JEE MAIN - Physics (2022 - 28th July Morning Shift - No. 10)
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 :
9 : 1
1 : 9
4 : 1
3 : 1
Explanation
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, A2 = $${{{A_1}} \over 3}$$
$$ \begin{aligned} & R_1=\frac{p l_1}{A_1} \\\\ & R_2=\frac{p l_2}{A_2}=\frac{p \times 3 l_1}{A_1 / 3}=9 \frac{p l_1}{A_1}=9 R_1 \\\\ & \therefore R_2: R_1=9: 1 \end{aligned} $$
So, if original length is $l_1$ then final length is $l_2$= $l_1$ + 2$l_1$ = 3$l_1$.
Then area becomes, A2 = $${{{A_1}} \over 3}$$
$$ \begin{aligned} & R_1=\frac{p l_1}{A_1} \\\\ & R_2=\frac{p l_2}{A_2}=\frac{p \times 3 l_1}{A_1 / 3}=9 \frac{p l_1}{A_1}=9 R_1 \\\\ & \therefore R_2: R_1=9: 1 \end{aligned} $$
Comments (0)
