A) 1
B) \[\frac{1}{2}\]
C) \[\frac{1}{4}\]
D) 2
Correct Answer: D
Solution :
| [d] Let \[({{\rho }_{A}},{{l}_{A}},{{r}_{A}},{{A}_{A}})\] and \[({{\rho }_{B}},{{l}_{B}},{{r}_{B}},{{A}_{B}})\] be specific resistances, lengths, radii and areas of wires A and B, respectively. |
| Resistance of\[A={{R}_{A}}=\frac{{{\rho }_{A}}{{l}_{A}}}{{{A}_{A}}}=\frac{{{\rho }_{A}}{{l}_{A}}}{\pi r_{A}^{2}}\] |
| Resistance of \[B={{R}_{B}}=\frac{{{\rho }_{B}}{{l}_{B}}}{{{A}_{B}}}=\frac{{{\rho }_{B}}{{l}_{B}}}{\pi r_{B}^{2}}\] |
| From given information, |
| \[{{\rho }_{B}}=2{{\rho }_{A}}\] |
| \[{{r}_{B}}=2{{r}_{A}}\] |
| and \[{{R}_{A}}={{R}_{B}}\] |
| \[\therefore \] \[\frac{{{\rho }_{A}}{{l}_{B}}}{\pi r_{A}^{2}}\,=\frac{{{\rho }_{B}}{{l}_{B}}}{\pi r_{B}^{2}}\] |
| \[\Rightarrow \] \[\frac{{{\rho }_{A}}{{l}_{A}}}{\pi r_{A}^{2}}\,=\frac{2{{\rho }_{A}}\times {{l}_{B}}}{\pi {{(2{{r}_{A}})}^{2}}}\] |
| \[\Rightarrow \] \[\frac{{{l}_{B}}}{{{l}_{A}}}=\frac{2}{1}=2:1\] |
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