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MGIMS WARDHA MGIMS WARDHA Solved Paper-2015

  • question_answer
    Two wires of same material and length are stretched by the same force. Their masses are in the ratio\[4:3\]. The ratio of their elongation will be

    A)  \[16:9\]                              

    B)  \[9:16\]  

    C)  \[4:3\]                                 

    D)  \[3:4\]

    Correct Answer: D

    Solution :

                    \[Y=\frac{Fl}{A\Delta l}\] Here\[y,l\]and F are constant So,            \[\Delta l\propto \frac{1}{A}\] Also            \[m=(V)\rho =(Al)\rho \] i.e.                          \[m\propto A\] \[\therefore \]                  \[\Delta l\propto \frac{1}{m}\] \[\Rightarrow \]               \[\frac{\Delta {{l}_{1}}}{\Delta {{l}_{2}}}=\frac{{{m}_{2}}}{{{m}_{1}}}=\frac{3}{4}\]                \[(\because {{m}_{1}}:{{m}_{2}}=4:3)\]


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