Question

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Answer 1

ANSWER

Given, diameter, d=0.5 mm

resistivity, ρ=1.6×10 ^−8 Ωm

Resistance, R=10 Ω

Let the length of wire be l.

A=πd ^2/4

R=ρl/A

= pl/πd^2/4

⟹ l= Rπd^2/4p

l= 10×3.14×(0.5×10 ^−3 ) ^2/

4×1.6×10 ^−8

l=122 m

R∝1/d ^2

If the diameter is doubled, resistance will be one-fourth.

Hence, new resistance =2.5Ω