CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    Compressibility of water is\[5\times {{10}^{-10}}{{m}^{2}}/N\]. The change in volume of 100 mL water subjected to\[15\times {{10}^{6}}Pa\]pressure will:

    A)  no change

    B)  increase by 0.75 mL

    C)  decrease by 1.50 mL

    D)  increase by 1.50 mL

    E)  decrease by 0.75 mL

    Correct Answer: E

    Solution :

    \[K=\frac{1}{B}=\frac{\Delta V}{V\Delta P}\] \[\therefore \]  \[5\times {{10}^{-10}}=\frac{\Delta V}{100\times {{10}^{-3}}\times 15\times {{10}^{6}}}\] \[\Delta V=5\times {{10}^{-10}}\times 100\times {{10}^{-3}}\times 15\times {{10}^{6}}\] \[=0.75\text{ }mL\] Since, pressure increases, so volume will decrease.


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