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AMU Medical AMU Solved Paper-1999

  • question_answer
    An elastic string has length\[\beta \]when subjected to 5 N tension. Its length is a when tension 4 N. When subjected to a tension of 9 N, its length will become

    A)  \[9(\beta -\alpha )\]

    B)  \[5\alpha -4\beta \]

    C)  \[5\beta -4\alpha \]

    D)  \[\beta -\alpha \]

    Correct Answer: C

    Solution :

    : \[k=\]force constant of spring = force/length. Increase in length due to force = force/\[k=\frac{F}{k}\]Original length\[=L\] \[\therefore \] Final length\[=L+\frac{F}{k}\] \[\therefore \]\[\beta =L+\frac{5}{k}\] \[\alpha =L+\frac{4}{k}\] \[\therefore \]\[\beta -\alpha =\frac{I}{k}\] ?..(i) \[\therefore \]\[\beta =L+5(\beta -\alpha )\] or \[\beta =L+5\beta -5\alpha \] or \[5\alpha -4\beta =L\] When force\[=9\text{ }N,\]final length be\[l\]. \[l=L+\frac{9}{k}=(5\alpha -4\beta )+9(\beta -\alpha )\]           \[=5\alpha -4\beta +9\beta -9\alpha =5\beta -4\alpha \]


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