AMU Medical AMU Solved Paper-2006

  • question_answer
    When a sphere is taken to bottom of sea 1 krr: deep, it contracts by 0.01%. The bulk modulus of elasticity of the material of sphere is \[\left( \text{Given }:\text{ Density of water }=\text{ 1 g}/\text{c}{{\text{m}}^{\text{3}}} \right)\]

    A)  \[\text{9}.\text{8}\times \text{1}{{0}^{\text{1}0}}\text{ N}/{{\text{m}}^{\text{2}}}\]   

    B)  \[\text{1}0.\text{2}\times \text{l}{{0}^{\text{1}0}}\text{ N}/{{\text{m}}^{\text{2}}}\]

    C)  \[0.\text{98}\times \text{1}{{0}^{\text{1}0}}\text{ N}/{{\text{m}}^{\text{2}}}\]               

    D)  \[\text{8}.\text{4}\times \text{l}{{0}^{\text{1}0}}\text{ N}/{{\text{m}}^{\text{2}}}\]

    Correct Answer: A

    Solution :

                     Bulk modulus \[K=\frac{\Delta p}{\frac{-\Delta V}{V}}\]                 \[v=\frac{\Delta \,V}{V}=\frac{0.01}{100}={{10}^{-4}}\] \[\Delta p=\] pressure of water                 \[=h\rho g\]                 \[={{10}^{3}}\times 1\times {{10}^{3}}\times 9.8\] \[\Delta \,p=9.8\times {{10}^{6}}\]                 \[K=\frac{9.8\times {{10}^{6}}}{{{10}^{-4}}}\]                 \[=9.8\times {{10}^{10}}N/{{m}^{2}}\]


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