Question

A uniform wire of resistance 100 ohm is melted and recast in to wire of length double that of the original.What would be the resistance of the new formed?

Answer 1

Given that, a uniform wire of resistance 100 ohm is melted and recast in to wire of length double that of the original.

We have to find the new resistance of the wire.

Now,

R = p l/A

Multiply and divide by length.

R = p l/A × l/l

R = p l²/V

Here; R = resistance, p = resistivity, l = length, A = area of cross-section and V = volume (area × length)

Given R = 100 ohm

Also, resistance is directly proportional to length of the wire i.e. R ∝ l. Similarly, R ∝ l².

Since, wire is melted and recast in to a wire of the length double that of the original wire.

So, here we will denote the resistance and length by R and l. Also, the new resistance and length by R' and l'.

So, we can write it like:

→ R/R' = l²/l'²

l' = 2l (given in question)

→ R/R' = l²/(2l)²

→ R/R' = l²/4l²

→ R/R' = 1/4

→ R' = 4R

As R is 100 ohm. So,

→ R' = 4(100)

→ R' = 400 ohm

Therefore, the new resistance of the wire is 400 ohm.

Answer 2

Correct Question -

A uniform wire of resistance 100 ohm is melted and recast in to wire of length double that of the original.

What would be the resistance of the new wire formed ?

Alternative SoLuTiOn -

In the above Question , the following information is given -

A uniform wire of resistance 100 ohm is melted and recast in to wire of length double that of the original.

We have to find the resistance of the new wire formed ...

Now, this wire can be considered as a Cylinder, which has a defined radius and height .

Here, the length of the wire , refers to the height of the cylinder .

The radius is the radius of the wire .

Here, as we can observe , the volume of the wire remains constant .

So,

Innitial Volume = Final Volume

$\sf{ => ( 1 / 3 ) × π × {r_{1}}^2 × l = ( 1 / 3 ) × π × {r_{2}}^2 × 2l }$

The like terms get cancelled .

So,

$\sf{ \dfrac{ r_{1} }{ r_{2} } = 2 }$

So, we can conclude that as the length of the wire is doubled, the cross Sectional area of the wire is reduced by half .

Now we know that -

Resistance = $\rho$ × ( L / A ) .......... { 1 }

Here ,

New Resistance = $\rho$ × ( l )( 4A ) .............. { 2 }

Dividing Equation { 1 } / { 2 } we get -

New resistance = 4 × Innitial Resistance .

=> New Resistance = 4 × 100 $\Omega$

=> New resistance = 400 $\Omega$

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