Question

A given wire of resistance 1 ohms is strechched to triple it's length , what will be the new resistance ? ​

Answer 1

Here is your answer mate .

R=1Ω,Length=L,Cross sectional area, CSA=A

The formula used is,

R=ρAL⟹ρAL=1..........(i)

The new length, L′=2L

And new CSA = A′

So we have, 

Volume of the wire= AL

Now after the act of stretching the volume remains constant.

Therefore, AL=A′L′

⟹AL=2LA′⟹A′=2A

Now the new resistance of the wire becomes

Rnew=ρA′L′⟹Rnew=ρ2A2L⟹Rnew=4ρAL

Rnew=4×1 by using equation (i)

Therefore the new resistance is 4Ω

Answer 2

Given: a wire of resistance 1Ωis stretched to double its length.

To find the new resistance

Solution:

Here,

R=1Ω,Length=L,Cross sectional area, CSA=A

The formula used is,

$\boxed{ \sf \: R = \rho \frac{L}{A} }$

$\sf \implies \: \rho \frac{L}{A} = 1$

Now ,

new length, L'= 3L

And new CSA' = A

So we have,

Volume of the wire= AL

Now after the act of stretching the volume remains constant.

$\sf \therefore \: AL=A'L' \\ \\ \sf \: AL=A'3L \\ \\ \sf \: A' = \frac{A}{3}$

Now the new resistance of the wire becomes

$\sf \: R_{new}= \rho \frac{ L'}{A'} \\ \\ R_{new}= \rho \frac{3 L}{ \frac{A}{3}} \\ \\ R_{new}= 9\rho \frac{ L}{A} \\ \\ \\ \\ \sf \: R_{new}= 9 \times R \: \: \: from \: eqn \: (1) \\ \\ \therefore \: new \: resistance \: is \: 9 ohm$