In order to increase the resistance of a given wire of uniform cross section to four times its value, a fraction of its length is stretched uniformly till the full length of the wire because `3/2` times the original length. What is the value of this fraction?

Options

A. `1//2`
B. `1//4`
C. `1//6`
D. `1//8`

Answer

D

Explanation

Let `l` be the of the wire A portion of the length of wire is stretched so the that the final length becomes

Let `R_(1) R_(2)` be the resistance of the wire below and after stretching. Let `A_(1)` be the `A_(2)` in the area of cross section of the original wire and `A_(2)` be the area of cross - section of the portion `(1.5 I = x)` of the wire On stretched the value of the wire remains unchanged so `l A_(1) = xA_(1) + (1.5I - x)A_(2)`
or `A_(2) = ((1 - x))/((1.5I - x)) A_(1)`
Also `R_(1) = (rho l)/(A_(1)) and R_(2) = (rho x)/(A_(1)) + (rho(1.5 l - x))/(A_(2))`
As per question `R_(2)= 4R_(1)`
`:. (4 rho l)/(A_(1)) = (rho x)/(A_(1)) + (rho(1.5l - x))/(A_(2)) = (rho x)/(A_(1)) + (rho(1.5l - x)^(2))/((1 - x)A_(1))`
On solving it we get, `(x)/(l) = (7)/(8)`
`:.` Fraction of wire stretched
`= ((1 - x))/(1) = (1 - (x)/(l)) = 1 - (7)/(8) = (1)/(8)`