BCECE Medical BCECE Medical Solved Papers-2011

  • question_answer
    A spring of constant \[5\times {{10}^{3}}\] N/m is stretched initially by 5 cm from the unstretched position. Then the work required to stretch it further by another 5 cm is

    A)  6.25 Nm        

    B)  12.5 Nm

    C)  18.75 Nm       

    D)  25.00 Nm

    Correct Answer: C

    Solution :

    Work, \[W=\frac{1}{2}\,k\,(x_{2}^{2}-x_{1}^{3})\]                 \[=\frac{1}{2}\times 5\times {{10}^{3}}\left[ {{\left( \frac{10}{100} \right)}^{2}}-{{\left( \frac{5}{100} \right)}^{2}} \right]\]                 \[=\frac{1}{2}\times \frac{5\times {{10}^{3}}}{{{10}^{4}}}(100-25)\] \[=\frac{75}{4}\]


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