Question

A wire 50 cm long and 1 mm² in cross-section
carries a current of 4 A when connected to a 2
V bettery. The resistivity of the wire is​

Answer 1

Explanation:

Resistance resistance of the wire

=(potential drop driving the current in the wire)/(current current flowing through the wire)

=2V / 4A

=0.5 ohm

l=50cm

a=1mm^2=0.01cm^2

We know,

resistance=(resistivity×length)/cross sectional area

resistivity= (R×a)/l

=(0.5 × 0.01)/50 ohm-cm

$= 1 \times {10}^{ - 4} ohm - cm$

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Answer 2

$\huge{\underline{\underline{\green{\mathfrak{Answer:}}}}}$

Given

length of the wire, L= 50cm = 0.5m

Area of cross-section, A = 1 mm²

=$\mathsf{1 × 10^{-6} m^2}$

current = 4A

Potential difference = 2V

We know that,

$\mathsf{Resistance = \frac{potential \:difference}{Current}}$

$\mathsf{\implies \: R = \frac{2}{4}}$

$\mathsf{\implies \: R = 0.5 \:Ω}$

Now,

$\fbox{\mathsf{\implies \:R = \rho \frac{L}{A}}}$

$\mathsf{\implies \:\rho = \frac{RA}{L}}$

$\mathsf{\implies \:\rho = \frac{0.5 × 1 × 10^{-6}}{0.5}}$

$\mathsf{\implies \:\rho = 1 × 10^{-6}}$

So, the resistivity of the wire = $\mathsf{1 × 10^{-6}}$Ωm