Question

The resistance R of a wire of length l and uniform area of cross-section A is given by R = rhol/A, where rho is the resistivity of the wire. You melt the wire and recast it to have a new length l' = 10l (keeping the volume V = Al of the wire constant). What is the new resistance of the wire, if the original resistance of the wire was 100.0 Ohm.

Answer 1

resistance of a wire R=rho*l/A

here,

rho---> is resistivity

A ---> cross sectional area,

l----> length of the wire,

resistance R1=100 ohms

now,

resistance of a wire R1=rho*l1/A1 ----(1)

and

resistance of a wire R2=rho*l2/A2 ----(2)

but,

l2=10*l1==> l2/l1=10

volume V1=V2

A1*l1=A2*l2

A1*l1=A2*(10*l1)

A1=10*A2 ===> A1/A2=10

and

from equation no (1) and (2)

R2/R1=(l2/l1)*(A1/A2)

R2/R1=10*10

R2=100*R1

R2=100*100

R2=10000 ohms

there fore,

new resistance R2=10000 ohms