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A constant wire has resistance $R$ . If the area of the cross-section of the wire is doubled, find the new resistance in terms of its original resistance.

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\frac{R}{2}

\frac{R}{4}

\frac{R}{2}

R

answer is B.

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Detailed Solution

A constant wire has resistance

R

. If the area of cross-section of the wire is doubled, then the new resistance will be

\frac{R}{4}

.
A wire's resistance is inversely proportional to its area and directly proportional to its length.
It means resistance

R \propto \frac{l}{A}

R = ρ \frac{l}{A}

where

l

is the length of the wire,

A

is the cross-sectional area of the wire

ρ

is the resistivity of the wire. Let initial resistance of the wire is

R = ρ \frac{l}{A}

When the wire is compressed to half of its original length without changing the volume, its length becomes,

l' = l / 2

But since the volume is same,

V' = V

\Rightarrow A' l' = Al

\Rightarrow A' = \frac{Al}{l'}

\Rightarrow A' = \frac{Al}{l / 2}

\Rightarrow A' = 2 A

Then the resistance

R = ρ \frac{\frac{l}{2}}{2 A}

That is new resistance

R = ρ \frac{l}{4 A}

\Rightarrow

R = \frac{R}{4}

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