MGIMS WARDHA MGIMS WARDHA Solved Paper-2011

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

    A)  no change

    B)  increase by 0.98 mL

    C)   decrease by 0.75 mL

    D)   increase by 1.50 mL

    Correct Answer: C

    Solution :

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


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