A uniform copper wire of mass 2.23X10^-3 Kg carries a current of 1A When 1.7V is applied across it. Calculate its length and area of cross-section. If the wire is uniformly stretched to double its length, Calculate the new resistance. Density of copper is 8.92 x 10^3 kg^-3 and resistivity is 1.7x10-8 ohm meter.
Answers
Hello Friend,
● Answer -
l = 5 m
A = 5×10^-8 m^2.
◆ Explaination -
# Given -
m = 2.23X10^-3 kg
d = 8.92×10^3 kg/m^3
ρ = 1.7×10^-8 Ω-m
I = 1 A
V = 1.7 V
# Solution -
Volume of the copper wire is given by -
v = m / d
lA = (2.23X10^-3) / (8.92×10^3)
lA = 25×10^-8 m^3 ...(1)
Resistance of the wire is given by -
R = V / I
ρl / A = 1.7 / 1
l/A = 1.7 / (1.7×10^-8)
l/A = 10^8 /m ...(2)
Solving (1) & (2), Putting l = 10^8 A
(10^8 A) A = 25×10^-8
A = 5×10^-8 m^2
l = 10^8 × 5×10^-8
l = 5 m
Hence, dimensions of the wire are l = 5 m & A = 5×10^-8 m^2.
Best luck for exams bud.
Answer:
L = 5 m
A = 5 * 10⁻⁸ m²
6.8 Ω
Explanation:
A uniform copper wire of mass 2.23X10^-3 Kg carries a current of 1A When 1.7V is applied across it. Calculate its length and area of cross-section. If the wire is uniformly stretched to double its length, Calculate the new resistance. Density of copper is 8.92 x 10^3 kg^-3 and resistivity is 1.7x10-8 ohm meter.
Density = Mass/Volume
8.92 x 10³ = 2.23X10⁻³ /Volume
=> Volume = 0.25 * 10⁻⁶ m³ = 25 * 10⁻⁸ m³
Volume = LA ( L = length , A = Cross sectional Area)
=> LA = 25 * 10⁻⁸ m³ - eq 1
R = ρ L/A
R = V/I
R = 1.7/1 = 1.7 Ω
1.7 = 1.7x10⁻⁸ (L/A)
=> L/A = 10⁸ - eq 2
Multiplying Eq 1 & 2
L² = 25
=> L = 5 m
A = 5 * 10⁻⁸ m²
If the wire is uniformly stretched to double its length
New length = 2L
Volume is same so new Area = A/2
Resistance will be 4 times = 4 * 1.7 = 6.8 Ω