solvarmedw

2022-09-30

A ribbon of material with a rectangular cross sectional area of (0.25 mm x 1.00 mm), length 21.9 cm and resistance 1.5 is used as a heating element in a toaster. What is the resistivity of the material?

$a)1.31\times {10}^{-6}\mathrm{\Omega}\cdot m\phantom{\rule{0ex}{0ex}}b)1.00\times {10}^{-6}\mathrm{\Omega}\cdot m\phantom{\rule{0ex}{0ex}}c)1.71\times {10}^{-6}\mathrm{\Omega}\cdot m\phantom{\rule{0ex}{0ex}}d)1.97\times {10}^{-6}\mathrm{\Omega}\cdot m\phantom{\rule{0ex}{0ex}}e)1.50\times {10}^{-6}\mathrm{\Omega}\cdot m$

$a)1.31\times {10}^{-6}\mathrm{\Omega}\cdot m\phantom{\rule{0ex}{0ex}}b)1.00\times {10}^{-6}\mathrm{\Omega}\cdot m\phantom{\rule{0ex}{0ex}}c)1.71\times {10}^{-6}\mathrm{\Omega}\cdot m\phantom{\rule{0ex}{0ex}}d)1.97\times {10}^{-6}\mathrm{\Omega}\cdot m\phantom{\rule{0ex}{0ex}}e)1.50\times {10}^{-6}\mathrm{\Omega}\cdot m$

Bleha7s

Beginner2022-10-01Added 11 answers

The resistance of a material is proportional to

$R\propto \frac{l}{A}\phantom{\rule{0ex}{0ex}}R=\frac{\rho l}{A}\text{}[\rho =resistivity]$

The expression for the resistivity of the material is

$\rho =\frac{RA}{l}$

Here, $\rho $ is the resistivity, R is the resistance, Ais the area of cross section, l is the length.

in this case,

$A=0.25\text{}nm\cdot 1.00\text{}mm\phantom{\rule{0ex}{0ex}}I=21.9\text{}cm\phantom{\rule{0ex}{0ex}}R=1.5\phantom{\rule{0ex}{0ex}}\rho =\frac{(1.5\mathrm{\Omega})((0.25\text{}mm\times 1.00\text{}mm)\times \frac{1m}{{10}^{6}\text{}mm})}{21.9\text{}cm\times \frac{1\text{}m}{100\text{}cm}}\phantom{\rule{0ex}{0ex}}=1.71\times {10}^{-6}\mathrm{\Omega}\text{}m$

Thus correct option is (c).

$R\propto \frac{l}{A}\phantom{\rule{0ex}{0ex}}R=\frac{\rho l}{A}\text{}[\rho =resistivity]$

The expression for the resistivity of the material is

$\rho =\frac{RA}{l}$

Here, $\rho $ is the resistivity, R is the resistance, Ais the area of cross section, l is the length.

in this case,

$A=0.25\text{}nm\cdot 1.00\text{}mm\phantom{\rule{0ex}{0ex}}I=21.9\text{}cm\phantom{\rule{0ex}{0ex}}R=1.5\phantom{\rule{0ex}{0ex}}\rho =\frac{(1.5\mathrm{\Omega})((0.25\text{}mm\times 1.00\text{}mm)\times \frac{1m}{{10}^{6}\text{}mm})}{21.9\text{}cm\times \frac{1\text{}m}{100\text{}cm}}\phantom{\rule{0ex}{0ex}}=1.71\times {10}^{-6}\mathrm{\Omega}\text{}m$

Thus correct option is (c).

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