Current Density And Resistivity

Resistance depends on geometry through (R = \rho \frac{L}{A}), where stretching a wire at constant volume couples length and area, since (V = AL) stays fixed. When the length is doubled to (2L), the constant-volume condition forces the cross-section to halve, (A' = A/2), because ((2L)(A/2) = LA). Substituting into the resistance formula, (R' = \rho \frac{2L}{A/2} = 4\rho \frac{L}{A} = 4R). Resistance therefore increases by a factor of four. The choice 2 times accounts only for the doubled length and ignores the thinning of the wire. The choice Half wrongly assumes the area grows. The choice Unchanged incorrectly treats resistance as a material constant independent of shape. This applies the NCERT geometric dependence of resistance, recognising that resistivity (\rho) is unchanged because the material is the same. A plausibility check confirms it: stretching makes a wire both longer and thinner, and both effects raise resistance, so a multiplicative factor of four (i.e. the square of the length factor) is exactly what the constant-volume scaling predicts. More generally, whenever a wire is drawn out to (k) times its original length at fixed volume, its resistance grows by a factor of (k^2), a useful shortcut that follows directly from the simultaneous lengthening and thinning and that students should recognise instantly rather than re-deriving each time.