Physics, asked by joeljohnsaji, 1 month ago

A copper wire has diameter 1 mm and resistivity of 1.6 × 10 –8 Ω m. What will be

the length of this wire to make its resistance 10 Ω? How much does the resistance

change if the diameter is doubled?​

Answers

Answered by nikhilrai27
0

Answer:

Given, diameter, d=0.5 mm

resistivity, ρ=1.6×10 −8Ωm

Resistance, R=10 Ω

Let the length of wire be l.

A=πd^ 2/4

R=ρl/A

= πd ^2 /4ρ

l⟹l= 4ρRπd 2

l=10×3.14×(0.5×10 −3 )^2/4×1.6×10^ −8

l=122R∝1/d^2

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

Hence, new resistance =2.5Ω

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