A wire of length 1m and radius 1mm

Asked by admin @ 31/08/2021 in Physics viewed by 350 People

A wire of length 1 m and radius 1mm is subjected to a load. The extension is x.The wire is melted and then drawn into a wire of square cross - section of side 2 mm Its extension under the same load will be

Answered by admin @ 31/08/2021

Final expansion = $\frac{\pi^{2}}{16} x$

Given:

A wire of length 1 m and radius 1mm is subjected to a load.

The extension is x.

The wire is melted and then drawn into a wire of square cross - section of side 2 mm.

To find:

Its extension under the same load will be.

Formula used:

dL = $\frac{FL}{AE}$

where dL = Change in length

F = Force applied

A = Area of cross section

E = young's modulus

Explanation:

$Volume of wire is $V=L \times \pi r^{2}$$=l \times \pi\left(10^{-3}\right)^{2}$$ = $\pi \times 10^-^6$ m³

$Area of square cross-section$=\left(2 \times 10^{-3}\right)^{2}$ =4 \times 10^{-6} m^{2}$$

$\text { Length of new wire }=\frac{\text { Volume }}{\text { Area }}=\frac{\pi \times 10^{-6}}{4 \times 10^{-6} }$ = $\frac{\pi }{4}$ meter.

Material is same so that young's modulus is same and force applied also same.

dL = $\frac{FL}{AE}$ By putting all the value we can get that.

Final expansion = $\frac{\pi^{2}}{16} x$

To learn more...

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A wire of length 1m and radius 1mm

Final expansion = $\frac{\pi^{2}}{16} x$

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