Final answer:
The resistance of a wire increases with length and decreases with cross-sectional area. When a cylindrical wire's length is doubled and its diameter is reduced, leading to the area being reduced by a factor of four, the resistance increases by a factor of four. Therefore, the resistance changes by 300% from its original value.
"the correct option is approximately option C"
Step-by-step explanation:
The question concerns the change in resistance of a cylindrical wire when its length is increased by 100% and how that affects the cross-sectional area and diameter. The resistance R of a cylindrical wire is given by R = ρ(L/A), where ρ is the resistivity of the material, L is the length, and A is the cross-sectional area. If the length of the wire doubles, its volume remains constant, leading to a decrease in its cross-sectional area by a factor of four because the area is proportional to the square of the diameter (given that the cross-sectional area A = π(d/2)^2, where d is the diameter).
As resistance is inversely proportional to the area, when the area reduces by a factor of four, the resistance increases by the same factor. Therefore, if the original resistance R is increased by 100%, the new resistance is 2R. Considering the decrease in area by a factor of four, the resistance then becomes 4(2R) which is 8R. The increase from R to 8R is a 700% increase, not covered by the options provided.
However, addressing the original options, if the length were increased by 100%, the original statement would be wrong as it states an increase of 100% would lead to a 300% change in resistance. The new resistance would be 4 times the original resistance which is a 300% increase (not 300% the original resistance).