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A wire of length L is drawn such that its diameter is reduced to half of its original diameter. If the initial resistance of the wire were 10 Ω, its new resistance would be
- a)40 Ω
- b)80 Ω
- c)120 Ω
- d)160 Ω
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
A wire of length L is drawn such that its diameter is reduced to half ...
The resistance of a wire is given by the formula:
R = ρL/A
where R is the resistance, ρ is the resistivity of the material, L is the length of the wire, and A is its cross-sectional area.
Since the wire is drawn such that its diameter is reduced to half of its original diameter, its cross-sectional area is reduced by a factor of 4 (since the area of a circle is proportional to its diameter squared). Therefore, the new cross-sectional area of the wire is A/4.
The length of the wire remains the same, so the new resistance can be calculated as:
R' = ρL/(A/4) = 4ρL/A
Since the initial resistance of the wire is 10 Ω, we can use this value to solve for ρ:
10 = ρL/A
ρ = 10A/L
Substituting this expression for ρ into the equation for R', we get:
R' = 4(10A/L)L/A = 40 Ω
Therefore, the new resistance of the wire is 40 Ω.
R = ρL/A
where R is the resistance, ρ is the resistivity of the material, L is the length of the wire, and A is its cross-sectional area.
Since the wire is drawn such that its diameter is reduced to half of its original diameter, its cross-sectional area is reduced by a factor of 4 (since the area of a circle is proportional to its diameter squared). Therefore, the new cross-sectional area of the wire is A/4.
The length of the wire remains the same, so the new resistance can be calculated as:
R' = ρL/(A/4) = 4ρL/A
Since the initial resistance of the wire is 10 Ω, we can use this value to solve for ρ:
10 = ρL/A
ρ = 10A/L
Substituting this expression for ρ into the equation for R', we get:
R' = 4(10A/L)L/A = 40 Ω
Therefore, the new resistance of the wire is 40 Ω.
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A wire of length L is drawn such that its diameter is reduced to half of its original diameter. If the initial resistance of the wire were 10 Ω, its new resistance would bea)40 Ωb)80 Ωc)120 Ωd)160 ΩCorrect answer is option 'D'. Can you explain this answer?
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