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J & K CET Medical J & K - CET Medical Solved Paper-2007

  • question_answer
    A particular force (F) applied on a wire increases its length by \[2\times {{10}^{-3}}m.\]To increase the wires length by \[4\times {{10}^{-3}}\] m the applied force will be

    A)  4F                                         

    B)  3F         

    C)  2F                                         

    D)  F

    Correct Answer: C

    Solution :

                     \[Y=\frac{F/A}{\Delta l/l}=\frac{F\times l}{A\times \Delta l}\](where V is Youngs modulus of elasticity) Since, V,\[l\]and A remain same.                 \[F\propto \Delta l\]                 \[\frac{{{F}_{1}}}{{{F}_{2}}}=\frac{\Delta {{l}_{1}}}{\Delta {{l}_{2}}}\] \[\Rightarrow \]               \[\frac{F}{{{F}_{2}}}=\frac{2\times {{10}^{-3}}}{4\times {{10}^{-3}}}\]                 \[{{F}_{2}}=2F\]


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