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A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water circular such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.


Answer :




Volume of the cylinder \(= \ \pi r^2 h \ = \ \frac{22}{7} \ × \ (60)^2 \ × \ 180 \)

\(= \ \frac{22 \ × \ 3600 \ × \ 180}{7} \ = \ \frac{14256000}{7} \) cm3

Volume of the solid = Volume of cone + Volume of hemisphere

\(= \ \frac{1}{3} \ × \ \frac{22}{7} \ × \ 60^2 \ × \ 120 \ + \ \frac{2}{3} \ × \ \frac{22}{7} \ × \ 60^3 \)

\(= \ \frac{3168000}{7} \ + \ \frac{3168000}{7} \ = \ \frac{6336000}{7} \) cm3

Volume of water left in the cylinder = Volume of the cylinder - Volume of the solid

\(= \ \frac{14256000}{7} \ - \ \frac{6336000}{7} \ = \frac{7920000}{7} \)

\(= \ 1131428.57142 \ = \ 1.131 \) m3

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