Question

Keeping volume constant, A wire of resistance R is pulled to five times of it`s actual length, then it`s new resistance will be

5R

10R

20R

25R
​

Answer 1

Given:

A wire of resistance R is stretched to 5 times its actual length provided the volume is kept constant.

To find:

The new resistance of the wire.

Concept:

Resistance any a conducting wire depends upon its dimensions and materialistic property. So , change in dimension of the wire will obviously lead to change in resistance.

Calculation:

Let new radius be r2.

Considering volume conservation :

$\therefore \: V1 = V2$

$= > \pi \times {r}^{2} \times l = \pi \times {(r2)}^{2} \times (5l)$

$= > {(r2)}^{2} = \dfrac{ {r}^{2} }{5}$

As per the question , the initial resistance is R :

$\therefore \: R = \rho \times \bigg( \dfrac{length}{area} \bigg)$

$= > \: R = \rho \times \bigg( \dfrac{l}{\pi {r}^{2} } \bigg)$

$= > \: R = \dfrac{ \rho l}{\pi {r}^{2} } \: \: .........(1)$

After stretching of wire , let new resistance be R2:

$\therefore \: R2 = \rho \times \bigg( \dfrac{length}{area} \bigg)$

$= > \: R2 = \rho \times \bigg \{ \dfrac{5l}{\pi {(r2)}^{2} } \bigg \}$

$= > \: R2 = \rho \times \bigg \{ \dfrac{5l}{\pi( \frac{ {r}^{2} }{5} ) } \bigg \}$

$= > \: R2 = \rho \times \bigg( \dfrac{25l}{\pi {r}^{2} } \bigg)$

$= > \: R2 = 25 \times \bigg( \dfrac{ \rho l}{\pi {r}^{2} } \bigg)$

Putting value from eq.(1) :

$= > \: R2 = 25 \times \bigg(R \bigg)$

$= > \: R2 = 25 R$

So , final answer is :

$\boxed{ \red{ \huge{ \bold{ R2 = 25 R }}}}$

Answer 2

Answer:

25R. thanks mark me brainlest please