Two exactly similar wires of steel and copper are stretched by equal forces. If the difference in their elongations is $0.5cm$, the elongation $(l)$ of each wire is $Y_{s}(steel) = 2.0 imes 10^{11}N/m^{2}$ $Y_{c}(copper) = 1.2 imes 10^{11}N/m^{2}$

Question

Two exactly similar wires of steel and copper are stretched by equal forces. If the difference in their elongations is $0.5cm$, the elongation $(l)$ of each wire is $Y_{s}(steel) = 2.0 	imes 10^{11}N/m^{2}$ $Y_{c}(copper) = 1.2 	imes 10^{11}N/m^{2}$

Dr. Abubacker · Accepted Answer

(a) $l \propto \frac{1}{Y} \Rightarrow \frac{Y_{s}}{Y_{c}} = \frac{l_{c}}{l_{s}} \Rightarrow \frac{l_{c}}{l_{s}} = \frac{2 	imes 10^{11}}{1.2 	imes 10^{11}} = \frac{5}{3}$ …(i)
Also $l_{c} - l_{s} = 0.5$ …(ii)
On solving (i) and (ii) $l_{c} = 1.25cm$ and $l_{s} = 0.75\ cm$

Q. Two exactly similar wires of steel and copper are stretched by equal forces. If the difference in their elongations is $0.5cm$, the elongation $(l)$ of each wire is $Y_{s}(steel) = 2.0 \times 10^{11}N/m^{2}$ $Y_{c}(copper) = 1.2 \times 10^{11}N/m^{2}$

Explanation

Reviewed by Dr. Abubacker

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