A) \[\frac{R}{4}\]
B) \[\frac{R}{8}\]
C) \[\frac{R}{16}\]
D) \[\frac{R}{32}\]
Correct Answer: C
Solution :
For a wire of length \[l,\]area of cross-section A, and specific resistance\[\rho ,\]the resistance is given by \[R=\frac{\rho l}{A}\] From above equation it is clear that resistance is directly proportional to length. When wire is cut into 4 pieces then resistance of each part is\[R'\,'\propto \frac{1}{4}\] \[\Rightarrow \] \[R'\,'\,=\frac{R}{4}\] Also, equivalent resistance for parallel combination is \[\frac{1}{R'}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}+\frac{1}{{{R}_{3}}}+\frac{1}{{{R}_{4}}}\] \[\therefore \] \[\frac{1}{R'}=\frac{4}{R'\,'}=\frac{4\times 4}{R}\] \[\Rightarrow \] \[R'=\frac{R}{16}\]
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