Question

resistance of a 50 cm long and 0.2 mm diameter wire is 0.8 ohm resistance of another wire of different material of 75 cm length and 0.1 mm diameter is 1.5 ohm compare the specific resistance of the material ​

Answer 1

answer is 3.2. okayyy

Answer 2

Therefore the relation between the specific resistance of both the materials is 'ρ₁ = 3.2ρ₂'.

Given:

Wire-1:

Length = L₁ = 50 cm

Diameter = D₁ = 0.2mm

Resistance = R₁ = 0.8 Ohm

Wire-2:

Length = L₂ = 75 cm

Diameter = D₂ = 0.1 mm

Resistance = R₂ = 1.5 Ohm

To Find:

Relation between the specific resistance of both wires.

Solution:

The given question can be solved very easily as shown below.

Wire-1:

Length = L₁ = 50 cm = 50 × 10⁻² m

Diameter = D₁ = 0.2mm = 0.2 × 10⁻³ m

Resistance = R₁ = 0.8 Ohm

Area of the wire A₁ = πD₁²/4 =  ( π × 0.04 × 10⁻⁶ )/4 = π × 0.01 × 10⁻⁶ = π ×  10⁻⁸ m²

⇒ Specific resistance ρ₁ = R₁A₁/L₁ = ( 0.8 × π ×  10⁻⁸ ) / 50 × 10⁻²

⇒ Specific resistance ρ₁ = 1.6π ×  10⁻⁸ Ohm-m

Wire-2:

Length = L₂ = 75 cm =  75 × 10⁻² m

Diameter = D₂ = 0.1 mm = 0.1 × 10⁻³ m

Resistance = R₂ = 1.5 Ohm

Area of the wire A₂ = πD₂²/4 =  ( π × 0.01 × 10⁻⁶ )/4 = 2.5π × 10⁻⁹ m²

⇒ Specific resistance ρ₂ = R₂A₂/L₂ = ( 1.5 × 2.5π ×  10⁻⁹ ) / 75 × 10⁻²

⇒ Specific resistance ρ₂ = 5π ×  10⁻⁹ Ohm-m

Now the relation between the specific resistance of both the materials is given by,

⇒ ρ₁ / ρ₂ = ( 1.6π ×  10⁻⁸ ) / ( 5π ×  10⁻⁹ )

⇒ ρ₁ / ρ₂ = 3.2

⇒ ρ₁ = 3.2ρ₂

Therefore the relation between the specific resistance of both the materials is 'ρ₁ = 3.2ρ₂'.

#SPJ2