Class 10 Volume of Combined Solids — practice questions with answers

Volume is additive: the volume of a combined solid is the sum of the volumes of its constituent solids.

V = $\frac{2}{3}$ pi r³ = ($\frac{2}{3}$)($\frac{22}{7}$)(3.5³) = ($\frac{2}{3}$)($\frac{22}{7}$)(42.875) = ($\frac{2}{3}$)(134.75) = 89.833 approx 89.83 cm³.

Bowl = $\frac{2}{3}$ pi (729) = 486 pi. Each bottle = pi (2.25)(4) = 9 pi. Number = $\frac{486\pi }{9\pi }$ = 54 bottles.

Base area $\pi r^2=\frac{22}{7}(7.84)=24.64$ cm². Cylinder = 24.64 × 12 = 295.68 cm³. Each cone = $\frac{1}{3}(24.64)(4.5)=36.96$ cm³, so both = 73.92 cm³. Wood = 295.68 − 73.92 = 221.76 cm³. 258.72 removes only one cone; 369.6 adds the cones instead of subtracting them; 73.92 is the wood taken out, not the wood left.

P = $\pi (36)(10)=\frac{22}{7}\times 360=1131.43$ cm³. For Q, cylinder = $\frac{22}{7}(36)(6)=678.86$ cm³ and the hemisphere below adds $\frac{2}{3}\pi r^3=\frac{2}{3}\times \frac{22}{7}\times 216=452.57$ cm³, giving 1131.43 cm³. They are equal, because a hemisphere of radius r holds as much as a cylinder of radius r and height $\frac{2r}{3}=4$ cm, and 6 + 4 = 10. The other options quote the hemisphere's volume, or half of it, as a difference that does not exist.