Question

4.A piece of wire of resistance 20 ohm is drawn out so that its length is increased to twice
its original length. Calculate the resistance of the wire in the new situation.​

Answer 1

$\large{\red{\frak{ Given}}}\begin{cases}\textsf{ A piece of wire has a resistance of 20 } \Omega . \\\textsf{ The lenght of the wire is increased to twice . } \end{cases}$

$\large{\red{\frak{ To \ Find}}}\begin{cases}\textsf{ The resistance in the new situation .}\end{cases}$

Here the wire is simply drawn twice of its original lenght . So here the Volume is constant. And the resistance of the wire is 20Ω . So we know that when the length of a wire is changed from $\sf L_1$ to $\sf L_2$ , having a resistance of R , then New Resistance is , Let the initial lenght be l , then the final lenght will be 2l .

$\purple{\bigstar}\underline{\boldsymbol{ According\ to \ Question :- }}$

$\sf:\implies \pink{Resistance_{(new)}= \bigg\lgroup \dfrac{L_2}{L_1}\bigg\rgroup ^2R }\\\\\sf:\implies R_{(new)}= \bigg\lgroup\dfrac{2l}{l}\bigg\rgroup^2 \times 20\Omega \\\\\sf:\implies R_{(new)}= 2^2 \times 20 \Omega \\\\\sf:\implies R_{(new)}= 4\times 20\Omega \\\\\sf:\implies\underset{\blue{\sf Required\ Answer}}{ \underbrace{\boxed{\pink{\frak{ Resistance_{(new)}= 80 \Omega }}}}}$