A) \[544\,\Omega \]
B) \[272\,\Omega \]
C) \[68\,\Omega \]
D) \[17\,\Omega \]
Correct Answer: A
Solution :
\[R=\rho \frac{l}{A}\] and mass m = volume (V) ´ density = (A l) d Since wires have same material so r and d is same for both. Also they have same mass Þ Al = constant Þ \[l\propto \frac{1}{A}\] Þ \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}}\times \frac{{{A}_{2}}}{{{A}_{1}}}={{\left( \frac{{{A}_{2}}}{{{A}_{1}}} \right)}^{2}}={{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{4}}\] Þ \[\frac{34}{{{R}_{2}}}={{\left( \frac{r}{2r} \right)}^{4}}\] Þ \[{{R}_{2}}=544\,\Omega \]
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