A) 1
B) \[\frac{1}{2}\]
C) \[\frac{1}{4}\]
D) 2
Correct Answer: D
Solution :
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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