Question

Q. 8. A given length of a wire is doubled and this
process is repeated once again. By what factor does
the resistance of the wire changes ? ​

Answer 1

Given :

• Length of a wire is Doubled.
• And this process is repeated once again .

To Find :

By what factor does the resistance of the wire changes .

Theory :

We know that

$\sf\:Resistance=\dfrac{\rho\:l}{A}$

Now multiply and divide by l

$\sf\implies\:Resistance=\dfrac{\rho\:l\times\:l}{A\times\:l}$

$\sf\implies\:Resistance=\dfrac{\rho\:l^2}{V}$

$\sf\:R\propto\:l$

$\sf\dfrac{R_2}{R_1}=(\dfrac{l_2}{l_1})^2$

Solution :

Let the Length of the be l and Resistance be R

$\sf\:Inital\:Resistance=R$

$\sf\:Inital\:length=l$

Now the length of the is dobuled

$\sf\:l_2=2l$

$\sf\dfrac{R_2}{R_1}=(\dfrac{l_2}{l_1})^2$

$\sf\implies\dfrac{R_2}{R}=(\dfrac{2l}{l})^2$

$\sf\implies\dfrac{R_2}{R}=\dfrac{4l}{l}$

$\sf\implies\dfrac{R_2}{R}=4$

$\sf\implies\:R_2=4R...(1)$

Now This process is repeated once again .

$\sf\:l_3=2l_2$

$\sf\implies\:l_3=4l$

$\sf\implies\dfrac{R_3}{R}=(\dfrac{4l}{l})^2$

$\sf\implies\dfrac{R_3}{R}=\dfrac{16l}{l}$

$\sf\implies\dfrac{R_3}{R}=16$

$\sf\implies\:R_3=16R$

Therefore,the resistance of the wire changes is increased by 16times .

Answer 2

✵ Given ✵

A given length of a wire is doubled and this  process is repeated once again.

$\rule{180}2$

✵ To Find ✵

By what factor does  the resistance of the wire changes.

$\rule{180}2$

✵ Solution ✵

We know,

$\boxed{\sf{\color{red}{R=\dfrac{\rho\ l}{A}}}}$

When the length is doubled,

$\sf{R'=\dfrac{\rho\ 2l}{A} }\\\\\displaystyle{\sf{\longrightarrow R'=2\bigg(\frac{\rho\ l}{A} \bigg)}}$

$\displaystyle{\sf{\longrightarrow R'=2R}}$

Again, when the length is doubled,

$\displaystyle{\sf{\longrightarrow R''=2R'}}\\\\\displaystyle{\sf{\longrightarrow R''=2\times2R}}\\\\\displaystyle{\sf{\longrightarrow R''=4R}}$

Therefore, at last, the resistance of the wire changes by a factor of 4.