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A copper wire has diameter 0.5 mm and resistivity of 1.6 x 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» - Resistivity (ρ) = 1.6 × 10-8 Ω mResistance (R) = 10 ΩDiameter (d) = 0.5 mmd = 5 × 10⁻⁴ mHence, we will get radiusRadius (r) = 0.25 mmr = 0.25 × 10⁻³ mr = 2.5 × 10⁻⁴ mWe need to find the area of cross-sectionA = πr2A = (22/7)(2.5 × 10⁻⁴)2A = (22/7)(6.25×10⁻⁸)A = 1.964 × 10-7 m2Find out :- We have to find the LENGTH of the wireLet the length of the wire be LFormula :- We know THATR = ρ (L) / (A)L = (R × A) / ρSubstituting the values in the above EQUATION we getL = (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 = 0.001mLet new resistance be RʹR = ρ (L) / (A)R’ = ρ (L) / (4A)R’ = ρ (L) X 1/(4A)Hence, if diameter doubles, resistance becomes 1/4 times.Answer :- Therefore, the length of the wire is 122.7 m and the new resistance becomes 1/4 times.Thank you . |
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