A uniform wire of resistance `=3Omega` and length `l` is stretched to double its length. Now it is bent to form a circular loop and two point `P & Q` lies on the loop such that they subtend `60^(@)` angle at centre. The equivalent resistance between two point P & Q is:

Options

A. `5/3 Omega`
B. `5/2 Omega`
C. `7/2 Omega`
D. `12/5 Omega`

Explanation

`R=3 Omega, R prop t^2`
`R'=2^2 times3=12 Omega` `R_1=(5pi)/(6pi) timesR'=10 Omega`
`R_2=pi/(6pi) times R'=2 Omega , R=(R_1R_2)/(R_1+R_2)=5/3 Omega`