A) 20\[\Omega \]
B) 80\[\Omega \]
C) 160\[\Omega \]
D) 320\[\Omega \]
Correct Answer: C
Solution :
[c] Volume of the wire remains constant. \[Al={{A}_{1}}{{l}_{1}}\] We have,\[{{l}_{1}}\]=\[2l\] and \[{{A}_{1}}=\frac{A}{2}\] Using \[{{R}_{1}}=\rho \frac{{{l}_{1}}}{{{A}_{1}}}\] \[=\rho \frac{2l}{{}^{A}/{}_{2}}\] \[=4\rho \frac{1}{A}\] \[=4\times R\] \[\therefore {{R}_{1}}=4\times 40\Omega \] \[=160\,\Omega \]
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