Question

If the diameter of a resistance wire is halved, then its resistance becomes:
(a) four times (b) half (c) one - fourth (d) two times

Answer 1

When we change the area of cross-section of a wire its length changes in such a way that the volume of the wire remains constant.

Let, A and L be the initial area of cross-section and length of the wire.

The resistance initially is, $R=\rho \frac{L}{A}$

When the area of cross-section is halved, then,

Volume of the wire = AL = $\frac{A}{2}\times L^{\prime }$ [$L^{\prime }$ is the new length]

Thus, $L^{\prime }=2L$

So, the new resistance is,

$R^{\prime }=\rho L^{\prime \prime }\times 2A$

$=\rho \times 2L\times \frac{A}{2}$

$=4\rho \frac{L}{A}=4R$

Thus, resistance becomes 4 times