Question

8. A copper wire has a diameter of 0.5 mm and resistivity of 1.6 x 10-8 2 m. What
will be the length of this wire to make its resistance 10 22 ? How much does the
resistance change if the diameter is doubled?​

Answer 1

Correct Question

A copper wire has a diameter of 0.5 mm and a resistivity of 1.6 x 10^-8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Solution

Given that, the diameter of the cooper wire is 0.5 mm and it's resistivity is 1.6 × 10^-8 Ωm.

{ Diameter = 0.5 mm

= 0.5/1000 m

= 5 × 10^-4 m }

Now,

2(Radius) = Diameter

Radius = Diameter/2

= (5 × 10^-4)/2

= 2.5 × 10^-4 m

Given resistivity (p) = 1.6 × 10^-8 Ωm.

R = p l/A

Also, A = πr²

A = 22/7 × (2.5 × 10^-4)^2

A = 22/7 × 6.25 × 10^-8

A = 1.964 × 10^-7 m^2

Using formula:

R = p l/A

→ l = (R × A)/p

{ R = 10 Ω (given) }

→ l = (10 × 1.964 × 10^-7)/(1.6 × 10^-8)

→ l = (1.964 × 10^-7 × 10^8)/(1.6)

→ l = 122.75 m

According to question,

Diameter is doubled.

D = 2D'

{ Diameter = 2(New diameter) }

So, R = 2R'

Also,

New Area = π(2r)²

= 4πr²

= 4 × (Area or Old Area)

Let's denote the new area by A' and old area by A

→ A' = 4A

Similarly,

New Resistance by R' and Old Resistance by R.

→ R' = p l/A'

→ R' = p l/4A

→ R' = 1/4 × p l/A

→ R'= 1/4 × R

If the diameter of the wire is doubled then the resistance becomes the one-fourth or 1/4.

Answer 2

Correct Question -

A copper wire has a diameter of 0.5 mm and resistivity of 1.6 x 10^(-8 )Ohm

What will be the length of this wire to make its resistance 10 Ohm ?

How much does the resistance change if the diameter is doubled?

Solution -

In the above Question, we have the following information -

A copper wire has a diameter of 0.5 mm and resistivity of 1.6 x 10^(-8) Ohm

So,

Resistivity Of The Copper wire $\rho = 1.6 \times {10}^{-8} \: Ohm$

Radius Of the wire = ( 0.5 / 2 ) mm = 2.5 mm = 2.5 × 10 ^ (-4) m .

Let the length of the wire be L m.

Area Of Cross Section -

$= > \pi {r}^{2} = \pi \: \times \: 2.5 \: \times \: {10}^{ - 4} \\ \\ \approx 1.964 \times {10}^{ - 7} \: {m}^{2}$

We know that -

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

Substituting the required Values-

$10 = 1.6 \times {10}^{-8} \dfrac{L}{ 1.964 \times {10}^{ - 7} } \\ \\ => L = 10 \times 1.964 \times {10}^{ - 7} \div { 10} ^ { -8 } \\ \\ => L \approx 122.7 m$

Now, the diameter is Doubled .

So, the cross Sectional area becomes 4 times the innitial area .

So the new resistance decreases 4 times the innitial resistance as resistance is inversely Proportional to the cross Sectional Area .