question_answer 1)

A steel wire of length 4.7 m and cross section \[=43\cdot \text{2 cm}\] stretches by the same amount as a copper wire of length 3.5 m and cross section \[0\cdot 5\times {{10}^{-2}}c{{m}^{2}}\]under a given load. What is the ratio of the Young's modulus of steel to that of copper?

Answer:

Given, For Steel wire, \[g=10m{{s}^{-2}};\]  \[Y=2\times {{10}^{11}}N{{m}^{-2}}.\] \[AC=CB=l=0\cdot 5m;\] \[m=100g=0\cdot 100kg\] For copper wire, \[AD=BD={{\left( {{l}^{2}}+{{x}^{2}} \right)}^{1/2}}\] \[\vartriangle l=\text{AD+DB-AB=2AD-AB}\] \[=2{{\left( {{l}^{2}}+{{x}^{2}} \right)}^{1/2}}-2l=2l{{\left( 1+\frac{{{x}^{2}}}{{{l}^{2}}} \right)}^{1/2}}\] \[-2l=2l\left[ 1+\frac{{{x}^{2}}}{2{{l}^{2}}} \right]-2l=\frac{{{x}^{2}}}{l}\] Let\[\therefore \] be the Young's modulus of steel wire and copper wire respectively.  \[=\frac{\vartriangle l}{2l}=\frac{{{x}^{2}}}{2{{l}^{2}}}\]         \[\text{2 T cos }\!\!\theta\!\!\text{  = mg}\]                ?. (1) And     \[\text{T=}\frac{\text{mg}}{\text{2 cos }\!\!\theta\!\!\text{ }}\]                                 ?.. (2) \[\text{cos }\!\!\theta\!\!\text{ =}\frac{\text{x}}{{{\left( {{\text{l}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}} \right)}^{\text{1/2}}}}\text{=}\frac{\text{x}}{\text{l}{{\left( \text{1+}\frac{{{\text{x}}^{\text{2}}}}{{{\text{l}}^{\text{2}}}} \right)}^{\text{1/2}}}}\text{=}\frac{\text{x}}{\text{l}\left( \text{1+}\frac{\text{1}{{\text{x}}^{\text{2}}}}{\text{2}{{\text{l}}^{\text{2}}}} \right)}\]         \[\text{As,xl,so1}\frac{\text{1}{{\text{x}}^{\text{2}}}}{\text{2}{{\text{l}}^{\text{2}}}}\] Hence    \[1+\frac{1{{x}^{2}}}{2{{l}^{2}}}\approx 1\]