RAJASTHAN ­ PET Rajasthan PET Solved Paper-2006

  • question_answer
    The compressibility of water is\[5\times {{10}^{-10}}{{m}^{2}}/\text{ }N\]. If the pressure at volume 100 ml is\[15\times {{10}^{6}}Pa,\]then the change of volume will be

    A)  zero

    B)  increase to 0.75 ml

    C)  decrease to 0.75 ml

    D)  increase to 1.50 ml

    Correct Answer: C

    Solution :

     \[K=\frac{\Delta V}{V\Delta p}\] \[\therefore \] \[5\times {{10}^{-10}}=\frac{\Delta V}{100\times {{10}^{-3}}\times 15\times {{10}^{6}}}\] \[\Delta V=5\times {{10}^{-10}}\times 100\times {{10}^{-3}}\times 15\times {{10}^{6}}\] \[=0.75\text{ }ml\] Therefore pressure increases and volume decreases.


You need to login to perform this action.
You will be redirected in 3 sec spinner