If the 0.145 mm diameter tungsten filament in a light bulb is to have a resistance of 0.161 Ω at 20.0°C, how long should it be? The resistiv

If the 0.145 mm diameter tungsten filament in a light bulb is to have a resistance of 0.161 Ω at 20.0°C, how long should it be? The resistivity of tungsten at 20.0°C is 5.60 10-8 Ω · m

diameteter (d) = 0.145 mm,

resistance (R) = 0.161 Ω,

resistivity (ρ) = 5.60 *10-8 Ω · m

$$\rho = \dfrac{RA}{L} \qquad \rm{A} = \dfrac{\pi\ d^2}{4} \\\\
\text{Length (L): } \qquad
\boxed{L=\dfrac{\pi\ d^2 }{4} \dfrac{R}{\rho}}$$

$$\small{\text{
$
L= \left(
\dfrac{ \pi * 0.145^2 *0.161 } { 4 * 5.60 \cdot 10^{-8} }
\right)
\left( \dfrac{mm^2\ \Omega}{\Omega\ m}
\right)
$
}}
\qquad \boxed{
1\, \dfrac{\Omega\ mm^2}{m} = 10^{-6}\ \Omega\ m \qquad
1\ \Omega\ m = \dfrac{10^6\ mm^2\ \Omega}{m}
}\\\\
\small{\text{
$
L= \left(
\dfrac{ \pi * 0.145^2 *0.161 } { 4 * 5.60 \cdot 10^{-8} }
\right)
\left( \dfrac{mm^2\ \Omega}{\dfrac{10^6\ mm^2\ \Omega}{m}}
\right)
$
}}\\\\\\
\small{\text{
$
L= \left(
\dfrac{ \pi * 0.145^2 *0.161 } { 4 * 5.60 \cdot 10^{-8} }
\right)
\left( \dfrac{m}{10^6}
\right)
$
}}\\\\\\
\small{\text{
$
L= \left(
\dfrac{ \pi * 0.145^2 *0.161 } { 4 * 5.60 \cdot 10^{-2} }
\right)
\ m
$
}}\\\\
\small{\text{
$
L=\dfrac{ \pi * 0.145^2 *0.161 \cdot 10^2} { 4 * 5.60 } \ m
$
}}\\\\
\small{\text{
$
L=0.04747486461 \ m
$
}}\\
\small{\text{
$
L=4.747486461 \ cm \approx 4.75 \ cm
$
}}$$