The resistance of a wire is 'R' ohm. If it is melted and stretched to 'n' times its original length, its new resistance will be

Question

The resistance of a wire is 'R' ohm. If it is melted and stretched to 'n' times its original length, its new resistance will be (A) (R/n) (B) n2R (C) (R/n2) (D) nR

Tardigrade Admin · Accepted Answer

We know that, R =(ρ l/A) or R =(ρ l2/ text Volume ) ⇒ textRal2 According to question l2=n l1 (R2/R1)=(n2 l12/l12) or (R2/R1)=n2 ⇒ R2=n2 R1

Q. The resistance of a wire is ' $R$ ' ohm. If it is melted and stretched to ' $n$ ' times its original length, its new resistance will be

$\frac{R}{n}$

$n^{2} R$

$\frac{R}{n ^{2}}$

$n R$

We know that, $R = \frac{ρl}{A}$
or $R = \frac{ρ l ^{2}}{Volume}$
$\Rightarrow Ral^{2}$
According to question $l_{2} = n l_{1}$
$\frac{R _{2}}{R _{1}} = \frac{n ^{2} l _{1}^{2}}{l _{1}^{2}}$
or $\frac{R _{2}}{R _{1}} = n^{2}$
$\Rightarrow R_{2} = n^{2} R_{1}$

Q. The resistance of a wire is 'R' ohm. If it is melted and stretched to 'n' times its original length, its new resistance will be

nR​

n2R

n2R​

nR