A piece of wire of resistance R is cut into f | vedclass

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(C) The resistance of a wire is directly proportional to its length $(R \propto l)$.When a wire of resistance $R$ is cut into five equal parts,the res

Vedclass · Accepted Answer

$25$

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(C) The resistance of a wire is directly proportional to its length $(R \propto l)$.When a wire of resistance $R$ is cut into five equal parts,the resistance of each part becomes $R/5$.When these five parts are connected in parallel,the equivalent resistance $R'$ is given by the formula:$1/R' = 1/R_1 + 1/R_2 + 1/R_3 + 1/R_4 + 1/R_5$Substituting $R_1 = R_2 = R_3 = R_4 = R_5 = R/5$:$1/R' = 5/R + 5/R + 5/R + 5/R + 5/R = 25/R$Therefore,$R/R' = 25$.

Material	Resistivity $(\Omega \,m)$
Silver	$1.6 \times 10^{-8}$
Copper	$1.62 \times 10^{-8}$
Aluminium	$2.63 \times 10^{-8}$
Tungsten	$5.20 \times 10^{-8}$
Nickel	$6.84 \times 10^{-8}$
Iron	$10.0 \times 10^{-8}$
Chromium	$12.9 \times 10^{-8}$

$A$ piece of wire of resistance $R$ is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is $R'$,then the ratio $R/R'$ is

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