A) \[\frac{1}{4}\]
B) \[\frac{1}{2}\]
C) \[1\]
D) \[2\]
Correct Answer: B
Solution :
In series connection the current in two wires will be same. \[{{i}_{1}}={{i}_{2}}\] or \[\frac{{{V}_{1}}}{{{R}_{1}}}=\frac{{{V}_{2}}}{{{R}_{2}}}\] \[\frac{{{R}_{2}}}{{{R}_{1}}}=\frac{{{V}_{2}}}{{{V}_{1}}}\Rightarrow \frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{V}_{1}}}{{{V}_{2}}}\] ?(i) Where \[{{V}_{1}}\]and \[{{V}_{2}}\]are the potential difference across the wires respectively \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{3V}{2V}=\frac{3}{2}\] given Since, both wires are of same material so their resistivity will be same and \[{{r}_{A}}\]and \[{{r}_{B}}\] are their radii, then \[{{R}_{1}}=\frac{\rho {{l}_{1}}}{{{A}_{12}}}=\frac{\rho {{l}_{1}}}{\pi r_{A}^{2}}\] \[{{R}_{2}}=\frac{\rho {{l}_{2}}}{{{A}_{2}}}=\frac{\rho {{l}_{2}}}{\pi r_{B}^{2}}\] \[\therefore \] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}}{{\left( \frac{{{r}_{B}}}{{{r}_{A}}} \right)}^{2}}\] but \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{6}{1}\] \[\therefore \] \[\frac{3}{2}=\left( \frac{6}{1} \right){{\left( \frac{{{r}_{B}}}{{{r}_{A}}} \right)}^{2}}\] ?(i) Hence, we have \[\frac{3}{2}=\left( \frac{6}{1} \right){{\left( \frac{{{r}_{B}}}{{{r}_{A}}} \right)}^{2}}\] \[\Rightarrow \] \[{{\left( \frac{{{r}_{B}}}{{{r}_{A}}} \right)}^{2}}=\frac{3}{2}\times \frac{1}{6}\] \[\Rightarrow \] \[{{\left( \frac{{{r}_{B}}}{{{r}_{A}}} \right)}^{2}}=\frac{1}{4}\] \[\frac{{{r}_{B}}}{{{r}_{A}}}=\frac{1}{4}\] \[{{r}_{B}}:{{r}_{A}}=1:2\]
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