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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2002

  • question_answer
    The resultant of two forces, one double the   other   in   magnitude,   is perpendicular to the smaller of the two forces. The angle between the two forces is:

    A) \[{{120}^{o}}\]                                 

    B) \[~{{60}^{o}}\]

    C) \[{{90}^{o}}\]                                    

    D) \[{{150}^{o}}\]

    Correct Answer: A

    Solution :

    Stress \[\propto \]strain \[\frac{F}{A}=\Upsilon \frac{\Delta l}{l}\] \[\Delta l=\]elongation \[F=\]load \[\Upsilon =\] Young's modulus \[A=\]area \[l=\]length of wire \[F\propto {{r}^{2}}\] (radius of wire). \[\Delta l\propto \frac{1}{{{r}^{2}}}\] \[\therefore \]Thin wire max. elongation with min. load. Graph is OA.


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