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JEE Main & Advanced Physics Elasticity Question Bank Youngs Modulus and Breaking Stress

  • question_answer
    The length of an elastic string is a metre when the longitudinal tension is 4 N and b metre when the longitudinal tension is 5 N. The length of the string in metre when the longitudinal tension is 9 N is           [EAMCET 2001]

    A)             \[a-b\]

    B)             \[5b-4a\]

    C)             \[2b-\frac{1}{4}a\]

    D)                      \[4a-3b\]

    Correct Answer: B

    Solution :

                    Let L is the original length of the wire and K is force constant of wire. Final length = initial length + elongation \[L'=L+\frac{F}{K}\] For first condition \[a=L+\frac{4}{K}\]                     ?(i) For second condition \[b=L+\frac{5}{K}\]   ?(ii) By solving (i) and (ii) equation we get \[L=5a-4b\] and \[K=\frac{1}{b-a}\] Now when the longitudinal tension is 9N, length of the string = \[L+\frac{9}{K}\]= \[5a-4b+9(b-a)\]\[x=5b-4a\].


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