Question 275: Current Electricity

Question

A wire of resistance R
1
is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is :

Accepted Answer

Length is increased by twice of its original length.
So, if original length is $l_1$ then final length is $l_2$= $l_1$ + 2$l_1$ = 3$l_1$.
Then area becomes, A
2
= $${{{A_1}} \over 3}$$
$$
\begin{aligned}
& R_1=\frac{p l_1}{A_1} \\
& R_2=\frac{p l_2}{A_2}=\frac{p 	imes 3 l_1}{A_1 / 3}=9 \frac{p l_1}{A_1}=9 R_1 \\
& 	herefore R_2: R_1=9: 1
\end{aligned}
$$

Answer

9 : 1

Answer

1 : 9

Answer

4 : 1

Answer

3 : 1

Current Electricity

A wire of resistance R 1 is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is :

Choose the correct answer: