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A solid is in the shape of a cylinder with a solid hemisphere attached to one of its ends as shown in the figure.
Let the height and diameter of the cylinder be same (say H). Prove that the volume of the solid shown in the figure is same as the volume of a cone of height H and diameter ‘2H'. Also find the total surface area of the solid in terms of H.
Let the height and diameter of the cylinder be same (say H). Prove that the volume of the solid shown in the figure is same as the volume of a cone of height H and diameter ‘2H'. Also find the total surface area of the solid in terms of H.
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(b) Radius of cylinder/hemisphere = $R = \frac{H}{2}$ (1/2 Mark)
Volume of the solid = Volume of cylinder + Volume of hemisphere
$= \pi R^2 H + \frac{2}{3} \pi R^3$ (1 Mark)
$= \pi (\frac{H}{2})^2 H + \frac{2}{3} \pi (\frac{H}{2})^3$ (1/2 Mark)
$= \frac{\pi H^3}{3}$ (1/2 Mark)
Volume of cone of height H and diameter 2H
$= \frac{1}{3} \pi \times H^2 \times H$ (1/2 Mark)
$= \frac{\pi H^3}{3}$ (1/2 Mark)
T.S.A. of the solid = C.S.A of cylinder + Area of circle +C.S.A of hemisphere
$= 2\pi \frac{H}{2} \times H + \pi \frac{H^2}{4} + 2\pi \frac{H^2}{4}$ (1/2 Mark)
$= \frac{7\pi}{4} H^2$ (1/2 Mark)
(b) Radius of cylinder/hemisphere = $R = \frac{H}{2}$ (1/2 Mark)
Volume of the solid = Volume of cylinder + Volume of hemisphere
$= \pi R^2 H + \frac{2}{3} \pi R^3$ (1 Mark)
$= \pi (\frac{H}{2})^2 H + \frac{2}{3} \pi (\frac{H}{2})^3$ (1/2 Mark)
$= \frac{\pi H^3}{3}$ (1/2 Mark)
Volume of cone of height H and diameter 2H
$= \frac{1}{3} \pi \times H^2 \times H$ (1/2 Mark)
$= \frac{\pi H^3}{3}$ (1/2 Mark)
T.S.A. of the solid = C.S.A of cylinder + Area of circle +C.S.A of hemisphere
$= 2\pi \frac{H}{2} \times H + \pi \frac{H^2}{4} + 2\pi \frac{H^2}{4}$ (1/2 Mark)
$= \frac{7\pi}{4} H^2$ (1/2 Mark)