JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Question Bank Mock Test - Work Power and Energy

  • 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 unstretched rubber band by \[L\] is,

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

    Correct Answer: A

    Solution :

    [a] Restoring force on rubber-band, \[F=ax+b{{x}^{2}}\]Work done in stretching the rubber-band by a small amount \[dx,dW=Fdx\] Net work done in stretching the rubber-band by L is \[W=\int{dW=\int\limits_{0}^{L}{Fdx=\int\limits_{0}^{L}{(ax+b{{x}^{2}})dx}}}\] \[\Rightarrow W={{\left[ a\frac{{{x}^{2}}}{2}+b\frac{{{x}^{3}}}{3} \right]}_{0}}^{L}=\frac{a{{L}^{2}}}{2}+\frac{b{{L}^{3}}}{3}\]


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