Question

aluminium wire of resistance R is pulled so that its length becomes 4 times its original length find the new resistance of the wire in terms of it original resistance

Answer 1

Aluminium wire of resistance R is pulled so that its length becomes 4 times its original length.

We have to find the new resistance of the wire in terms of its original resistance.

We know, Resistance of a wire depends on length and cross sectional area of it. it is given by,

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

where ρ is resistivity, l length and A is cross sectional area of wire.

if volume remains constant.

i.e., Al = constant

$\implies R=\rho\frac{l^2}{Al}$

then, it is clear that, R ∝ l²

∴ R₁/R₂ = (l₁/l₂)²

here R₁ = initial resistance = R

R₂ = new resistance

l₁ = initial length of wire = l (let)

l₂ = new length of wire = 4l

∴ R/R₂ = (l/4l)² = 1/16

⇒R₂ = 16R

Therefore the new resistance of the wire in term of its original resistance is 16R.