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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2005

  • question_answer
    Two wires of the same material have lengths in the ratio 1 : 2 and their radii are in the ratio 1: \[\sqrt{2}\]. If they are streched by applying equal forces, the increase in their lengths will be in the ratio

    A)  2:\[\sqrt{2}\]                                   

    B)  \[\sqrt{2}\]:2

    C)  1:1                                        

    D)  1 : 2

    Correct Answer: C

    Solution :

    Young's modulus of elasticity is given by \[Y=\frac{mgL}{\pi {{r}^{2}}l}\] For two wires of same material and stretched by same force                 \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{{{L}_{1}}}{{{L}_{2}}}\frac{r_{2}^{2}}{r_{1}^{2}}\] Given    \[\frac{{{L}_{1}}}{{{L}_{2}}}=\frac{1}{2},\]   \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{1}{\sqrt{2}}\] \[\therefore \]  \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{1}{2}\times \frac{2}{1}=1:1\]


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