1. A copper wire has diameter 1 mm and resistivity of 1.5 x 10 m. What will be the length of this wire to make its resistance 5 ? How much does the resistance change if the diameter is half?(3 marks)
Answers
Answer:
Resistivity (ρ) = 1.6 × 10-8 Ω m
Resistance (R) = 10 Ω
Diameter (d) = 0.5 mm
d = 5 × 10⁻⁴ m
Hence, we will get radius
Radius (r) = 0.25 mm
r = 0.25 × 10⁻³ m
r = 2.5 × 10⁻⁴ m
We need to find the area of cross-section
A = πr2
A = (22/7)(2.5 × 10⁻⁴)2
A = (22/7)(6.25×10⁻⁸)
A = 1.964 × 10-7 m2
Explanation:
mula
We know that
R = ρ (L) / (A)
L = (R × A) / ρ
Substituting the values in the above equation we get
L = (10 × 1.964 × 10⁻⁷) / 1.6 × 10⁻⁸ m
L = 1.964×10-6 /1.6 × 10-8
L = 122.72 m
If 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.
Answer
Therefore, the length of the wire is 122.7 m and the new resistance becomes 1/4 times.
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Given
- Resistivity (ρ) = 1.6 × 10-8 Ω m
- Resistance (R) = 10 Ω
- Diameter (d) = 0.5 mm
- d = 5 × 10⁻⁴ m
Hence, we will get radius
- Radius (r) = 0.25 mm
- r = 0.25 × 10⁻³ m
- r = 2.5 × 10⁻⁴ m
We need to find the area of cross-section
- A = πr2
- A = (22/7)(2.5 × 10⁻⁴)2
- A = (22/7)(6.25×10⁻⁸)
- A = 1.964 × 10-7 m2
Find out
We have to find the length of the wire
Let the length of the wire be L
Formula
We know that
- R = ρ (L) / (A)
- L = (R × A) / ρ
Substituting the values in the above equation we get
- L = (10 × 1.964 × 10⁻⁷) / 1.6 × 10⁻⁸ m
- L = 1.964×10-6 /1.6 × 10-8
- L = 122.72 m
If 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.
Answer
Therefore, the length of the wire is 122.7 m and the new resistance becomes 1/4 times
Explanation:
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