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AIIMS AIIMS Solved Paper-2006

  • 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 2 m and 3 m 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 :

    The average power per unit area that is incident perpendicular to the direction of propagation is called the intensity, i.e., \[I=\frac{P}{4\pi {{r}^{2}}}\] or  \[I\propto \frac{1}{{{r}^{2}}}\] or   \[\frac{{{I}_{1}}}{{{I}_{2}}}={{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{2}}\] Here, \[{{r}_{1}}=2m,{{r}_{2}}=3m\] \[\therefore \]  \[\frac{{{I}_{1}}}{{{I}_{2}}}={{\left( \frac{3}{2} \right)}^{2}}=\frac{9}{4}\] 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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