A wire of length L is drawn such that its diameter is reduced to half of its original diameter. If the initial resistance of the wire were ` 10 Omega` , its new resistance would be .

Options

A. `40 Omega`
B. `80 Omega`
C. `120 Omega`
D. `160 Omega`

Answer

D

Explanation

Let the original diameter of the wire be D. therefore the new diameter is D/2
Original area of cross - section is `(piD^2)/(4)`
and the final area of cross - section is `(piD^2)/(16)`
The new length of the wire is given by
`Lxx(piD^2)/(4) rArr L'=16/4L=4L`
Now, we known that the resistance is given by `R=rhoL/A`
`therefore R'=rho(L')/(A')=rho(4L)/(A//4)=16R`.
`[therefore A'=(piD^2)/(16)=A/4]`
`therefore R'=16xx10=160Omega`