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WB JEE Medical WB JEE Medical Solved Paper-2006

  • question_answer
    A wire can be broken by applying a load of 200 N. The force required to break another wire of the same length and same material, but double in diameter, is :

    A)  200 N                                   

    B)  400 N

    C)  600 N                                   

    D)  800 N

    Correct Answer: D

    Solution :

                     \[Y=\frac{FL}{Al}\] or            \[F=\frac{YAl}{L}\] or            \[F\propto A\,or\,F\propto {{r}^{2}}\,or\,F\propto {{d}^{2}}\] \[\therefore \]  \[\frac{{{F}_{1}}}{{{F}_{2}}}=\frac{d_{1}^{2}}{d_{2}^{2}}\] Given, di \[{{d}_{1}}=d,\,{{d}_{2}}=2d,\,{{F}_{1}}=200\,N\] \[\therefore \]  \[\frac{200}{{{F}_{2}}}=\frac{{{(d)}^{2}}}{{{(2d)}^{2}}}=\frac{1}{4}\] or            \[{{F}_{2}}=4\times 200=800\,N\]


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