A wire of 5Ω resistance is drawn out so that its new length is two times the original length. If the resistivity of the wire remains the same and the cross-sectional area is halved, the new resistance is
- 40Ω
-
20Ω
- 10Ω
- 5Ω
The correct answer is: B
Explanation
Let the original length = L1Let the original resistance = R1 = 5Ω
Let the original resistivity = P1
Let the original area = a1
Let the new length = L2 = 2L1
let the new area = a2 = 1/(2a2)
Let the new resistance = R2
Let the new resistivity = P2
But since the resistivity remains the same,
=> P1 = P2
| ∴ P1 = | R1 a1 |
| _L1 |
| = P2 = | R2 a2 |
| _L2 |
But a2 = 1/2a1; and L2 = 2L1
| ∴ | R1 a1 |
| _L1 |
| = | R2 2a1/2 |
| _2L1 |
| => | 5 x a1 |
| _L1 |
| = | R2 x a1 |
| _4L1 |
| ∴R2 = | 5 x a1 x 4L1 |
| _a1 x L1 |
= 20Ω