A wire is broken into four equal parts. A pac | vedclass

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(D) Let the resistance of the original wire be $R$. Since resistance is directly proportional to length $(R \propto l)$,each of the four equal parts w

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$\frac{1}{16}$

Vedclass · Answer

(D) Let the resistance of the original wire be $R$. Since resistance is directly proportional to length $(R \propto l)$,each of the four equal parts will have a resistance of $R' = R/4$.When these four wires are bundled together,they are connected in parallel.The equivalent resistance $R_{eq}$ of $n$ resistors in parallel,each of resistance $R'$,is given by $R_{eq} = R'/n$.Here,$n = 4$ and $R' = R/4$.Therefore,$R_{eq} = \frac{R/4}{4} = \frac{R}{16}$.Thus,the resistance of the packet is $\frac{1}{16}$ of the original resistance.

$A$ wire is broken into four equal parts. $A$ packet is formed by keeping the four wires together. The resistance of the packet in comparison to the resistance of the original wire will be:

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