if a wire of resistance R is fold n times so that it's length becomes
[tex] \frac{1}{n} [/tex]
th of it's initial length then its new resistance becomes_____?
(a).nR
(b).
[tex] {n}^{2} [/tex]
(c).R/n
(d).
[tex] \frac{r}{ {n}^{2} } [/tex]
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if a wire of resistance R is fold n times so that it's length becomes
[tex] \frac{1}{n} [/tex]
th of it's initial length then its new resistance becomes_____?
(a).nR
(b).
[tex] {n}^{2} [/tex]
(c).R/n
(d).
[tex] \frac{r}{ {n}^{2} } [/tex]
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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,
[tex]\bf{R=\rho\:\frac{L}{A}\:\:....eqn(1)}[/tex]
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,
[tex]\bf{R_2 =\rho\:\frac{L}{n}\times\frac{1}{nA} }[/tex]
[tex]\bf{R_2 = \rho\:\frac{L}{A}\times\frac{1}{n^2}}[/tex]
using eqn (1)
[tex]\bf{R_2=R\times\frac{1}{n^2}}[/tex]
[tex]\boxed{\bf{\red{R_2=\frac{R}{n^2}}}}[/tex]
Hence,
Correct option will be (d) R / n² .
Answer:
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