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 becoes `3/2` times the original length. What is the value of this fraction?

Options

A. `1//8`
B. `1//6`
C. `1//10`
D. `1//4`

Answer

A

Explanation

Since, new Resistance = 4R
`4R = (rho(l-x))/A + (rho(x + 0.5l))/A^(.)`
where `A. = x/(0.5 l + x) A`
`therefore 4 xx (rho l)/A = (rho (l-x)/A) + (rho (x +0.5l)^(2))/(xA)`
`rArr 4l = xl + xl + 0.25l^(2)`
`rArr x/l = 1/4 xx 1/2 = 1/8 rArr 1/8 rArr x/l = 1/8`