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When a metallic wire is stretched with a tension $T_1$ its length is $l_1$ and with a tension $T_2$ its length is $l_2$. The original length of the wire is

$\begin {array} {1 1} (1)\;\large\frac{l_1T_2-l_2T_1}{T_2-T_1} \\ (2)\;\large\frac{l_1T_2+l_2T_1}{T_1+T_2} \\ (3)\;\sqrt{l_1l_2} \\(4)\;\large\frac{l_1+l_2}{2} \end {array}$

 

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1 Answer

(1) $ \large\frac{l_1T_2-l_2T_1}{T_2-T_1}$
answered Nov 7, 2013 by pady_1
 

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