Given
Height of the first cylinder `h_(1)`=220 cm
Radius of the first cylinder `r_(1)`=`(24)/(2)= 12cm`
Height of the second cylinder, `h_(2)`= 60 cm
and radius of the second cylinder , `r_(2)` 8cm
volume of iron pole = volume of first cylinder + volume of second cylinder
`pir_(1^(2))+pir_(2^(2))h_(2)=pir_(1^(2))h_(1)+r_(2^(2))`
=3.14(144`xx`220+64`xx`60
`=3.14`(31680+3840)=3.14`xx`355520 `cm^(2)`
1`cm^(3)` of iron has appproximaterly mass =g=`(8)/(1000)`kg
`therefore` (3.14`xx`35520)`m^(3)` of iron has approximately mass
=3.14`xx`284.160 kg=892.26 kg