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A uniform wire of resistance 100 Ω is melted and recasted into a wire of length twice that of the original. The resistance of the new wire is _____.

A. 100 Ω
B. 200 Ω
C. 300 Ω
D. 400 Ω

1 Answer

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The resistance of the new wire is D. 400 Ω

How to find the resistance ?

In this case, the resistivity and the cross-sectional area remain the same because the wire is melted and recasted from the same material. However, the length is doubled (L' = 2L).

Therefore:

R' = ρ * L' / A²

= ρ * (2L) / A ²

= 2 * (ρ * L / A ²)

= 2 * R

Since the wire is doubled in length but presumably not doubled in width or thickness (assuming it was originally cylindrical), the cross-sectional area will be reduced by half. This will further increase the resistance.

Accounting for both the increase in length and the decrease in area, the new resistance becomes:

R' = ρ * (2L) / (A/2) ²

= 4 * (ρ * L / A ²)

= 4 * R

= 4 x 100

= 400 Ω

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