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NEET Sample Paper NEET Sample Test Paper-71

  • 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, \[{{r}_{1}}=2\,\,m,\,\,{{r}_{2}}=3\,m\] \[{{r}_{1}}=2\,\,m,\,\,{{r}_{2}}=3\,\,m\] Note: As amplitude \[A\propto \sqrt{I}\], 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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