# A tightly coiled spring having 75 coils,each 3.5 cm in diameter is made of insulated metal wire 3.25 mm in diameter. An ohm meter connected across its opposite ends reads 1.74ohm . what is the reseptivity of metal ?

Md Mushahid
21 Points
3 years ago
let us consider given 75 coils
diameter of coiled spring  dc = 3.50cm = $3.5\times 10^{-2}$m
diameter of insulated metal wire dm = 3.25mm =$3.25\times 10^{-3}$m
$R =1.74\Omega$
$\rho = \frac{R\times A}{L}$
First we need to find the value of  L,A
$L=75\times \pi \times 3.50\times 10^{-2}$
L =8247m
$A = \pi \times r ^{2} = \frac{\pi }{4}\times d ^{2}$
$A = \frac{3.142 }{4}\times\left ( 3.25\times 10^{-3} \right )^{2}$
$A = 8.296\times10 ^{-6} m$

$\rho = \frac{R\times A}{L}$
$\rho = \frac{1.74\times 8.296\times 10^{-6}}{8.247}$

$\rho = 1.7540\times 10^{-6} \Omega m$