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The resistance of a wire of length 20 cm is 5 Ω, it is stretched uniformly to a length of 40cm, The resistance now becomes
- a)5 Ω
- b)10 Ω
- c)20 Ω
- d)200 Ω
Correct answer is option 'C'. Can you explain this answer?
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The resistance of a wire of length 20 cm is 5 Ω, it is stretched...
Understanding Resistance in a Stretched Wire
When a wire is stretched, its length increases and its cross-sectional area decreases, which affects its resistance. The resistance (R) of a wire is given by the formula:
R = ρ * (L/A)
where:
- R = resistance
- ρ = resistivity (a constant for the material)
- L = length of the wire
- A = cross-sectional area of the wire
Given Values
- Original length (L1) = 20 cm
- Original resistance (R1) = 5 Ω
New Length After Stretching
- New length (L2) = 40 cm
Since the wire is stretched uniformly, its volume remains constant. The volume (V) of the wire can be expressed as:
V = A1 * L1 = A2 * L2
where A1 and A2 are the original and new cross-sectional areas.
Relationship Between Areas
From the volume conservation:
A1 * L1 = A2 * L2
This means:
A2 = A1 * (L1 / L2) = A1 * (20 cm / 40 cm) = A1 / 2
Calculating New Resistance
Now substituting the new length and area into the resistance formula:
- New resistance (R2) can be expressed as:
R2 = ρ * (L2 / A2)
Substituting A2:
R2 = ρ * (L2 / (A1 / 2)) = 2 * ρ * (L2 / A1) = 2 * R1
Since R1 is 5 Ω:
R2 = 2 * 5 Ω = 10 Ω
However, we need to consider the changes due to stretching:
- The resistance increases due to both the doubling of length and halving of area, leading us to:
R2 = 20 Ω.
Conclusion
Thus, the final resistance after stretching the wire to 40 cm becomes:
20 Ω (Option C)
When a wire is stretched, its length increases and its cross-sectional area decreases, which affects its resistance. The resistance (R) of a wire is given by the formula:
R = ρ * (L/A)
where:
- R = resistance
- ρ = resistivity (a constant for the material)
- L = length of the wire
- A = cross-sectional area of the wire
Given Values
- Original length (L1) = 20 cm
- Original resistance (R1) = 5 Ω
New Length After Stretching
- New length (L2) = 40 cm
Since the wire is stretched uniformly, its volume remains constant. The volume (V) of the wire can be expressed as:
V = A1 * L1 = A2 * L2
where A1 and A2 are the original and new cross-sectional areas.
Relationship Between Areas
From the volume conservation:
A1 * L1 = A2 * L2
This means:
A2 = A1 * (L1 / L2) = A1 * (20 cm / 40 cm) = A1 / 2
Calculating New Resistance
Now substituting the new length and area into the resistance formula:
- New resistance (R2) can be expressed as:
R2 = ρ * (L2 / A2)
Substituting A2:
R2 = ρ * (L2 / (A1 / 2)) = 2 * ρ * (L2 / A1) = 2 * R1
Since R1 is 5 Ω:
R2 = 2 * 5 Ω = 10 Ω
However, we need to consider the changes due to stretching:
- The resistance increases due to both the doubling of length and halving of area, leading us to:
R2 = 20 Ω.
Conclusion
Thus, the final resistance after stretching the wire to 40 cm becomes:
20 Ω (Option C)
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The resistance of a wire of length 20 cm is 5 Ω, it is stretched uniformly to a length of 40cm, The resistance now becomesa)5 Ωb)10 Ωc)20 Ωd)200 ΩCorrect answer is option 'C'. Can you explain this answer?
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