Current Electricity — Question 10

Given,

Resistance (R) = 3 Ω

Length (l) = 10 cm

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

Then,

R = 3 Ω

length = 10 cm,

new length = 30 cm,

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

From relation

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

Initial resistance 3 = ρ $\dfrac{10}{a}$    [Equation 1]

New resistance R₂ = ρ $\dfrac{30}{\dfrac{a}{3}}$ = ρ $\dfrac{90}{a}$    [Equation 2]

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

$\dfrac{3}{R_2} = \dfrac{ρ \dfrac{10}{a}}{ ρ \dfrac{90}{a}} \\[0.5em] \dfrac{3}{R_2} = \dfrac{10}{90} \\[0.5em] \dfrac{3}{R_2} = \dfrac{1}{9} \\[0.5em] \Rightarrow R_2 = 3 \times 9 \\[0.5em] \Rightarrow R_2 = 27 \varOmega \\[0.5em]$

Hence, the new resistance = 27 Ω.