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BCECE Engineering BCECE Engineering Solved Paper-2007

  • question_answer
    Compressional wave pulses are sent to the bottom of sea from a ship and the echo is heard after 2 s. If bulk modulus of elasticity of water is \[2\times {{10}^{9}}\text{ }N/{{m}^{2}}\]and mean temperature is \[4{{\,}^{o}}C,\] the depth of the sea will be

    A)  1014 m                                

    B)  1414 m

    C)         2828 m                

    D)         none of these

    Correct Answer: B

    Solution :

    The speed of sound (longitudinal waves) in water is given by \[v=\sqrt{\frac{B}{d}}\] where B is bulk modulus of water and d is density. Given, \[B=2\times {{10}^{9}}N/{{m}^{2}},d={{10}^{3}}\,kg/{{m}^{3}}\] \[\therefore \]  \[v=\sqrt{\frac{2\times {{10}^{9}}}{{{10}^{3}}}}=1.414\times {{10}^{3}}\]                 \[=1414\,m/s\] When sound travels back to the observer, it covers twice the distance. So, time of echo.                                 \[t=\frac{2d}{v}\]                 \[\therefore \]  \[d=\frac{tv}{2}=\frac{1414\times 2}{2}=1414\,m\]


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