J & K CET Engineering J and K - CET Engineering Solved Paper-2005

  • question_answer
    The intensity level of sound A is 30 dB greater than of B. How many times more intense is the sound A than B?

    A)  \[10\]          

    B)  \[100\]

    C)  \[1000\]        

    D)  \[2\]

    Correct Answer: C

    Solution :

    The intensity level in sound is given by \[L={{\log }_{10}}\frac{I}{{{I}_{0}}}\] where \[{{I}_{0}}\] is initial intensity. Given,       \[{{L}_{A}}=30+{{L}_{B}}\] \[\therefore \] \[{{\log }_{10}}\frac{{{I}_{A}}}{{{I}_{0}}}=30+{{\log }_{10}}\frac{{{I}_{B}}}{{{I}_{0}}}\] \[\Rightarrow \] \[{{\log }_{10}}\frac{{{I}_{A}}}{{{I}_{B}}}=3\] \[\Rightarrow \] \[\frac{{{I}_{A}}}{{{I}_{B}}}={{10}^{3}}\] \[\Rightarrow \] \[{{I}_{A}}=1000{{I}_{B}}\] Hence, sound A is 1000 times more intense than sound B.                                   


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