A copper wire of resistance R is cut into ten | vedclass

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(D) The resistance of a wire is directly proportional to its length,$R \propto l$.Since the wire is cut into ten equal parts,the resistance of each pa

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$\frac{R}{25}$

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(D) The resistance of a wire is directly proportional to its length,$R \propto l$.Since the wire is cut into ten equal parts,the resistance of each part is $r = \frac{R}{10}$.When two such pieces are connected in series,the resistance of one combination is $r_s = r + r = 2r = 2 	imes \frac{R}{10} = \frac{R}{5}$.We have five such combinations connected in parallel. The equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{r_s} + \frac{1}{r_s} + \frac{1}{r_s} + \frac{1}{r_s} + \frac{1}{r_s} = \frac{5}{r_s}$.Therefore,$R_{eq} = \frac{r_s}{5} = \frac{R/5}{5} = \frac{R}{25}$.

$A$ copper wire of resistance $R$ is cut into ten parts of equal length. Two pieces each are joined in series and then five such combinations are joined in parallel. The new combination will have a resistance

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