A) 2:1
B) 1:2
C) 16:1
D) 4:1
Correct Answer: C
Solution :
Mass of 1st wire \[=(\pi r_{1}^{2}{{l}_{1}})\sigma \] Mass of Und wire \[=(\pi r_{2}^{2}{{l}_{2}})\sigma \] \[\therefore \] \[(\pi r_{1}^{2}{{l}_{1}})\sigma =(\pi r_{2}^{2}{{l}_{2}})\sigma \] \[\Rightarrow \] \[\frac{{{l}_{1}}}{{{l}_{2}}}={{\left(\frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{2}}\] \[\therefore \] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{\rho \frac{{{l}_{1}}}{{{A}_{1}}}}{\rho \frac{{{l}_{2}}}{{{A}_{2}}}}\] \[=\frac{{{l}_{1}}}{{{l}_{2}}}\times \frac{{{A}_{2}}}{{{A}_{1}}}=\frac{{{l}_{1}}}{{{l}_{2}}}\times {{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{2}}\] \[={{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{2}}={{\left( \frac{1}{1/2} \right)}^{4}}\] \[=16:1\]
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