A wire of 9 Ω resistance having 30 cm length is tripled on itself . what is its new resistance?

Initially,

Resistance, R = 9 Ω

Length, L = 30 cm = 0.3 m

Let A be the area of cross-section and ρ be the resistivity.

When the length is stretched to 3L, the area of cross-section should decrease in such a way that the volume of the wire remain constant.

So, Volume = AL = A ^/ (3L)

=> A ^/ = A/3

Now,

The new resistance is,

R ^/ = ρ(3L)/A ^/ = ρ(3L)/(A/3) = 9R = (9)(9) = 81 Ω