Question

A uniform wire of resistance R is stretched until its radius becomes half of the initial value. Then the change in resistance is
a) 4R
b) 16R
c) 3R
d) 15R
ANSWER IS 15 R,
why?

Answer 1

Given:

A uniform wire of resistance R is stretched until its radius becomes half of the initial value.

To find:

Change in resistance ?

Calculation:

Initial resistance will be :

$R = \dfrac{ \rho \times l}{a}$

• 'l' is length and 'a' is area.

Now, when stretched, the volume of wire remains constant.

$\implies \: V1 = V2$

$\implies \: \pi {r}^{2} \times l = \pi { (\dfrac{r}{2} )}^{2} \times l_{2}$

$\implies \: \pi {r}^{2} \times l = \dfrac{\pi { r}^{2}}{4} \times l_{2}$

$\implies \: l_{2} = 4l$

• Also, since radius becomes half, the area will be πr²/4 (i.e. it will become a/4)

Now, new resistance:

$R_{2} = \dfrac{ \rho \times (4l)}{ \dfrac{a}{4} } = 16 \times \dfrac{ \rho \times l}{a} = 16R$

Now, change in resistance is :

∆R = 16R - R = 15R

• OPTION d) is ✔️