A piece of silver has a resistace of `2 Omega`. What will be the resistance of constantan wire of two-third length and one-third diameter, if the specific resistance of constantan is 30 times that of silver.

Explanation

Let `l,D` be the length and diameter of silver wire respectively. Then resistance of silver is
`R= (rho l)/((pi D^(2)//4)) = (4 rhol)/(pi D^(2))=2`....(i)
For constant wire,
`l'=2/3l , D'=D/3, rho'=30 rho`
`R'= (4rho' l')/(pi D'^(2))=(4xx 30 rho xx (2//3)l)/(pi (D//3)^(2))`
`=180 xx (4 rho l)/(pi D^(2))=180 xx 2=360 Omega`