Question

constant to wire of length 10 metre and cross section of the wire is 0.5 mm has resistivity of 4.9 into 10 ^ - 7 ohm find the resistance of the wire​

Answer 1

A constant wire of length 10 m (l = 19 m) and area of cross-section is 0.5 mm (A = 0.5 mm) has resistivity of 4.9 × 10^-7 ohm (p = 4.9 × 10^-7 ohm).

We have to find the resistance (R) of the wire.

We know that,

R = p l/A

Before that, convert area of cross-section into m. To convert mm into m, divide the given value by 1000.

Area of cross-section = 0.5 × 10^-3 m

Now,

Substitute the known in the above formula,

→ R = (4.9 × 10^-7 × 19)/(0.5 × 10^-3)

→ R = (4.9 × 19 × 10^(-7+3))/0.5

→ R = (93.1 × 10^-4)/0.5

→ R = 186.2 × 10^-4

→ R = 1.862 × 10^-2

Therefore, the resistance of the wire is 1.86 × 10^-2 ohm.

The resistance of the conductor (wire) is directly proportional to the length of wire

i.e. R ∝ l ..................(1st equation)

Inversely proportional to the area of cross-section of wire.

i.e. R ∝ 1/A .............(2nd equation)

From (1st equation) & (2nd equation) we have,

R ∝ l/A

If we remove the sign of proportionality we use a constant i.e. p (rho).

So, R = p l/A

Answer 2

GIVEN:

• A = 0.5mm = 0.5 × 10-³m²
• I = 10m
• ρ = 4.9 × 10-⁷

TO FIND:

• Resistance of wire

SOLUTION:

Using

R = ρI/A

Where

• ρ = 4.9 × 10-⁷
• I = 10m
• A = 0.5 × 10-³ m²
• R = ?

→ R = [(4.9 × 10-⁷)( 19 )]/0.5 × 10-³

→ R = 1.862 × 10-² ( ANSWER )

DERIVATION OF FORMULA:

We know that

R ∝ I …… ( i )

Substituting I = 1/A

R ∝ 1/A ……( ii )

From eq i & ii

R ∝ I/A

Removing the proportionality symbol we get a constant rho ( ρ )

R = ρI/A