Resistivity And Wire Geometry

Question

A uniform wire of resistance R is stretched so that its length doubles while its volume stays constant, so what is the new resistance?

RankGuru · Accepted Answer

Resistance depends on geometry through the relation $R = \rho \frac{l}{A}$, where $\rho$ is resistivity, $l$ is length, and $A$ is cross-sectional area, as established in NCERT Class 12, Chapter 3 (Current Electricity). When a wire is stretched at constant volume, $V = Al$ stays fixed, so doubling the length to $2l$ forces the area to halve to $A/2$. The new resistance becomes $R' = \rho \frac{2l}{A/2} = 4\rho \frac{l}{A} = 4R$. More generally, at constant volume the area can be written as $A = V/l$, so resistance becomes $R = \rho l^2 / V$, meaning resistance scales as the square of the length; thus a length factor of 2 gives a resistance factor of 4. The option 2R accounts only for the doubling of length and forgets that the area shrinks to half simultaneously, so it underestimates the change. The option R/2 wrongly treats the cross-sectional area as increasing, which cannot happen when volume is fixed and length grows. The option R/4 inverts the correct dependence entirely, as if lengthening the wire lowered its resistance. A consistency check confirms the reasoning: since $R \propto l^2$ under constant volume, $2^2 = 4$, and physically the resistance can only increase when a wire is made both longer and thinner, so the fourfold rise in magnitude is entirely sensible.

Resistivity And Wire Geometry

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