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.

`=(rho l)/(pi(R^(2) -r^(2))`
`R_(1) =(rho l)/(pi(5^(2) -4^(2))) =(rho l)/(9 pi) =k/9`
Final resistance
`R_(2) =(rho l)/(pi(5^(2))) =(rho l)/(25 pi) =k/25`
Percentage of change `=(R_(2)-R_(1))/R_(1) xx 100`
`=(k/25-k/9)/(k/9) xx 100 = -64%`