\dfrac{1}{{{{\left( {{\text{Distance}}} \right)}^2}}}\]
Complete step by step solution:
We know that the points P and Q are at distances of 2m and 3m respectively from the source. Consider \[{d_1} = 2{\text{ m,}}{d_2} = 3{\text{ m}}\]. Here, \[{d_1}\]and \[{d_2}\] are the distances of the points P and Q from the source respectively.
Let us find the ratio of their intensities. Now, suppose that \[{I_1}\] and \[{I_2}\] are the intensities of the waves at points P and Q respectively.
We know that \[Intensity{\text{ }}\alpha {\text{ }}\dfrac{1}{{{{\left( {{\text{Distance}}} \right)}^2}}}\]
\[{I_1}{\text{ }}\alpha {\text{ }}\dfrac{1}{{{{\left( 2 \right)}^2}}}\] and \[{I_2}{\text{ }}\alpha {\text{ }}\dfrac{1}{{{{\left( 3 \right)}^2}}}\]
Thus, we get
\[{I_1}{\text{ }} = k{\text{ }}\dfrac{1}{{{{\left( 2 \right)}^2}}}\] and \[{I_2}{\text{ }} = k{\text{ }}\dfrac{1}{{{{\left( 3 \right)}^2}}}\]
Here, \[k\] is the proportionality constant.
So, we get
\[\dfrac{{{I_1}{\text{ }}}}{{{I_2}{\text{ }}}} = {\text{ }}\dfrac{{\dfrac{k}{{{{\left( 2 \right)}^2}}}}}{{\dfrac{k}{{{{\left( 3 \right)}^2}}}}}\]
\[\dfrac{{{I_1}{\text{ }}}}{{{I_2}{\text{ }}}} = {\text{ }}\dfrac{k}{{{{\left( 2 \right)}^2}}} \times \dfrac{{{{\left( 3 \right)}^2}}}{k}\]
\[\dfrac{{{I_1}{\text{ }}}}{{{I_2}{\text{ }}}} = {\text{ }}\dfrac{{{{\left( 3 \right)}^2}}}{{{{\left( 2 \right)}^2}}}\]
By simplifying, we get
\[\dfrac{{{I_1}{\text{ }}}}{{{I_2}{\text{ }}}} = {\text{ }}\dfrac{9}{4}\]
Hence, the ratio of the intensities of the waves at P and Q is 9 : 4.
Therefore, the correct option is (A).
Note: Many students make mistakes in understanding the relation between intensity and the distance of a point from the source. For understanding this relationship, we must know the formula of intensity of sound. If so, then only we get the desired result.
Formula used:
The intensity of a sound is given below.
\[I = \dfrac{P}{{4\pi {r^2}}}\]
Here, \[I\] is the intensity of sound, \[P\] is the power of a source and \[r\] is the distance of a point from the source.
That means the intensity of the wave is inversely proportional to the distance of a point from the source. Mathematically, it is denoted by \[Intensity{\text{ }}\alpha {\text{ }}\dfrac{1}{{{{\left( {{\text{Distance}}} \right)}^2}}}\]
Complete step by step solution:
We know that the points P and Q are at distances of 2m and 3m respectively from the source. Consider \[{d_1} = 2{\text{ m,}}{d_2} = 3{\text{ m}}\]. Here, \[{d_1}\]and \[{d_2}\] are the distances of the points P and Q from the source respectively.
Let us find the ratio of their intensities. Now, suppose that \[{I_1}\] and \[{I_2}\] are the intensities of the waves at points P and Q respectively.
We know that \[Intensity{\text{ }}\alpha {\text{ }}\dfrac{1}{{{{\left( {{\text{Distance}}} \right)}^2}}}\]
\[{I_1}{\text{ }}\alpha {\text{ }}\dfrac{1}{{{{\left( 2 \right)}^2}}}\] and \[{I_2}{\text{ }}\alpha {\text{ }}\dfrac{1}{{{{\left( 3 \right)}^2}}}\]
Thus, we get
\[{I_1}{\text{ }} = k{\text{ }}\dfrac{1}{{{{\left( 2 \right)}^2}}}\] and \[{I_2}{\text{ }} = k{\text{ }}\dfrac{1}{{{{\left( 3 \right)}^2}}}\]
Here, \[k\] is the proportionality constant.
So, we get
\[\dfrac{{{I_1}{\text{ }}}}{{{I_2}{\text{ }}}} = {\text{ }}\dfrac{{\dfrac{k}{{{{\left( 2 \right)}^2}}}}}{{\dfrac{k}{{{{\left( 3 \right)}^2}}}}}\]
\[\dfrac{{{I_1}{\text{ }}}}{{{I_2}{\text{ }}}} = {\text{ }}\dfrac{k}{{{{\left( 2 \right)}^2}}} \times \dfrac{{{{\left( 3 \right)}^2}}}{k}\]
\[\dfrac{{{I_1}{\text{ }}}}{{{I_2}{\text{ }}}} = {\text{ }}\dfrac{{{{\left( 3 \right)}^2}}}{{{{\left( 2 \right)}^2}}}\]
By simplifying, we get
\[\dfrac{{{I_1}{\text{ }}}}{{{I_2}{\text{ }}}} = {\text{ }}\dfrac{9}{4}\]
Hence, the ratio of the intensities of the waves at P and Q is 9 : 4.
Therefore, the correct option is (A).
Note: Many students make mistakes in understanding the relation between intensity and the distance of a point from the source. For understanding this relationship, we must know the formula of intensity of sound. If so, then only we get the desired result.
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