A) 12 ohm
B) 3.0 ohm
C) 1.5 ohm
D) None of the above
Correct Answer: C
Solution :
Same mass, same material i.e., volume is the same of Al = constant Also, \[R=\rho \frac{l}{A}\Rightarrow \frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}}\times \frac{{{A}_{2}}}{{{A}_{1}}}={{\left( \frac{{{A}_{2}}}{{{A}_{1}}} \right)}^{2}}{{\left( \frac{{{d}_{2}}}{{{d}_{1}}} \right)}^{4}}\] \[\Rightarrow \] \[\frac{24}{{{R}_{2}}}={{\left( \frac{d}{d/2} \right)}^{4}}=16\Rightarrow {{R}_{1}}=1.5\,\Omega \]
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