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​The resistivity of the material of wire having resistance 0.7Ω0.7Ω​, length 1 m and diameter 0.2 mm is:​​
Question

​The resistivity of the material of wire having resistance 0.7Ω0.7Ω​, length 1 m and diameter 0.2 mm is:​​

A.

2.2×108Ωm2.2 \times 10^{-8} \Omega m​​

B.

22×108Ωm22 \times 10^{-8} \Omega m​​

C.

0.22×108Ωm0.22 \times 10^{-8} \Omega m​​

D.

220×108Ωm220 \times 10^{-8} \Omega m​​

Correct option is A

​​The correct answer is(A) ​2.2×10−8Ωm​​.

To find the electrical resistivity of the material of the wire, we can use the formula for resistance:

R=ρLAR=\frac{ρL}{A}

Where:
- R is the resistance (in ohms),
- ρ is the resistivity (in ohm-meters),
- L is the length of the wire (in meters),
- A is the cross-sectional area of the wire (in square meters).

Step 1: Convert Length and Diameter to Appropriate Units
The length of the wire is given as 100 cm, which we need to convert to meters:
L=100cm=1mL=100cm=1m

The diameter of the wire is given as 0.2 mm, which we also need to convert to meters:Diameter=2.0mm=2.0×104m{Diameter}={2.0mm}={2.0}\times{10}^{-4}m

Step 2: Calculate the Radius
The radius r is half of the diameter:r=Diameter2=2.0×1042=1.0×104mr=\frac{Diameter}{2}=\frac{{2.0}\times{10}^{−4}}{2}={1.0}\times{10}^{−4}m

Step 3: Calculate the Cross-Sectional Area
The cross-sectional area A of the wire can be calculated using the formula for the area of a circle:
A=πr2
Substituting the radius:A=π(1.0×104)2=π1.0×108A={π} {({1.0}\times{10}^{−4})}^{2}={π} {1.0}×{10}^{−8}

Using π≈3.14:A3.14×108m2A≈{3.14}×{10}^{−8}{m}^2

Step 4: Rearranging the Resistance Formula
We need to find the resistivity ρ. Rearranging the formula gives us:
ρ=RALρ=\frac{R⋅A}L

Step 5: Substitute the Known Values
Now we can substitute the known values into the equation. The resistance R is given as 0.7 ohms:ρ=0.7(3.14×108)1ρ=\frac{{0.7}⋅ {(3.14}\times{10}^{−8)}}{1}

Step 6: Calculate the Resistivity
Calculating the above expression:ρ0.7×3.14×108ρ≈{0.7}\times{3.14}\times{10}^{−8}
ρ2.198×108ohmmetersρ≈{2.198}\times{10}^{−8}{ohmmeters}ρ2.198×108ohmmetersρ≈{2.198}\times{10}^ {−8}ohmmeters

Final Answer
Thus, the electrical resistivity of the material is approximately:ρ2.2×108ohmmetersρ≈{2.2}\times{10}^{−8}ohmmeters

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