JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Afternoon)

  • question_answer
    A small speaker delivers 2 W of audio output. At what distance from the speaker will one detect 120 dB intensity sound? [Given reference intensity of sound as \[{{10}^{-12}}W/{{m}^{2}}\]] [JEE Main 12-4-2019 Afternoon]

    A) 10 cm             

    B) 30 cm

    C) 40 cm

    D) 20 cm

    Correct Answer: C

    Solution :

    Loudness of sound is given by \[dB=10\log \frac{I}{{{I}_{0}}}\]\[\left( \begin{align}   & \text{I is intensity of sound} \\  & {{\text{I}}_{0}}\text{ is reference intensity of sound} \\ \end{align} \right)\] \[\therefore \]\[120=10\log \left( \frac{I}{{{I}_{0}}} \right)\]\[\Rightarrow I=1W/{{m}^{2}}\] Also\[I=\frac{P}{4\pi {{r}^{2}}}=\frac{2}{4\pi {{r}^{2}}}\] \[\therefore \]\[r=\sqrt{\frac{2}{4\pi }}=\sqrt{\frac{1}{2\pi }}m=0.399m\approx 40cm\]             


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