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:

Answer

1.67

Explanation

`R = 3 Omega, R prop l^(2)`
`R' = 2^(2) xx 3 = 12 Omega`
`R_(1) = (5 pi)/(6 pi) xx R' = 10 Omega`
`R_(2) = (pi)/(6 pi) xx R' = 2 Omega`
`R = (R_(1) R_(2))/(R_(1) + R_(2)) = (5)/(3) Omega`