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A copper wire has a diameter of 0.5mm . Its resistivity is 1.6×10^-8 ohm . 1. What will be the length of this wire to make its resistance 10 ohm .2. How much does the resistance change when the diameter is doubled . |
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Answer» The LENGTH of the wire is 122.7 m and the new resistance becomes 1/4 times.Explanation:Given,Resistivity (ρ) = 1.6 × 10-8 Ω mResistance (R) = 10 ΩDiameter (d) = 0.5 mmd = 5 × 10⁻⁴ mHence, we will GET radius.=> Radius (r) = 0.25 mmr = 0.25 × 10⁻³ mr = 2.5 × 10⁻⁴ mWe need to find the area of cross-section,=> A = πr2A = (22/7)(2.5 × 10⁻⁴)2A = (22/7)(6.25×10⁻⁸)A = 1.964 × 10-7 m2We have to find the length of the wire.Let the length of the wire be LFormulaWe KNOW that,R = ρ (L) / (A)L = (R × A) / ρSubstituting the values in the above equation we get,L = (10 × 1.964 × 10⁻⁷) / 1.6 × 10⁻⁸ mL = 1.964×10-6 /1.6 × 10-8L = 122.72 MIF the diameter of the wire is doubled, the new diameter = 2 × 0.5 = 1mm = 0.001m.Let new resistance be Rʹ,R = ρ (L) / (A)R’ = ρ (L) / (4A)R’ = ρ (L) X 1/(4A)HENCE, if diameter doubles, resistance becomes 1/4 times.hopefully its helped u dear :) |
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