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:

Question

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: (A) 9: 1 (B) 1: 9 (C) 4: 1 (D) 3: 1

Tardigrade Admin · Accepted Answer

R 1=ρ ( L 1/ A 1) R 2=ρ((3 L 1/ A 1 / 3))=9 ρ ( L 1/ A 1) ∴ ( R 2/ R 1)=9

Q. 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:

$9 : 1$

$1 : 9$

$4 : 1$

$3 : 1$

$R_{1} = ρ \frac{L _{1}}{A _{1}}$
$R_{2} = ρ (\frac{3 L _{1}}{A _{1} /3}) = 9 ρ \frac{L _{1}}{A _{1}}$
$∴ \frac{R _{2}}{R _{1}} = 9$

Q. 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