Two exactly similar wires of steel and copper are stretched by equal forces. If difference in their

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8.Mechanical Properties of Solids

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

${l_s} = 0.75\,cm,\;{l_c} = 1.25\,cm$

${l_s} = 1.25\,cm,\;{l_c} = 0.75\,cm$

${l_s} = 0.25\,cm,\;{l_c} = 0.75\,cm$

${l_s} = 0.75\,cm,\;{l_c} = 0.25\,cm$

Solution

(a) $l \propto \frac{1}{Y} \Rightarrow \frac{{{Y_s}}}{{{Y_c}}} = \frac{{{l_c}}}{{{l_s}}} \Rightarrow \frac{{{l_c}}}{{{l_s}}} = \frac{{2 \times {{10}^{11}}}}{{1.2 \times {{10}^{11}}}} = \frac{5}{3}$…(i)

Also ${l_c} – {l_s} = 0.5$…(ii)

On solving (i) and (ii) ${l_c} = 1.25\;cm$ and ${l_s} = 0.75\;cm$.

Standard 11

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

Solution

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