Question

When a piece of aluminium wire of finite length is drawn through a series of dies to reduce its diameter to half its original value, its resistance will become:-

• A. Two times
• B. Four times
• C. Eight times
• D. Sixteen times

Answer 1

Answer: (D)

The correct option is D Sixteen times
In stretching wire,

We know that, $R\alpha \frac{1}{r^4}$

$R=K\frac{1}{r^4}$

Where the diameter is rediced to $1/2$

We have $r^1=\left(\frac{r^1}{2}\right)$

$R=K\frac{1}{(\frac{r}{2}{)}^4}R=K\frac{1}{\frac{r^4}{16}}R=16K\frac{1}{r^4}R=16\left(K\frac{1}{r^4}\right)$

$\left[d\right]\therefore$ The resistance will be $16$ times.