Question

A resistance R is stretched to four times its length. Its new resistance will be

Answer 1

Answer:

it remain same

Explanation:

because it will not depend on length

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Answer 2

$\tt \: \blue{Resistnace \: of \: the \: original \: wire } \\ \bf \: R = \frac{ ρL}{A}, \: \: \sf \: where \: A= \frac{V}{L} \\ \\ \bf ⟹ R = \frac{ρL²}{V} \\ \\ \cal \small { Now \: its \: length \: is \: stretched \: such \: that \: new \: length }\\ \rm L′ = 4L \\ \\ \tt \: {Thus \: resistance \: of \: the \: wire} \\ \rm \: R′ = \frac{ρ(4L)²}{V} = 16R \\ \\ \red{ \small{ \bf Hence \: new \: resistance \: becomes \: 16 \: times \: the \: initial \: resistance.}}$