A tightly coiled spring having 75 coils, each 3.50 cm in diameter, is made of insulated metal wire 3.25 mm in diameter. An ohm meter connected across its opposite ends reads 1.74 Ω. What is the resistivity of the metal?

A tightly coiled spring having 75 coils, each 3.50 cm in diameter, is made of insulated metal wire 3.25 mm in diameter. An ohm meter connected across its opposite ends reads 1.74 $\mathrm{\Omega }$. What is the resistivity of the metal?
Solution:
Given,
number of turns, $n$ = 75
diamter of spring, $d_1$ = 3.50 cm
radius of spring, $r_1$ = 1.75 $\times$ 10${}^{-2}$ cm
diameter of metal wire, $d_2$ = 3.25 mm
radius of metal wire, $r_2$ = 1.625 $\times$ 10${}^{-3}$ m
resistance of metal wire, $R$ = 1.74 $\mathrm{\Omega }$
resistivity of metal wire, $\rho$ = ?
Let, $l$ be the length of metal wire.

Now,
$$\begin{array}{rl}\text{length of metal wire}& =\text{length of spring coil}\\ & =n\times 2\pi r_1\\ & =75\times 2\pi \times 1.75\times {10}^{-2}\\ \therefore l& =8.25m\end{array}$$ Again,
$$\begin{array}{rl}R& =\frac{\rho l}{A}\\ \rho & =\frac{RA}{l}\\ & =\frac{1.74\times \pi r_2^2}{8.25}\\ & =\frac{1.74\times \pi \times (1.625\times {10}^{-3}{)}^2}{8.25}\\ \therefore \rho & =1.75\times {10}^{-6}\mathrm{\Omega }m\end{array}$$ Hence, the resistivity of the metal wire is $1.74\times {10}^{-6}$ $\mathrm{\Omega }m$.
Detailed explanation of this numerical problem can be found in the video below. ( Skip to 11:08 ).