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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2005

  • question_answer
        A spring of force constant 800 N/m has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is:

    A)  16 J                                       

    B)  8 J

    C)  32 J                                       

    D)  24 J   

    Correct Answer: B

    Solution :

                    The work is stored as the P.E. of the body and is given by, \[U=\int_{{{x}_{1}}}^{{{x}_{2}}}{{{F}_{external}}}dx\] Or           \[U=\int_{{{x}_{1}}}^{{{x}_{2}}}{kx}\,dx\]                 \[=\frac{1}{2}k(x_{2}^{2}-x_{1}^{2})\]                 \[=\frac{800}{2}[{{(0.15)}^{2}}-{{(0.05)}^{2}}]\]                                                 \[[k=800(given)]\]                 \[=400[0.2\times 0.1]\]                 \[=8\,joule\]


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