The radius of the iron-sphere `-(4)/(2)`cm=2cm.
`therefore` the volume of the iron-sphere `=(4)/(3)pixx2^(3)cc.=(32pi)/(3)cc.`
Let the radius of the iron-plate be r cm.
`therefore` the area of the iron-plate `=pir^(2)` sq-cm.
Since the iron-plate is of thickness `(2)/(3)` cm, its volume =`(2)/(3)pir^(2)` cu.cm.
As per question `(2)/(3)pir^(2)=(32pi)/(3)rArr r^(2)=16rArr r=sqrt(16)=4`
Hence the radius of the circular iron-plate=4 cm.
