Question

What will be the resistance of a wire of 0.65m length and 0.2 mm in radius, and resistivity 3*10^-6 Ωm

Answer 1

Answer:

Given –

• Resistivity = 3 × 10^{-6} Ωm.
• Length of the wire = 0.65 m.
• Radius of the wire = 0.2 mm.

To Find –

• Resistance ?

Solution –

Radius = 0.2 mm = 0.0002 m.

Area of cross section = π (0.0002)²

➸ π (0.0000004)

➸ 3.14 × (0.00000004)

➸ 0.0000001256 m² = 1.256 × 10^{-7} m²

Formula : R = rho*l/A

➸ R = 3 × 10^{-6} × (0.65/1.256 × 10^{-7})

➸ R = 3 × 10^{-6-(-7)} × 0.65/1.256

➸ R = 3 × 10^{-6+7} × 0.65/1.256

➸ R = 10¹ × 1.95/1.256

➸ R = (10 × 1000 × 1.95)/100 × 1256

➸ R = (1000 × 195)/1256 × 10

➸ R = (100 × 195)/1256

➸ R = (100 × 0.155)

➸ R = 15.5 Ω

So, the resistance of the wire is 15.5 Ω.

Answer 2

GIVEN :-

• Length(l) = 0.65 m

• Radius(r) = 0.2 mm

• Resistivity(ρ) = 3 × 10⁻⁶ Ω-m

TO FIND :-

The resistance.

FORMULA TO BE USED :-

Resistance = Resistivity × Length/Area of cross section

$\sf{\implies R=\dfrac{\rho l}{A} }$

SOLUTION :-

Radius = 0.2 mm = 0.2 ÷ 1000 = 0.0002 m

Area of cross-section(A) = πr² = π(0.0002)²

⇒Area of cross-section = π(0.0000004)

⇒Area of cross-section = 3.14 × 0.0000004

⇒Area of cross-section = 0.0000001256

⇒Area of cross-section = 1.256 × 10⁻⁷

$\rule{200}{1.8}$

Now, using the formula,

$\sf{R=\dfrac{3\times10^{-6}\times0.65}{1.256\times10^{-7}} }$

$\sf{\implies R=\dfrac{3\times0.65\times10^{-6}\times 10^7}{1.256} }\\\\\sf{\implies R=\dfrac{3\times0.65\times10^{-6+7}}{1.256} }\\\\\sf{\implies R=\dfrac{1.95\times 10^{1}}{1.256} }\\\\\sf{\implies R=\dfrac{19.5}{1.256} }\\\\\boxed{\sf{\implies R=15.52\ \Omega}}$

Therefore, the resistance of the wire is 15.52 Ω.