VMMC VMMC 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)                                         2 F                        

    C)        \[\frac{F}{4}\]   

    D)         4F

    Correct Answer: C

    Solution :

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


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