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Answer :

Volume of the air contained in the model = Volume of the cylindrical portion of the model + Volume of its two conical ends.

\( = \ \pi r^2 h_1 \ + \ 2 \ × \ \frac{1}{3} \pi r^2 h_2 \ \)

\( = \ \pi r^2(h_1 \ + \ \frac{2}{3} h_2) \)

where \( r \ = \ \frac{3}{2} \) cm , \(h_1 \ = \ 8 \) cm and \(h_2 \ = \ 2 \) cm

\( = \ \frac{22}{7} \ × \ ( \frac{3}{2})^2 \ × \ (8 \ + \ \frac{2}{3} \ × \ 2) \) \( = \ \frac{22}{7} \ × \ \frac{9}{4} \ × \ \frac{24 \ + \ 4}{3} \)

\(= \ \frac{22}{7} \ × \ \frac{9}{4} \ × \ \frac{28}{3} \) \( = \ 66 \) cm^{3}

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