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Volume Strain

A slightly co...

Question

A slightly conical wire of length L and end radii $r_{1}$ and $r_{2}$ is stretched by two forces F and F applied parallel to the length in opposite directions and normal to the end faces. If Y denotes Young's modulus, then the extension produced is:

\frac{F L}{π r_{1} r_{2} Y}

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\frac{F L Y}{π r_{1} r_{2}}

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\frac{F L}{π r_{1} Y}

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\frac{F L}{π r_{1}^{2} Y}

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Solution

The correct option is C $\frac{F L}{π r_{1} r_{2} Y}$
If $R$ be the rate of increase of radius per length, $R = \frac{r_{2} - r_{1}}{L}$
Consider a element of thickness $d x$ at a distance $x$ from narrow end of the wire.
The radius of the element is $y = r_{1} + R x$ .
The extension in the element is $d l = \frac{F d x}{(π y^{2}) Y}$
thus, $l = \frac{F}{π Y} \int_{0}^{L} \frac{d x}{r_{1} + R x}$
let $r_{1} + R x = p, R d x = d p$
or $l = \frac{F}{π Y} \int_{0}^{r_{1} + R L} \frac{d p}{R p^{2}} = \frac{F}{π Y R} [\frac{1}{r_{1}} - \frac{1}{r_{1} + R L}] = \frac{F}{π Y R} [\frac{r_{1} + R L - r_{1}}{r_{1} (r_{1} + R L)}]$
or $l = \frac{F L}{π Y (r_{2} - r_{1})} . \frac{r_{2} - r_{1}}{r_{1} (r_{1} + r_{2} - r_{1})} = \frac{F L}{π r_{1} r_{2} Y}$

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A slightly conical wire of length L and end radii r1 and r2 is stretched by two forces F and F applied parallel to the length in opposite directions and normal to the end faces. If Y denotes Young's modulus, then the extension produced is:

A slightly conical wire of length L and end radii $r_{1}$ and $r_{2}$ is stretched by two forces F and F applied parallel to the length in opposite directions and normal to the end faces. If Y denotes Young's modulus, then the extension produced is: