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Manipal Engineering Manipal Engineering Solved Paper-2012

  • 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                 \[\frac{F}{\left( \frac{\pi {{D}^{2}}}{4} \right)}=\frac{F}{\pi {{D}^{2}}}\]                          \[F=4F\]


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