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The adjoining figure represents a solid consisting of a right circular cylinder with a hemisphere at one end and a cone at the other. Their common radius is 7 cm. The height of the cylinder and the cone are each of 4 cm. Find the volume of the solid.

The adjoining figure represents a solid consisting of a right circular cylinder with a hemisphere at one end and a cone at the other. Their common radius is 7 cm. The height of the cylinder and the cone are each of 4 cm. Find the volume of the solid. Mensuration, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Mensuration

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Answer

Given, common radius (r) = 7 cm,

Height of cone (h) = 4 cm,

Height of cylinder (H) = 4 cm.

Volume of solid (V) = Volume of cone + Volume of cylinder + Volume of hemisphere.

V=13πr2h+πr2H+23πr3=πr2(h3+H+2r3)=227×(7)2×(43+4+2×73)=22×7×(43+4+143)=154×(4+12+143)=154×303=154×10=1540 cm3.V = \dfrac{1}{3}πr^2h + πr^2H + \dfrac{2}{3}πr^3 \\[1em] = πr^2\Big(\dfrac{h}{3} + H + \dfrac{2r}{3}) \\[1em] = \dfrac{22}{7} \times (7)^2 \times \Big(\dfrac{4}{3} + 4 + \dfrac{2 \times 7}{3}) \\[1em] = 22 \times 7 \times \Big(\dfrac{4}{3} + 4 + \dfrac{14}{3}) \\[1em] = 154 \times \Big(\dfrac{4 + 12 + 14}{3}\Big) \\[1em] = 154 \times \dfrac{30}{3} \\[1em] = 154 \times 10 \\[1em] = 1540 \text{ cm}^3.

Hence, the volume of solid = 1540 cm3.

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