Current Electricity — Question 15

9R

Reason — When the wire is stretched to thrice it's length, it's area of cross section becomes 1/3 and it's length becomes 3l.

Let, a be the area of initial cross section and ρ be the specific resistance of the material of wire.

Then,

length = l,

Resistance = R Ω,

new length = 3l,

new area = $\dfrac{a}{3}$

From relation

R = ρ $\dfrac{l}{a}$ = ρ $\dfrac{l}{πr^2}$

Initial resistance R₁ = R = ρ $\dfrac{l}{a}$    [Equation 1]

New resistance R_n = ρ $\dfrac{3l}{\dfrac{a}{3}}$ = ρ $\dfrac{9l}{a}$    [Equation 2]

On dividing eqn (ii) by (i), we get,

$\dfrac{R_n}{R} = \dfrac{ ρ \dfrac{9l}{a}}{ρ \dfrac{l}{a}} \\[1em] \Rightarrow \dfrac{R_n}{R} = 9 \\[1em] \Rightarrow R_n = 9R$

Hence, the new resistance = 9R.