A uniform wire of resistance R is uniformly compressed along its length, unitl its radius becomes n times the original radius. Now resistance of the wire becomes.

Options

A. `(R)/(n^(4))`
B. `(R)/(n^(2))`
C. `(R)/(n)`
D. nR

Answer

A

Explanation

Given : Initial resistance of wire `(R_(1)) = R `
Initial radius of wire `(r_(1)) = r ` and
Final radius of wire `(r_(2))` = nr
Since volumn of the wire remains same after compression
therefore `l_(1) * A_(1) = l_(2)* A_(2)`
`(l_(1))/(l_(2)) = (A_(2))/(A_(1)) = (pir_(2)^(2))/(pir_(1)^(2) )=(pi(nr)^(2))/(pi(r^(2)))=n^(2)`
We know that resistance of the wire
`(R) = pxx p xx (l)/(A) prop (1)/(A)`
` :. (R_(1))/(R_(2)) = (l_(1))/(l_(2)) xx(A_(2))/(A_(1)) = n^(2) xx n^(2) = n^(4)`
(or) `R_(2) = (R_(1))/(n^(4)) = (R)/(n^(4))`