Question

Two copper wire of length l and 2l have radii r and 2r respectively.The ratio of their specific resistance is

Answer 1

Two copper wire of length l and 2l have radii r and 2r respectively.The ratio of their specific resistance is

Resistance = ρ * Length / Area

Wire 1 :  Length = L    Radii = R

Area = pie × R²

Specific Resistivity = ρ

Resistance of Wire 1 = ρ × Length / Area

=> Resistance of Wire 1 = ρ × L / (Pie × R²)

Wire 2 :  Length = 2L     Radii = 2R

Area = pie × (2R)²

Specific Resistivity = ρ  same as of wire 1 as material is same

Resistance of Wire 2 = ρ × Length / Area

=> Resistance of Wire 2= ρ × 2L / (Pie × (2R)²)

Resistance of Wire 1 /Resistance of Wire 2  =( ρ × L / (Pie × R²) )/ (ρ × 2L / (Pie × (2R)²))

=> Resistance of Wire 1 /Resistance of Wire 2 =    (L × (2R)² ) / (R²× 2L)

=> Resistance of Wire 1 /Resistance of Wire 2 =    (L × 4R² ) / (R²× 2L)

=> Resistance of Wire 1 /Resistance of Wire 2 =   2/1

Ratio of resistances of wire = 2:1

Answer 2

answer 1 : 1

we know,

$\quad R=\rho\frac{l}{A}$

where R is resistance, l is length of wire, A is cross sectional area and $\rho$ is specific resistance.

you should remember that, for same material specific resistance remains same. it is material property. specific resistance doesn't change when we change length and cross sectional area of same material wire.

so, ratio of specific resistance is 1 : 1.

[ note :- I know, you will think if R = $\rho\frac{l}{A}$ then, $\rho=\frac{RA}{l}$, but it doesn't apply in case of same material. you should apply for different material ]