Q133 / 160 · AP EAMCET 2015 · Physics

The length of a metal wire is $l_1$ when the tension in it is $F_1$ and $l_2$ when the tension is $F_2$. Then, original length of the wire is

  • A
    $\frac{l_1 F_1+l_2 F_2}{F_1+F_2}$
  • B
    $\frac{l_2-l_1}{F_2-F_1}$
  • Correct answer
    C
    $\frac{l_1 F_2-l_2 F_1}{F_2-F_1}$
  • D
    $\frac{l_1 F_1-l_2 F_2}{F_2-F_1}$

Solution

We have, $F_1 \propto$ $\left(l_1-l\right)$, where $\ell=$ original length Similarly, $F_2 \propto \quad\left(l_2-l\right)$, The ratio, $\frac{F_1}{F_2}=\frac{l_1-l}{l_2-l}$ Solving, we get $\Rightarrow \quad l=\frac{F_2 l_1-F_1 l_2}{F_2-F_1}$

Hence, the correct answer is option: 'C'.

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