Final answer:
The resistance of the new cylindrical conductor with one fourth the original length of the initial conductor is R/16, because while the length is reduced by a factor of 4, the cross-sectional area increases by a factor of 4, resulting in a resistance 16 times smaller.
Step-by-step explanation:
The resistance (R) of a cylindrical conductor is given by the formula:
R = ρ(L/A)
where ρ (rho) is the resistivity of the material, L is the length, and A is the cross-sectional area. When the conductor is melted down and formed into a new cylindrical conductor of one fourth the original length (L/4), the volume of the conductor remains the same since the same amount of material is used.
The original cross-sectional area (A) is related to the radius (r) by A = πr2.
Because the new conductor has the same volume but a shorter length, its new cross-sectional area must be greater. Specifically, if the length is reduced by a factor of 4, the area must increase by a factor of 4 to keep the volume constant (since volume = length × area).
New Area (A') = 4A
The new resistance (R') of the shorter cylinder with a larger area now is:
R' = ρ(L/4)/(4A)
Which simplifies to:
R' = ρL/(16A) = R/16
Therefore, the correct answer is that the resistance of the new conductor is R/16.