A) 12.6 : 2
B) 1.31 : 1
C) 4.65 : 3
D) 2.69 : 4
Correct Answer: B
Solution :
Let \[{{Y}_{1}}\] and \[{{Y}_{2}}\] be the Youngs modulus of copper and iron wires respectively. \[\therefore \] \[{{Y}_{1}}=110\times {{10}^{9}}N{{m}^{-2}}\] \[{{Y}_{2}}=190\times {{10}^{9}}N{{m}^{-2}}\] Also, let \[{{A}_{1}}\] and \[{{A}_{2}}\] be the areas of cross-section of copper and iron wires respectively. If \[{{d}_{1}}\] and \[{{d}_{2}}\] be their respective diameters, then \[{{A}_{1}}=\frac{\pi d_{1}^{2}}{4}\] and \[{{A}_{2}}=\frac{\pi d_{2}^{2}}{4}\] \[\therefore \] \[\frac{{{A}_{1}}}{{{A}_{2}}}=\frac{d_{1}^{2}}{d_{2}^{2}}={{\left( \frac{{{d}_{1}}}{{{d}_{2}}} \right)}^{2}}\] Let \[\Delta l\] be the extension produced in each wire. Let F = tension produced in each wire \[\therefore \] From the relation, \[Y=\frac{stress}{strain}\], we get Strain for copper wire \[=\frac{F/{{A}_{1}}}{{{Y}_{2}}}\] and Strain for iron wire \[=\frac{F/{{A}_{2}}}{{{Y}_{2}}}\] As the bar is supported symmetrically, the two strains are equal \[\frac{F}{{{A}_{1}}{{Y}_{1}}}=\frac{F}{{{A}_{2}}{{Y}_{2}}}\] \[\Rightarrow \] \[{{A}_{1}}{{Y}_{1}}={{A}_{2}}{{Y}_{2}}\Rightarrow \frac{{{A}_{1}}}{{{A}_{2}}}=\frac{{{Y}_{2}}}{{{Y}_{1}}}\] \[\Rightarrow \] \[\frac{\pi d_{1}^{2}/4}{\pi d_{2}^{2}/4}=\frac{{{Y}_{2}}}{{{Y}_{1}}}\] \[\Rightarrow \] \[\frac{{{d}_{1}}}{{{d}_{2}}}=\sqrt{\frac{{{Y}_{2}}}{{{Y}_{1}}}}=\sqrt{\frac{190\times {{10}^{9}}}{110\times {{10}^{9}}}}\] \[\frac{{{d}_{1}}}{{{d}_{2}}}=1.31\] \[\Rightarrow \] \[{{d}_{1}}:{{d}_{2}}=1.31:1\]
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