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AMU Medical AMU 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)  2 F

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

    D)  4 F

    Correct Answer: C

    Solution :

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


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