The following four wires are made of the same | vedclass

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Question

Young's modulus,

$Y = \frac{{FL}}{{A\Delta L}} = \frac{{4FL}}{{\pi {D^2}\Delta L}}\,\,or\,\,\Delta L = \frac{{4FL}}{{\pi {D^2}Y}}$

Where $F$

Vedclass · Accepted Answer

length $ =50 \;cm$,diameter $=0.5\; mm$

Vedclass · Answer

Young's modulus,

$Y = \frac{{FL}}{{A\Delta L}} = \frac{{4FL}}{{\pi {D^2}\Delta L}}\,\,or\,\,\Delta L = \frac{{4FL}}{{\pi {D^2}Y}}$

Where $F$ is the force applied, $L$ is the length, $D$ is the diameter and $\Delta L$ is the extension of the wire respectively. 

As each wire is made up of same material therefore their $Young's$ modulus is same for each wire.

Foe all the four wires, $Y,F\,(=tension)$ are the same.

$	herefore \Delta L \propto \frac{L}{{{D^2}}}$

$In\,\,\left( a ight)\,\,\frac{L}{{{D^2}}} = \frac{{200\,cm}}{{{{\left( {0.2\,cm} ight)}^2}}} = 5 	imes {10^3}\,c{m^{ - 1}}$

$In\,\,\left( b ight)\,\,\,\frac{L}{{{D^2}}} = \frac{{300\,cm}}{{{{\left( {0.3\,cm} ight)}^2}}} = 3.3 	imes {10^3}\,c{m^{ - 1}}$

$In\,\,\left( c ight)\,\,\frac{L}{{{D^2}}} = \frac{{50\,cm}}{{{{\left( {0.05\,cm} ight)}^2}}} = 20 	imes {10^3}\,c{m^{ - 1}}$

$In\,\,\left( d ight)\,\,\frac{L}{{{D^2}}} = \frac{{100\,cm}}{{{{\left( {0.1\,cm} ight)}^2}}} = 10 	imes {10^3}\,\,c{m^{ - 1}}$

The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?

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