A copper rod with length $1.4$ m and area of cross-section of $2 c m^{2}$ is fastened to a steel rod with length L and cross-sectional area $1 c m^{2}$ . The compound rod is subjected to equal and opposite pulls of magnitude $6.00 \times 10^{4}$ N at its ends.
Find the length L of the steel rod if the elongation of the two rods are equal.

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Solution

$Δ l_{C U} = Δ l_{s t e e l}$
$\frac{F l_{C}}{A_{C} Y_{C}} = \frac{F l_{S}}{A_{S} Y_{S}}$
$\Rightarrow l_{S} = (\frac{A_{S}}{A_{C}}) (\frac{Y_{S}}{Y_{C}}) l_{C}$
or, $L = (\frac{1.0}{2.0}) (\frac{2 \times 10^{11}}{1.1 \times 10^{11}}) 1.40 = 1.27$ m.

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ΔlCU=ΔlsteelFlCACYC=FlSASYS⇒lS=(ASAC)(YSYC)lCor, L=(1.02.0)(2×10111.1×1011)1.40=1.27m.

$Δ l_{C U} = Δ l_{s t e e l}$
$\frac{F l_{C}}{A_{C} Y_{C}} = \frac{F l_{S}}{A_{S} Y_{S}}$
$\Rightarrow l_{S} = (\frac{A_{S}}{A_{C}}) (\frac{Y_{S}}{Y_{C}}) l_{C}$
or, $L = (\frac{1.0}{2.0}) (\frac{2 \times 10^{11}}{1.1 \times 10^{11}}) 1.40 = 1.27$ m.