A copper wire has length L and cross-sectional area A. Its resistance is R. If we halved the length and halved the diameter of wire, then what will be the resistance of this wire?

Explanation

The correct answer is 2R. To understand why, we first consider that when the length of the wire is halved, the resistance is halved as well. Therefore, if the original resistance is R, halving the length gives R/2.Next, when the diameter of the wire is halved, the cross-sectional area decreases to a quarter of its original size (since area is proportional to the square of the diameter). This increase in resistance due to the diameter change causes the resistance to quadruple.Thus, the new resistance due to the change in diameter would be 4 times the reduced value from the halved length, which results in (4 * (R/2)) = 2R.The other options are incorrect because they do not account correctly for the effects of both changes on resistance. 4R and 3R incorrectly assume a different relationship between diameter and resistance, while R disregards the halving of the length.