A) \[25.2\text{ }\Omega \]
B) \[0.6\text{ }\Omega \]
C) \[26.7\text{ }\Omega \]
D) \[\text{0}\text{.8 }\Omega \]
Correct Answer: A
Solution :
\[R=\rho \frac{l}{A}\] Given, \[{{R}_{1}}=4.2\Omega ,\,\,{{l}_{1}}=1m,\,{{A}_{1}}=\pi r_{1}^{2}=\pi {{\left( \frac{0.31}{2} \right)}^{2}}\] \[{{R}_{2}}=?,\,\,{{I}_{2}}=1.5{{A}_{2}}=\pi r_{2}^{2}=\pi {{\left( \frac{0.155}{2} \right)}^{2}}\] \[\therefore \] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{l}_{1}}}{{{A}_{1}}}\times \frac{A{{ & }_{2}}}{{{l}_{2}}}\] \[\frac{4.2}{R}=\frac{1}{\pi {{(0.31)}^{2}}}\times \frac{\pi {{(0.155)}^{2}}}{1.5}\] \[\Rightarrow \] \[R=\frac{4.2\times 0.31\times 0.31\times 1.5}{0.155\times 0.155}=25.2\Omega \]
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