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NEET NEET SOLVED PAPER 2015 (C)

  • question_answer
    The approximate depth of an ocean is 2 700 m. The compressibility of water is \[45.4\times {{10}^{-11}}P{{a}^{-1}}\] and density of water is\[{{10}^{3}}kg/{{m}^{3}}\]. What fractional compression of water will be obtained at the bottom of the ocean?

    A)  \[0.8\times {{10}^{-2}}\]             

    B)  \[1.0\times {{10}^{-2}}\]             

    C)         \[1.2\times {{10}^{-2}}\]         

    D)         \[1.4\times {{10}^{-2}}\]

    Correct Answer: C

    Solution :

                    Given d = 2700m \[\rho ={{10}^{3}}kg/{{m}^{3}}\] Compressibility \[=45.4\times {{10}^{-11}}\] per pascal The pressure at the bottom of ocean is given by \[p=\rho gd\] \[={{10}^{3}}\times 10\times 2700=27\times {{10}^{6}}\,\text{Pa}\] So, fractional compression = compressibility                                                             \[\times \] pressure \[=45.4\times {{10}^{-11}}\times 27\times {{10}^{6}}=1.2\times {{10}^{-2}}\]


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