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NEET AIPMT SOLVED PAPER SCREENING 2005

  • question_answer
                    A point source emits sound equally in all directions in a non-absorbing medium. Two points P and Q are at distance 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

    Correct Answer: A

    Solution :

                              Key Idea: The average power per unit area that is incident perpendicular to the direction of propagation is called the intensity.                 Intensity of sound                 \[I=\frac{P}{4\pi {{r}^{2}}}\]                         \[orI\,\propto \,\frac{1}{{{r}^{2}}}\]                         \[or\frac{{{I}_{1}}}{{{I}_{2}}}={{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{2}}\]                 Here, \[{{r}_{1}}=2\,m,\,{{r}_{2}}=3\,m\]                 Substituting the values, we have                 \[\frac{{{I}_{1}}}{{{I}_{2}}}={{\left( \frac{3}{2} \right)}^{2}}=\frac{9}{4}\]                 Note:    As amplitude \[A\propto \sqrt{l}\], a spherical harmonic wave emanating from a point source can therefore, be written as:                         \[y(r,\,t)=\frac{A}{r}\sin (kr-\omega t)\]


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