A hollow copper cylinder is of inner radius 4 cm outer radius 5 cm. Now hollow portion is completely filled with suitable copper wires. Find percentage change in its electric resistance.

Explanation

A hollow cylinder of inner radius `r` and outer radius `R` has specific resistance `rho` if its length is `l` then its resistance
`=(rhol)/(pi(R^(2)-r^(2))`
`R_(1)=(rhol)/(pi(5^(2)-4^(2)))=(rhol)/(9pi)=(k)/(9)`
Final resistance
`R_(2)=(rhol)/(pi(5^(2)))=(rhol)/(25pi)=(k)/(25)`
Percentage of change`=(R_(2)-R_(1))/(R_(1))xx100`
`=((k)/(25)-(k)/(9))/((k)/(9))xx100=-64%`