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The resistance of a 10 m long wire is 10 Ω. Its length is increased by 25% by stretching the wire uniformly. The resistance of wire will change to (approximately)
- a)12.5 Ω
- b)14.5 Ω
- c)15.6 Ω
- d)16.6 Ω
Correct answer is option 'C'. Can you explain this answer?
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The resistance of a 10 m long wire is 10 Ω. Its length is increa...
Ω. If the wire is cut into two equal pieces, each piece will have a resistance of 20 Ω.
Explanation:
The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area. When a wire is cut into two equal pieces, the length of each piece is halved. Therefore, the resistance of each piece will double because resistance is directly proportional to length.
Let's use the formula for resistance:
R = ρL/A
where R is resistance, ρ is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area of the wire.
If we rearrange the formula to solve for A, we get:
A = ρL/R
Now let's plug in the values for the original wire:
R = 10 Ω
L = 10 m
ρ (for copper) = 1.68 x 10^-8 Ωm
A = (1.68 x 10^-8 Ωm)(10 m)/(10 Ω)
A = 1.68 x 10^-7 m^2
This is the cross-sectional area of the original wire. When the wire is cut into two equal pieces, each piece will have half the length and half the resistance. Therefore, the resistance of each piece will be:
R = 2(10 Ω) = 20 Ω
We can use the same formula to calculate the cross-sectional area of each piece:
A = (1.68 x 10^-8 Ωm)(5 m)/(20 Ω)
A = 4.2 x 10^-8 m^2
Since each piece has the same cross-sectional area, they are equal in diameter and length.
Explanation:
The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area. When a wire is cut into two equal pieces, the length of each piece is halved. Therefore, the resistance of each piece will double because resistance is directly proportional to length.
Let's use the formula for resistance:
R = ρL/A
where R is resistance, ρ is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area of the wire.
If we rearrange the formula to solve for A, we get:
A = ρL/R
Now let's plug in the values for the original wire:
R = 10 Ω
L = 10 m
ρ (for copper) = 1.68 x 10^-8 Ωm
A = (1.68 x 10^-8 Ωm)(10 m)/(10 Ω)
A = 1.68 x 10^-7 m^2
This is the cross-sectional area of the original wire. When the wire is cut into two equal pieces, each piece will have half the length and half the resistance. Therefore, the resistance of each piece will be:
R = 2(10 Ω) = 20 Ω
We can use the same formula to calculate the cross-sectional area of each piece:
A = (1.68 x 10^-8 Ωm)(5 m)/(20 Ω)
A = 4.2 x 10^-8 m^2
Since each piece has the same cross-sectional area, they are equal in diameter and length.
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The resistance of a 10 m long wire is 10 Ω. Its length is increased by 25% by stretching the wire uniformly. The resistance of wire will change to (approximately)a)12.5 Ωb)14.5 Ωc)15.6 Ωd)16.6 ΩCorrect answer is option 'C'. Can you explain this answer?
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