Choice D is correct because:

The resistance R of a conductor is given by the formula: R=ρAL, where ρ is the resistivity of the material, L is the length, and A is the cross-sectional area.
The volume V of the cylindrical piece is given by V=A×L. Since the volume remains constant, A1L1=A2L2.
Initially, the resistance is R=ρA1L1.
The length is doubled, so L2=2L1. Since the volume is constant, A1L1=A2(2L1), which means A2=2A1.
The new resistance R′ is given by R′=ρA2L2=ρ2A12L1=ρA12L1×2=4ρA1L1.
Since R=ρA1L1, the new resistance R′=4R.

Choice A is incorrect because:

If the new resistance were 2R, it would imply that either the length decreased or the area increased significantly. Since the length doubled and the volume remained constant, the area decreased, leading to an increase in resistance, not a decrease.
R′=2R would mean ρA2L2=21ρA1L1. Substituting L2=2L1, we get ρA22L1=21ρA1L1, which simplifies to A22=2A11, implying A2=4A1. This contradicts the condition that the volume remains constant when the length is doubled.

Choice B is incorrect because:

If the new resistance were R, it would imply that the ratio of length to area remained constant. However, the length doubled, and to maintain constant volume, the area must have halved. Therefore, the resistance cannot remain the same.
R′=R would mean ρA2L2=ρA1L1. Substituting L2=2L1, we get ρA22L1=ρA1L1, which simplifies to A22=A11, implying A2=2A1. This contradicts the condition that the volume remains constant when the length is doubled.

Choice C is incorrect because:

If the new resistance were 2R, it would imply that the ratio of length to area doubled. Since the length doubled and the volume remained constant, the area halved. This would result in a resistance four times the original, not twice.
R′=2R would mean ρA2L2=2ρA1L1. Substituting L2=2L1, we get ρA22L1=2ρA1L1, which simplifies to A22=A12, implying A2=A1. This contradicts the condition that the volume remains constant when the length is doubled.

Resistance and Resistivity | 2005 - M/J - 9702/1 | Q33