Two exactly similar wires of steel (YS = 20 × 1011 dyne/cm2) and copper (YC=12×1011dyne/cm2) are stretched by equal forces. If the total elongation is 1cm elongation of copper wire is

YS=20×1011 dyne /cm2YC=12×1011 dyne /cm2eS+eCu=1 cm....(1)e∝1Y⇒eSeCu=YCuYS=12×101120×1011⇒eS=35eCu ...…(2)From (1) and (2): 35eCu+eCu=1 ⇒8eCu=5⇒eCu=58 cm

Two exactly similar wires of steel (YS = 20 × 1011 dyne/cm2) and copper (YC=12×1011dyne/cm2) are stretched by equal forces. If the total elongation is 1cm elongation of copper wire is

YS=20×1011 dyne /cm2YC=12×1011 dyne /cm2eS+eCu=1 cm....(1)e∝1Y⇒eSeCu=YCuYS=12×101120×1011⇒eS=35eCu ...…(2)From (1) and (2): 35eCu+eCu=1 ⇒8eCu=5⇒eCu=58 cm

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Two exactly similar wires of steel (Y_S = 20 × 10¹¹ dyne/cm²) and copper (Y_C=12×10¹¹dyne/cm²) are stretched by equal forces. If the total elongation is 1cm elongation of copper wire is

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3/5 cm

5/3 cm

3/8 cm

5/8 cm

answer is D.

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Detailed Solution

$Y_{S} = 20 \times 10^{11} dyne / {cm}^{2}$
$Y_{C} = 12 \times 10^{11} dyne / {cm}^{2}$

$e_{S} + e_{Cu} = 1 cm . . . . (1)$
$e \propto \frac{1}{Y} \Rightarrow \frac{e_{S}}{e_{C u}} = \frac{Y_{C u}}{Y_{S}} = \frac{12 \times 10^{11}}{20 \times 10^{11}}$
$\Rightarrow e_{S} = \frac{3}{5} e_{C u} . . . \dots (2)$
From (1) and (2):