A wire of resistance R1 is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is :
Solution
Length is increased by twice of its original length. <br/><br/>So, if original length is $l_1$ then final length is $l_2$= $l_1$ + 2$l_1$ = 3$l_1$.
<br/><br/>Then area becomes, A<sub>2</sub> = ${{{A_1}} \over 3}$
<br/><br/>$$
\begin{aligned}
& R_1=\frac{p l_1}{A_1} \\\\
& R_2=\frac{p l_2}{A_2}=\frac{p \times 3 l_1}{A_1 / 3}=9 \frac{p l_1}{A_1}=9 R_1 \\\\
& \therefore R_2: R_1=9: 1
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Current Electricity · Topic: Ohm's Law and Resistance
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