A) \[40\Omega \]
B) \[80\Omega \]
C) \[60\Omega \]
D) \[180\Omega \]
Correct Answer: D
Solution :
Key Idea: When length is increased volume remains same and specific resistance constant. The resistance of a wire of length \[l\] is given by \[R=\rho \frac{l}{A}\] ... (i) When wire is stretched, volume remains constant, hence, \[V=Al=\]constant ? (ii) Multiply and divide Eq. (i) by\[l\], we get \[R=\frac{\rho {{l}^{2}}}{Al}\] \[\therefore \] \[\frac{{{R}_{2}}}{{{R}_{1}}}={{\left( \frac{{{l}_{2}}}{{{l}_{1}}} \right)}^{2}}\] Given, \[{{l}_{2}}=3{{l}_{1}}\] \[\frac{{{R}_{2}}}{20}={{\left( \frac{3{{l}_{1}}}{{{l}_{1}}} \right)}^{2}}=9\] \[\Rightarrow \] \[{{R}_{2}}=20\times 9=180\,\,\Omega \]
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