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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 13

  • question_answer
    An isotropic point source S of sound emits constant power. Two points A and B separated by a distance r are situated near the source as shown in figure. The difference of the intensity level of sound at the points A & B is about

    A)         \[3dB\]                         

    B) \[2dB\]

    C) \[6dB\]             

    D)        \[12dB\]

    Correct Answer: C

    Solution :

    Intensity at A, \[{{I}_{A}}=\frac{P}{4\pi {{r}^{2}}};\] intensity at B, \[{{I}_{B}}=\frac{P}{4\pi \,{{(2r)}^{2}}}\]
    Sound level at A,  \[{{S}_{A}}=10\log \frac{{{I}_{A}}}{{{I}_{0}}}\]
    Sound level at B, \[{{S}_{B}}=10\log \frac{{{I}_{B}}}{{{I}_{0}}}\]
    Difference of sound level at A and B is \[{{S}_{A}}-{{S}_{B}}=10\log \frac{{{I}_{A}}}{{{I}_{0}}}-10\log \frac{{{I}_{B}}}{{{I}_{0}}}=10\log \left( \frac{{{I}_{A}}}{{{I}_{B}}} \right)\]
    \[=10\log 4=20\log 2\approx 6\,dB\]


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