A wire of resistance 9 Ohm having length 30 cm is tripled on itself. What is it's new resistance?

Given,

Resistance (R) = 9 Ω

When the wire is tripled on itself, it's area of cross section becomes thrice and it's length becomes $\dfrac{l}{3}$.

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

Then,

length = 30 cm,

new length = $\dfrac{l}{3}$ = $\dfrac{30}{3}$ = 10 cm,

new area a_n = 3a

From relation

R = ρ $\dfrac{l}{a}$

Initial resistance 9 = ρ $\dfrac{30}{a}$    [Equation 1]

New resistance R_n = ρ $\dfrac{10}{3a}$    [Equation 2]

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

$\dfrac{R$n}{9} = \dfrac{ρ \dfrac{10}{3a}}{ ρ \dfrac{30}{a}} \\[0.5em] \dfrac{Rn}{9} = {\dfrac{10}{90}} \\[0.5em] \dfrac{Rn}{9} = {\dfrac{1}{9}} \\[0.5em] Rn = 1 \varOmega \\[0.5em]

Hence, the new resistance = 1 Ω.