Question

A given piece of wire of length L radius r and resistance R is stretched uniformly to a wire of radius $\frac{r}{2}$. The new resistance is:

• A. 16 R
• B. 4 R
• C. $\frac{R}{4}$
• D. $\frac{R}{16}$

Answer 1

The correct option is A

16 R

As R = ρ$\frac{L}{A}$ and volume = L x A = Constant.

Area becomes $\frac{1}{4}$th of the initial value and length becomes 4 times the initial value.

So, resistance increases by 16 times.