Question

if a wire of resistance R is fold n times so that it's length becomes
$\frac{1}{n}$
th of it's initial length then its new resistance becomes_____?

(a).nR
(b).
${n}^{2}$
(c).R/n
(d).
$\frac{r}{ {n}^{2} }$



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Answer 1

Question :-

If a wire of resistance R is folded n times so that it's length becomes  1 / n   th  of it's initial length then its new resistance becomes ____ ?

Options :

(a) n R

(b) n²

(c) R / n

(d) R / n²  

Solution :-

Initial resistance =  R

initial length of wire = L

initial area of cross section of wire = A

then,

$\bf{R=\rho\:\frac{L}{A}\:\:....eqn(1)}$

Now,

Let, new resistance of Wire = R₂

∵ given that length of wire will become 1 / n  th of initial length

therefore,

new length of wire = L / n

and, new area of cross section = n A

then,

$\bf{R_2 =\rho\:\frac{L}{n}\times\frac{1}{nA} }$

$\bf{R_2 = \rho\:\frac{L}{A}\times\frac{1}{n^2}}$

using eqn (1)

$\bf{R_2=R\times\frac{1}{n^2}}$

$\boxed{\bf{\red{R_2=\frac{R}{n^2}}}}$

Hence,

Correct option will be  (d)    R / n² .

Answer 2

Answer:

your solution could be like this