Question

A given wire of resistance 1 ohm is stretched to double its length what will be the new resistance

Answer 1

R=ρ$\frac{l}{a}$

New length = 2l

New area of cross section = a/2

Therefore new resistance R' = ρ$\frac{2l}{a/2}$

                                             = 4R

∴ R' = 4R

Answer 2

Answer: The new resistance will be 4 ohm.

Explanation:

Given that,  

Resistance R = 1 ohm

Length L = 2l

We know that,

$R = \rho \dfrac{L}{A}$.....(I)

Where, A = area

$\rho$ = resistivity

$A = \rho l$.....(II)

After stretch,

The new resistance is

$R'=\rho \dfrac{l'}{A'}$.....(III)

The volume is constant after and before stretch

So, $Al= A'l'$

$Al= A'2l$

$A'= \dfrac{A}{2}$

Now , put the value off A' and l in equation (III)

$R' = \rho\dfrac{2l}{\dfrac{A}{2}}$

$R' = \rho\dfrac{4l}{A}$.....(IV)

Now, divided equation (IV) by equation (I)

$\dfrac{R'}{R} = \dfrac{\rho 4lA}{A\rho l}$

$R' = 4 ohm$

Hence, the new resistance will be 4 ohm.