A slightly conical wire of length ‘L’ and radii $r_{1}$ and $r_{2}$ is stretched by two forces of magnitude F applied parallel to length in opposite directions and normal to end faces. If ‘Y’ denotes the young’s modulus then the elongation in the wire is $\frac{KFL}{π (r_{1} r_{2}) Y}$ , the value of ‘K’ is__.

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answer is 1.

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Detailed Solution

$\tan θ = \frac{d r}{d y} = \frac{r_{1} - r_{2}}{L}$

$\begin{array}{l} d y = \frac{L d r}{r_{1} - r_{2}} \\ d e = \frac{F d Y}{π r^{2} Y} = \frac{F}{π r^{2} Y} (\frac{L d r}{r_{1} - r_{2}}) = \frac{F L}{π Y (r_{1} - r_{2})} \frac{d r}{r^{2}} \end{array}$

$\int d e = \frac{F L}{π Y (r_{1} - r_{2})} \int_{r_{1}}^{r_{2}} \frac{d r}{r^{2}}$

$\begin{array}{l} \int d e = \frac{F L}{π Y (r_{1} - r_{2})} \int_{r_{1}}^{r_{2}} \frac{d r}{r^{2}} \\ = \frac{F L}{π Y (r_{1} - r_{2})} {[- \frac{1}{r}]}_{r_{1}}^{r_{2}} \\ = \frac{F L}{π Y (r_{1} - r_{2})} [- \frac{1}{r_{1}} + \frac{1}{r_{2}}] \\ e = \frac{F L}{π Y (r_{1} - r_{2})} [\frac{r_{1} - r_{2}}{r_{1} r_{2}}] \\ e = \frac{F L}{π Y r_{1} r_{2}} = \frac{K F L}{π Y r_{1} r_{2}} \\ K = 1 \end{array}$