A) 1 : 1
B) 1 : 2
C) 4:1
D) 1 : 4
Correct Answer: C
Solution :
As we know that, \[{{R}_{1}}=\rho \frac{l}{a}=\rho \frac{{{l}^{2}}}{V}\] where,\[l=\] length of wire a = area of cross-section of the wire and V = volume of the wire \[{{R}_{1}}\propto {{l}^{2}}\] \[\Rightarrow \] \[\frac{{{R}_{1}}}{{{R}_{2}}}={{\left( \frac{{{l}_{1}}}{{{l}_{2}}} \right)}^{2}}.{{\left( \frac{1}{2} \right)}^{2}}\] \[\Rightarrow \] \[{{R}_{2}}:{{R}_{1}}=4:1\]You need to login to perform this action.
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