A) \[2:1\]
B) \[1:4\]
C) \[1:8\]
D) \[8:1\]
Correct Answer: C
Solution :
We know that Young?s modulus \[Y=\frac{F}{\pi {{r}^{2}}}\times \frac{L}{{{\ell }^{2}}}\] Since Y, F are same for both the wires, we have, \[\frac{1}{r_{1}^{2}}\frac{{{L}_{1}}}{{{\ell }_{1}}}=\frac{1}{r_{2}^{2}}\frac{{{L}_{2}}}{{{\ell }_{2}}}\] \[\frac{{{\ell }_{1}}}{{{\ell }_{2}}}=\frac{r_{2}^{2}\times {{L}_{1}}}{r_{1}^{2}\times {{L}_{2}}}=\frac{{{({{D}_{2}}/2)}^{2}}\times {{L}_{1}}}{({{D}_{1}}/{{2}^{2}})\times {{L}_{2}}}\] \[\frac{{{\ell }_{1}}}{{{\ell }_{2}}}=\frac{D_{2}^{2}\times {{L}_{1}}}{D_{1}^{2}\times {{L}_{2}}}=\frac{D_{2}^{2}}{{{\left( 2{{D}_{2}} \right)}^{2}}}\times \frac{{{L}_{2}}}{2{{L}_{2}}}=\frac{1}{8}\] \[{{\ell }_{1}}:{{\ell }_{2}}=1:8\]
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