Question

A wire of length 1 and resistance R is stretched to get the radius of cross section halved. What is the new resistance of the wire after stretching?

Answer 1

Given:

A wire of length l and resistance R is stretched to get the radius of cross section halved.

To find:

New resistance of wire.

Calculation:

Since the wire has been stretched , its volume will remain constant;

$\therefore$ V1 = V2

$= > \pi {r}^{2} l = \pi {( \dfrac{r}{2} )}^{2} l_{2}$

$= > l = \dfrac{ l_{2} }{4}$

$= > l_{2} = 4l$

Initial resistance:

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

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

Final resistance:

$R_{2}= \rho \times ( \dfrac{4l}{a_{2}} )$

$= > R_{2}= \rho \times \bigg \{\dfrac{4l}{\pi {( \frac{r}{2}) }^{2} } \bigg \}$

$= > R_{2}= \rho \times \bigg \{\dfrac{4l}{ \frac{\pi {r}^{2} }{4} } \bigg \}$

$= > R_{2}= \rho \times \bigg \{\dfrac{16l}{ \pi {r}^{2} } \bigg \}$

$= > R_{2}= 16\times \bigg \{ \rho \times \dfrac{l}{ \pi {r}^{2} } \bigg \}$

$= > R_{2}= 16\times \bigg \{R \bigg \}$

$= > R_{2}= 16R$

So, final answer is:

$\boxed{ \sf{ \red{R_{2}= 16R }}}$

Answer 2

Explanation:

THE ABOVE ANSWER IS CORRECT