JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति JEE PYQ - Work Energy Power and Collision

  • question_answer
    When a rubber-band is stretched by a distance x, it exerts a restoring force of magnitude \[F=ax+b{{x}^{2}}\]where a and b are constants. The work done in stretching the un stretched rubber-band by L is:         [JEE MAIN 2014]

    A) \[\frac{a{{L}^{2}}}{2}+\frac{b{{L}^{3}}}{3}\]

    B) \[\frac{1}{2}\left[ \frac{a{{L}^{2}}}{2}+\frac{b{{L}^{3}}}{3} \right]\]

    C) \[a{{L}^{2}}+b{{L}^{3}}\]

    D) \[\frac{1}{2}(a{{L}^{2}}+b{{L}^{3}})\]

    Correct Answer: A

    Solution :

    [a] \[w=\int\limits_{0}^{L}{F.}dx\Rightarrow \omega =\frac{a{{L}^{2}}}{2}+\frac{b{{L}^{3}}}{3}\]


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