How many number of turns of nichrome wire of specific resistance `10^(-6)Omegam` and diameter 2mm that should be wound on a cylinder of diameter 5 cm to obtain a resistance of `40Omega`?

Explanation

If R is the radius of the cylinder
`r` is the radius of the wire
`N` is the number of turns
then `R^(')=(rhol)/(A)becauseR^(')=(rho(2piR)N)/(pir^(2))`
`40=(10^(-6)(2xx2.5xx10^(-2)xxN))/(1xx10^(-6))=thereforeN=800`