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BVP Medical BVP Medical Solved Paper-2008

  • question_answer
    The breaking force for a wire of diameter D of a material is F. The breaking force for a wire of the same material of radius D is

    A)  F                                            

    B) 2F

    C) \[\frac{F}{4}\]                                   

    D) 4F

    Correct Answer: C

    Solution :

                    Breaking stress \[=\frac{breaking\,force}{area}=\]constant \[\frac{F}{\left( \frac{\pi {{D}^{2}}}{4} \right)}=\frac{F}{\pi {{D}^{2}}}\] \[\frac{F}{\pi {{D}^{2}}}=4\frac{F}{\pi {{D}^{2}}}\] \[F=4F\] \[F=\frac{F}{4}\]              


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