Let the radius of the end of the wire be r cm.
`:.` Volume of the wire `= (22)/(7) xx r^(2) xx 56` cc
Also, the volume of iron `= 11 ` cc
`:.(22)/(7) xx r^(2) xx56=11` or, `r^(2) = (11 xx 7)/(22xx56)= (1)/(2xx8)=(1)/(16)` or, `r = sqrt((1)/(16))=1/4=0.25`
Hence the radius of the end of the wire is 0.25 cm.