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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2011

  • question_answer
    Two wires made of same material having lengths \[L\]and \[2L\] and radii \[r\] and \[2r\] respectively are clamped rigidly at their one end and are stretched by applying forces equal to F and 2F respectively. If extension in length of first wire is I, the extension in second wire is

    A)  \[l\]                                      

    B)  \[2l\]    

    C)  \[\frac{1}{2}l\]                                 

    D)  \[4l\]

    Correct Answer: A

    Solution :

    Young's modulus of elasticity of material of wire \[Y=\frac{FL}{\pi {{r}^{2}}\Delta l}\] \[\therefore \] Increase in length \[\Delta l=\frac{FL}{\pi {{r}^{2}}Y}\]                 \[\frac{\Delta {{l}_{1}}}{\Delta {{l}_{2}}}=\frac{{{F}_{1}}}{{{F}_{2}}}\times \frac{{{L}_{1}}}{{{L}_{2}}}\times {{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{2}}\]                 \[\frac{l}{\Delta {{l}_{2}}}=\frac{F}{2F}\times \frac{L}{2L}\times {{\left( \frac{2r}{r} \right)}^{2}}\] So,          \[\Delta {{l}_{2}}=l\]


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