A conductor of radius r,length $\ell$ and resistivity $\rho$has resistance R.It is melted down and formed into a new conductor,also cylindrical,with one fourth the length of the original conductor.The resistance of the new conductor is
A) $\frac { 1 } { 16 }$ R
B) $\frac { 1 } { 4 }$ R
C)R
D)4R
E)16R

Question

Quizplus · Accepted Answer

Resistance of a conductor is given by $$R=\frac { \rho \ell } { A }$$Where $\rho$ is the resistivity of the conductor, $\ell$ is the length of the conductor and A is the area of cross-section of the conductor.If the conductor is melted down and formed into a new conductor with one fourth the length of the original conductor, then the new length will be $\frac { \ell } { 4 }$.The new area of cross-section will be $$A'=\frac { \pi r^2 } { 4 }$$Where r is the radius of the conductor.Substituting these values into the equation for resistance, we get:$$R'=\frac { \rho \frac { \ell } { 4 } } { \frac { \pi r^2 } { 4 } }$$$$R'=\frac { \rho \ell } { \pi r^2 }$$$$R'=\frac { 1 } { 4 } R$$Therefore, the resistance of the new conductor is $\frac { 1 } { 4 }$ times the resistance of the original conductor.

A Conductor of Radius R,length ℓ\ellℓ And Resistivity ρ\rhoρ Has Resistance R 116\frac { 1 } { 16 }161​

A Conductor of Radius R,length $\ell$ And Resistivity $\rho$ Has Resistance R $\frac { 1 } { 16 }$