Question

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

Plz if u know only give the answer.​

Answer 1

✯ Length = 122.7 m ✯

✯ Resistance = 2.5 Ω ✯

Explanation:

Given:

• Diameter of the copper wire = 0.5 mm
• Resistivity of the copper wire $(\rho)$= 1.6 × $\sf{10^{-8}}$ Ω m
• Resistance (R) = 10 Ω

To find:

• Length of the wire to make its resistance 10 Ω.
• Change of resistance if the diameter is doubled.

Solution:

• Diameter = 0.5 mm

Then,

Radius of the wire (r) = 0.5/2 mm

$\implies$ Radius (r) = 0.25 mm

$\implies$ Radius (r) = 2.5 × $\sf{10^{-4}}$ m

We know that,

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

To find l , we need to find A ( area of cross section).

$\implies\sf{A=\pi\:r^2}$

$\implies\sf{A=3.14\times\:(2.5\times\:10^{-4})^2}$

$\implies\sf{A=3.14\times\:(2.5)^2\times\:(10^{-4})^2}$

$\implies\sf{A=3.14\times\:6.25\times\:10^{-8}}$

$\implies\sf{A=19.625\times\:10^{-8}}$

Therefore, area of the cross section is $\sf{19.625\times\:10^{-8}}$ m².

Now find the length (l) of the wire.

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

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

• Put values.

$\implies\sf{l=\dfrac{10\times\:19.625\times\:10^{-8}}{1.6\times\:10^{-8}}}$

$\implies\sf{l=\dfrac{10\times\:19.625}{1.6}}$

$\implies\sf{l=122.65}$

$\implies\sf{l=122.7}$

Therefore, the length of the wire is 122.7 m.

_________________

If the diameter of the wire is doubled,

New diameter = 2× old diameter

New radius = 2× old radius = 2r m

Then,

New area(A') = πr²

→ New area(A') = π(2r)²

→ New area(A')= 4πr²

→ New area(A')= 4 × old area

→ New area = 4A

Let new resistance be Rʹ

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

$\implies$R’ = ρ ( l ) / ( 4A )

$\implies$R’ = ρ (l) X 1/( 4A )

Hence if diameter doubles, resistance becomes 1/4 times.

Therefore new resistance= 2.5 Ω.