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Haryana PMT Haryana PMT Solved Paper-2011

  • question_answer
    A steel wire is 1 m long and has \[1\text{ }m{{m}^{2}}\]area of   cross-section. If it takes 200 N to stretch thin      wire by 1 mm, how much force will be required to stretch a wire of same material and diameter         from its normal length of 10 m to a length of 10.02 m?

    A)                  1000 N                                

    B)  200 N

    C)                  2000 N                                

    D)  400 N

    Correct Answer: D

    Solution :

                    \[Y\,=\frac{F}{A}.\frac{L}{l}\]                 \[\therefore \]\[F\propto \frac{l}{K}\] ( Y and A are same for both wires)                 \[\therefore \]\[\frac{{{F}_{2}}}{{{F}_{1}}}=\frac{{{l}_{2}}}{{{l}_{1}}}\times \frac{{{L}_{1}}}{{{L}_{2}}}=\frac{0.02}{1\times {{10}^{-3}}}\times \frac{1}{10}=2\]                 \[{{F}_{2}}=2{{F}_{1}}=2\times 200=400N\]


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