Two wires, one made of copper and other of steel are joined end to end (as shown in figure). The area of cross-section of copper wire is twice that of steel wire. [image:9_image6.jpeg] They are placed under compressive force of magnitudes F. The ratio for their lengths such that change in lengths of both wires are same is ($Y_{s} = 2 imes 10^{11}Nm^{- 2}\ and\ Y_{c} = 1.1 imes 10^{11}\ Nm^{- 2})$

Question

Two wires, one made of copper and other of steel are joined end to end (as shown in figure). The area of cross-section of copper wire is twice that of steel wire. [image:9_image6.jpeg] They are placed under compressive force of magnitudes F. The ratio for their lengths such that change in lengths of both wires are same is ($Y_{s} = 2 	imes 10^{11}Nm^{- 2}\ and\ Y_{c} = 1.1 	imes 10^{11}\ Nm^{- 2})$

Dr. Abubacker · Accepted Answer

(b) $Y_{s} = \frac{Fl_{s}}{A_{s}\mathrm{\Delta}L_{s}}$
And $Y_{C} = \frac{FL_{C}}{A_{C}\mathrm{\Delta}L_{C}}\ $
$	herefore\ \ \frac{L_{C}}{L_{S}}\  = \ \frac{\frac{Y_{C}A_{C}\mathrm{\Delta}L_{C}}{F}}{\frac{Y_{S}A_{S}\mathrm{\Delta}L_{S}}{F}} = \left( \frac{Y_{C}}{Y_{S}} ight)\left( \frac{A_{C}}{A_{S}} ight)\left( \frac{\mathrm{\Delta}L_{C}}{\mathrm{\Delta}L_{S}} ight)$
Here, $\frac{A_{C}}{A_{S}} = 2,\frac{\mathrm{\Delta}L_{C}}{\mathrm{\Delta}L_{S}} = 1,\frac{Y_{C}}{Y_{S}} = \frac{1.1}{2}\ $
$	herefore\ \ \frac{L_{C}}{L_{S}} = \frac{1.1}{2}\  	imes 2\  	imes 1 = 1.1$

Explanation

Reviewed by Dr. Abubacker

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