Question

Resistance of a metal wire of length 1 m is 26 22 at 20°C. the
diameter of the wire is 0.3 mm, what will be the resistivity of the
metal at that temperature? Using Table 12.2, predict the material
of the wire.​

Answer 1

R=26 Ω
l=1m
d−0.3mm

r=
2
d
​
=
2
0.3×10
−3

​
=0.15×10
−3
m

A=πr
2
=3.14×(0.15×10
−3
)
2
=0.071×10
−6
m
2


R=
A
ρL
​


or ρ=
l
RA
​
=
1
26×0.071×10
−6

​
=1.84×10
−6
Ωm

Answer 2

Answer:

Correct Question:

Resistance of a metal wire of length 1 m is 26 Ω at 20°C. The diameter of the wire is 0.3 mm, what will be the resistivity of metal at that temperature?

Using Table 12.2, predict the material of the wire.

Given:

Resistance (R) = 26 Ω

Diameter of wire (d) = 0.3 mm = $\sf 3 \times 10^{-4}$

Length of wire (l) = 1 m

To Find:

➡Resistivity of metal at 20°C \sf (\rho)(ρ)

➡Material of the wire

Resistivity of metallic wire is given as:

$\boxed{ \boxed{ \bf{ \rho = \dfrac{RA}{l} }}}$

A → Area of cross-section of wire $\sf \dfrac{\pi d^2}{4}$

So,

$\bf \rho = \dfrac{R\pi {d}^{2} }{4l}$

By substituting values we get:

$\begin{gathered}\rm \implies \rho = \dfrac{26 \times 3.14 \times {(3 \times {10}^{ - 4}) }^{2} }{4 \times 1} \\ \\ \rm \implies \rho = \dfrac{26 \times 3.14 \times 9\times {10}^{ - 8}}{4 } \\ \\ \rm \implies \rho = \dfrac{ 734.76 \times {10}^{ - 8}}{4 } \\ \\ \rm \implies \rho = 183.69 \times {10}^{ - 8} \\ \\ \rm \implies \rho = 1.84 \times {10}^{ - 6} \: \Omega \: m\end{gathered}$

$\therefore\boxed{\mathfrak{Resistivity \ of \ metal \ at \ 20\degree C \ (\rho) = 1.84 \times 10^{-6} \ \Omega \ m}}$

Resistivity of metal at 20°C (ρ)=1.84×10

From Table 12.2 of Class - 10 NCERT Chapter - 12 Electricity (Attached)

$\boxed{\mathfrak{Material \ of \ the \ wire = Manganese}}$