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JEE Main & Advanced Sample Paper JEE Main Sample Paper-24

  • question_answer
    A wire fixed at the upper end stretches by length 1 by applying a force F. The work done in stretching is:

    A)  \[\frac{F}{2\,l}\]                                   

    B)  \[Fl\]

    C)  \[2F\,l\]                                    

    D)  \[\frac{F\,l}{2}\]

    Correct Answer: D

    Solution :

    Work done in stretching the wire = potential energy stored \[=\frac{1}{2}\,\times Stress\,\times Strain\times \,Volume\] \[=\frac{1}{2}\,\times \frac{F}{A}\,\times \frac{l}{L}\,\times AL\,=\frac{1}{2}\,Fl\]


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