Question

A uniform wire of resistance 100 Ω is melted and recast as a wire whose length is double that of the original. What would be the resistance of the wire?

Answer 1

$\huge{\text{\underline{Question:-}}}$

A uniform wire of resistance 100Ω is melted and recast into wire of length double that of the original. What would be the resistance of the wire?

$\huge{\text{\underline{Solution:-}}}$

★ Given:-

• R = 100 Ω

★ Where:-

• l = l'
• R = 100s \ ωs
• l' = 2l’

★ Using formula:-

$\large{\boxed{\text{R = pl / A}}}$

$\implies$R = p(2l) / A

$\implies$R = 2(pl / A)

$\implies$R' = 2R’

★ Resistance is doubled:-

R = 4 X 100 Ω = 400 Ω

$\large{\boxed{\text{= 400 ohm}}}$

Therefore the resistance of the wire is 400 Ω.

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Answer 2

Answer:

Explanation:

Given :-

R = 100 Ω

l' = 2l

To Find :-

Resistance of a wire

Solution :-

Area × l = Area' × 2l

Area' = Area/2

f = Specific resistance

⇒ R' = fl/2

⇒ 100 = fl2/Area/2

⇒ 100 = 4fl/2

⇒ 100 = 4 × R

⇒ R = 100 × 4

⇒ R = 400 Ohm

Hence, the resistance of the wire is 400 ohm.