A) \[45\,\Omega \]
B) \[15\,\Omega \]
C) \[5/3\,\,\Omega \]
D) \[5\,\,\Omega \]
Correct Answer: A
Solution :
Since, volume is constant, i.e., \[\pi _{1}^{2}{{l}_{1}}=\pi r_{2}^{2}{{l}_{2}}\] \[\Rightarrow \] \[{{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}=\left( \frac{{{l}_{2}}}{{{l}_{1}}} \right)=\left( \frac{3l}{l} \right)=3\] Now, \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{\rho \frac{{{l}_{1}}}{{{A}_{1}}}}{\rho \frac{{{l}_{2}}}{{{A}_{2}}}}\] \[=\frac{{{l}_{1}}/r_{1}^{2}}{{{l}_{2}}/r_{2}^{2}}=\frac{{{l}_{1}}}{{{l}_{2}}}\times \frac{r_{2}^{2}}{r_{1}^{2}}\] \[\Rightarrow \] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{1}{3}\times \frac{1}{3}\] \[\Rightarrow \] \[{{R}_{2}}=9{{R}_{1}}=9\times 5=45\,\Omega \]
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