Question 19 of 31intermediate🔧 ApplyNumerical5 marks

A copper wire has diameter 0.5 mm and resistivity of 1.6×101.6 \times 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?

Correct Answer

(length) 122.72 m (approximately)

(resistance_change) Resistance decreases from 10 Ω to 2.5 Ω, i.e., it becomes one-fourth of the original resistance.

Exercise: EXERCISES | Q: 6 | (Page Page 23)
For More Understanding

Explanation

This numerical problem applies the formula R = ρl/A derived from the relationship that resistance is directly proportional to length and inversely proportional to area of cross-section. When diameter is doubled, the area becomes four times (since A = πd²/4), causing resistance to become one-fourth.

Solution Steps

  1. Given: Diameter (d) = 0.5 mm = 5×105 \times 10⁻⁴ m; Resistivity (ρ) = 1.6×101.6 \times 10⁻⁸ Ω m; Resistance (R) = 10 Ω

  2. Step 1: Calculate area of cross-section: A = πd²/4 = π × (5×105 \times 10⁻⁴)²/4 = π × 25×1025 \times 10⁻⁸/4 = 19.63×1019.63 \times 10⁻⁸ m² (or 1.963×101.963 \times 10⁻⁷ m²)

  3. Step 2: Using formula R = ρl/A, rearrange to find length: l = RA/ρ

  4. Step 3: Substitute values: l = (10×1.96310 \times 1.963 × 10⁻⁷)/(1.6×101.6 \times 10⁻⁸) = 122.72 m

  5. Step 4: When diameter is doubled, new diameter = 2d, so new area A' = π(2d)²/4 = 4 × πd²/4 = 4A

  6. Step 5: New resistance R' = ρl/A' = ρl/4A = R/4 = 10/4 = 2.5 Ω

  7. Conclusion: The resistance decreases from 10 Ω to 2.5 Ω, becoming one-fourth of the original value.

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