A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8g mass. (Use π = 3.14)

Solution

Let h cm be the height and r cm be the radius of the first cylinder. Then

h = 220 cm and r = $24 over 2$ = 12 cm
Now, Volume = πr² h
= (π x 12 x 12 x 220) cm³= 31680 π
Let H cm be the height and R cm be the radius of the second cylinder. ThenH = 60 cm and R = 8 cm
Now, Volume = $πR squared straight H$
$open parentheses straight pi space straight x space 8 space straight x space 8 space straight x space 60 close parentheses space c m cubed equals space 3840 space straight pi$
Total volume of the pole
= 31680 $straight pi$ + 3840 $straight pi$
= 35520 $straight pi$ = (35520 x 3.14) cm³
= 111532.8 cm³
Now, Required weight
= $open parentheses fraction numerator 111532.8 space straight x space 8 over denominator 1000 end fraction close parentheses space kg equals space 892.26 space kg$