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?
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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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