Premium Online Home Tutors
3 Tutor System
Starting just at 265/hour
Answer :
Volume of the solid iron pole = Volume of the cylindrical portion + Volume of the other cylindrical portion
\( = \ \pi (r_1)^2 h_1 \ + \ \pi (r_2)^2 h_2 \)
\(= \ 3.14 \ × \ (12)^2 \ × \ 220 \ + \ 3.14 \ × \ (8)^2 \ × \ 60 \)
\(= \ 3.14 \ × \ 144 \ × \ 220 \ + \ 3.14 \ × \ 64 \ × \ 60 \)
\(= \ 99475.2 \ + \ 12057.6 \)
\(= \ 111532.8 \) cm3
\(\therefore \) The mass of the pole
\(= \ 111532.8 \ × \ 8 \) grams
\(= \ \frac{111532.8 \ × \ 8}{1000} \ = \ 892.26 \) kg