A) 2.0
B) 2.3
C) 1.75
D) 5.0
Correct Answer: B
Solution :
\[Y=\frac{Fl}{A\Delta l}=\frac{4F}{\pi {{D}^{2}}}\times \frac{l}{\Delta l}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,D=\frac{4Fl}{\pi Y\Delta l}\,\,\,\,\Rightarrow \,\,\,\,D\propto \frac{1}{\sqrt{Y}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\frac{{{D}_{Cu}}}{{{D}_{Al}}}=\sqrt{\frac{{{Y}_{Al}}}{{{Y}_{Cu}}}}\] \[{{D}_{Cu}}\,\,=\,\,3\sqrt{\frac{7\times {{10}^{10}}}{12\times {{10}^{10}}}}\,\,=\,\,2.3\,\,mm\]You need to login to perform this action.
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